Equivalent Conditions for Differentiability - USM · 2010. 8. 30. · Cauchy-Riemann EquationsPolar...

Cauchy-Riemann Equations Polar Coordinates Analytic Functions Harmonic Functions

Equivalent Conditions for Differentiability

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Introduction

1. Although all computations for differentiability work very wellwith complex numbers, it would be nice to have an idea whatcomplex differentiability means in terms of real functions of realnumbers.

2. The Cauchy-Riemann equations provide just that.3. As such, we obtain a new way to look at complex differentiable

functions and a new way to look at certain real differentiablefunctions.

4. This will allow us to prove the complex differentiability ofcertain functions (such as the exponential function and thelogarithm function). Proving these facts with differencequotients would be very painful with the tools we have at hand.

Introduction1. Although all computations for differentiability work very well

with complex numbers

, it would be nice to have an idea whatcomplex differentiability means in terms of real functions of realnumbers.

with complex numbers, it would be nice to have an idea whatcomplex differentiability means in terms of real functions of realnumbers.

2. The Cauchy-Riemann equations provide just that.

3. As such, we obtain a new way to look at complex differentiablefunctions and a new way to look at certain real differentiablefunctions.

functions

and a new way to look at certain real differentiablefunctions.

4. This will allow us to prove the complex differentiability ofcertain functions (such as the exponential function and thelogarithm function).

Proving these facts with differencequotients would be very painful with the tools we have at hand.

Theorem.

Cauchy-Riemann Equations. Let f (z) = u(x,y)+ iv(x,y)be a function on an open domain with continuous partial derivativesin the underlying real variables. Then f is differentiable at z = x+ iyif and only if

∂u∂x

(x,y) =∂v∂y

(x,y) and∂u∂y

(x,y) =−∂v∂x

(x,y).

These equations are called the Cauchy-Riemann Equations.Moreover, we have

f ′(z) =∂u∂x

(z)+ i∂v∂x

Theorem. Cauchy-Riemann Equations.

Let f (z) = u(x,y)+ iv(x,y)be a function on an open domain with continuous partial derivativesin the underlying real variables. Then f is differentiable at z = x+ iyif and only if

∂u∂x

(x,y) =∂v∂y

(x,y) and∂u∂y

(x,y) =−∂v∂x

(x,y).

f ′(z) =∂u∂x

(z)+ i∂v∂x

Theorem. Cauchy-Riemann Equations. Let f (z) = u(x,y)+ iv(x,y)be a function on an open domain with continuous partial derivativesin the underlying real variables.

Then f is differentiable at z = x+ iyif and only if

∂u∂x

(x,y) =∂v∂y

(x,y) and∂u∂y

(x,y) =−∂v∂x

(x,y).

f ′(z) =∂u∂x

(z)+ i∂v∂x

Theorem. Cauchy-Riemann Equations. Let f (z) = u(x,y)+ iv(x,y)be a function on an open domain with continuous partial derivativesin the underlying real variables. Then f is differentiable at z = x+ iy

if and only if

∂u∂x

(x,y) =∂v∂y

(x,y) and∂u∂y

(x,y) =−∂v∂x

(x,y).

f ′(z) =∂u∂x

(z)+ i∂v∂x

Theorem. Cauchy-Riemann Equations. Let f (z) = u(x,y)+ iv(x,y)be a function on an open domain with continuous partial derivativesin the underlying real variables. Then f is differentiable at z = x+ iyif and only if

∂u∂x

(x,y) =∂v∂y

and∂u∂y

(x,y) =−∂v∂x

(x,y).

f ′(z) =∂u∂x

(z)+ i∂v∂x

∂u∂x

(x,y) =∂v∂y

(x,y) and∂u∂y

(x,y) =−∂v∂x

(x,y).

f ′(z) =∂u∂x

(z)+ i∂v∂x

∂u∂x

(x,y) =∂v∂y

(x,y) and∂u∂y

(x,y) =−∂v∂x

(x,y).

These equations are called the Cauchy-Riemann Equations.

Moreover, we have

f ′(z) =∂u∂x

(z)+ i∂v∂x

∂u∂x

(x,y) =∂v∂y

(x,y) and∂u∂y

(x,y) =−∂v∂x

(x,y).

f ′(z) =∂u∂x

(z)+ i∂v∂x

Proof of “⇒.”

∂u∂x

(x,y) = limt→x

u(t,y)−u(x,y)t− x

t+iy→x+iy

f (t + iy)− f (x+ iy)(t + iy)− (x+ iy)

)= ℜ

(f ′(z)

)= ℑ

(if ′(z)

)= ℑ

(i lim

x+it→x+iy

f (x+ it)− f (x+ iy)(x+ it)− (x+ iy)

)= ℑ

x+it→x+iyi

f (x+ it)− f (x+ iy)i(t− y)

)= ℑ

(limt→y

f (x+ it)− f (x+ iy)t− y

)= lim

v(x, t)− v(x,y)t− y

=∂v∂y

The other equation is proved similarly.

Proof of “⇒.”∂u∂x

= limt→x

t+iy→x+iy

f (t + iy)− f (x+ iy)(t + iy)− (x+ iy)

)= ℜ

(f ′(z)

)= ℑ

(if ′(z)

)= ℑ

(i lim

x+it→x+iy

f (x+ it)− f (x+ iy)(x+ it)− (x+ iy)

)= ℑ

x+it→x+iyi

)= ℑ

(limt→y

)= lim

=∂v∂y

(x,y) = limt→x

t+iy→x+iy

f (t + iy)− f (x+ iy)(t + iy)− (x+ iy)

)= ℜ

(f ′(z)

)= ℑ

(if ′(z)

)= ℑ

(i lim

x+it→x+iy

f (x+ it)− f (x+ iy)(x+ it)− (x+ iy)

)= ℑ

x+it→x+iyi

)= ℑ

(limt→y

)= lim

=∂v∂y

(x,y) = limt→x

t+iy→x+iy

f (t + iy)− f (x+ iy)(t + iy)− (x+ iy)

= ℜ(f ′(z)

)= ℑ

(if ′(z)

)= ℑ

(i lim

x+it→x+iy

f (x+ it)− f (x+ iy)(x+ it)− (x+ iy)

)= ℑ

x+it→x+iyi

)= ℑ

(limt→y

)= lim

=∂v∂y

(x,y) = limt→x

t+iy→x+iy

f (t + iy)− f (x+ iy)(t + iy)− (x+ iy)

)= ℜ

(f ′(z)

= ℑ(if ′(z)

)= ℑ

(i lim

x+it→x+iy

f (x+ it)− f (x+ iy)(x+ it)− (x+ iy)

)= ℑ

x+it→x+iyi

)= ℑ

(limt→y

)= lim

=∂v∂y

(x,y) = limt→x

t+iy→x+iy

f (t + iy)− f (x+ iy)(t + iy)− (x+ iy)

)= ℜ

(f ′(z)

)= ℑ

(if ′(z)

(i lim

x+it→x+iy

f (x+ it)− f (x+ iy)(x+ it)− (x+ iy)

)= ℑ

x+it→x+iyi

)= ℑ

(limt→y

)= lim

=∂v∂y

(x,y) = limt→x

t+iy→x+iy

f (t + iy)− f (x+ iy)(t + iy)− (x+ iy)

)= ℜ

(f ′(z)

)= ℑ

(if ′(z)

)= ℑ

(i lim

x+it→x+iy

f (x+ it)− f (x+ iy)(x+ it)− (x+ iy)

x+it→x+iyi

)= ℑ

(limt→y

)= lim

=∂v∂y

(x,y) = limt→x

t+iy→x+iy

f (t + iy)− f (x+ iy)(t + iy)− (x+ iy)

)= ℜ

(f ′(z)

)= ℑ

(if ′(z)

)= ℑ

(i lim

x+it→x+iy

f (x+ it)− f (x+ iy)(x+ it)− (x+ iy)

)= ℑ

x+it→x+iyi

(limt→y

)= lim

=∂v∂y

(x,y) = limt→x

t+iy→x+iy

f (t + iy)− f (x+ iy)(t + iy)− (x+ iy)

)= ℜ

(f ′(z)

)= ℑ

(if ′(z)

)= ℑ

(i lim

x+it→x+iy

f (x+ it)− f (x+ iy)(x+ it)− (x+ iy)

)= ℑ

x+it→x+iyi

)= ℑ

(limt→y

= limt→y

=∂v∂y

(x,y) = limt→x

t+iy→x+iy

f (t + iy)− f (x+ iy)(t + iy)− (x+ iy)

)= ℜ

(f ′(z)

)= ℑ

(if ′(z)

)= ℑ

(i lim

x+it→x+iy

f (x+ it)− f (x+ iy)(x+ it)− (x+ iy)

)= ℑ

x+it→x+iyi

)= ℑ

(limt→y

)= lim

=∂v∂y

(x,y) = limt→x

t+iy→x+iy

f (t + iy)− f (x+ iy)(t + iy)− (x+ iy)

)= ℜ

(f ′(z)

)= ℑ

(if ′(z)

)= ℑ

(i lim

x+it→x+iy

f (x+ it)− f (x+ iy)(x+ it)− (x+ iy)

)= ℑ

x+it→x+iyi

)= ℑ

(limt→y

)= lim

=∂v∂y

(x,y) = limt→x

t+iy→x+iy

f (t + iy)− f (x+ iy)(t + iy)− (x+ iy)

)= ℜ

(f ′(z)

)= ℑ

(if ′(z)

)= ℑ

(i lim

x+it→x+iy

f (x+ it)− f (x+ iy)(x+ it)− (x+ iy)

)= ℑ

x+it→x+iyi

)= ℑ

(limt→y

)= lim

=∂v∂y

Proof of “⇐.”

lim∆z→0

f (z+∆z)− f (z)∆z

= lim∆x,∆y→0

f (x+∆x,y+∆y)− f (x,y)∆x+ i∆y

= lim∆x,∆y→0

f (x+∆x,y+∆y)− f (x,y+∆y)∆x+ i∆y

+f (x,y+∆y)− f (x,y)

∆x+ i∆y

= lim∆x,∆y→0

u(x+∆x,y+∆y)−u(x,y+∆y)∆x+ i∆y

+ iv(x+∆x,y+∆y)− v(x,y+∆y)

∆x+ i∆y

+u(x,y+∆y)−u(x,y)

∆x+ i∆y+ i

v(x,y+∆y)− v(x,y)∆x+ i∆y

= lim∆x,∆y→0

∂u∂x (x+ξu,y+∆y)∆x

∆x+ i∆y+ i

∂v∂x (x+ξv,y+∆y)∆x

∆x+ i∆y

+∂u∂y (x,y+ηu)∆y

∆x+ i∆y+ i

∂v∂y (x,y+ηv)∆y

∆x+ i∆y

Proof of “⇐.”

lim∆z→0

f (z+∆z)− f (z)∆z

= lim∆x,∆y→0

f (x+∆x,y+∆y)− f (x,y)∆x+ i∆y

= lim∆x,∆y→0

f (x+∆x,y+∆y)− f (x,y+∆y)∆x+ i∆y

+f (x,y+∆y)− f (x,y)

∆x+ i∆y

= lim∆x,∆y→0

+ iv(x+∆x,y+∆y)− v(x,y+∆y)

∆x+ i∆y

∆x+ i∆y+ i

= lim∆x,∆y→0

∆x+ i∆y+ i

∆x+ i∆y

∆x+ i∆y+ i

∆x+ i∆y

Proof of “⇐.”

lim∆z→0

f (z+∆z)− f (z)∆z

= lim∆x,∆y→0

f (x+∆x,y+∆y)− f (x,y)∆x+ i∆y

= lim∆x,∆y→0

f (x+∆x,y+∆y)− f (x,y+∆y)∆x+ i∆y

+f (x,y+∆y)− f (x,y)

∆x+ i∆y

= lim∆x,∆y→0

+ iv(x+∆x,y+∆y)− v(x,y+∆y)

∆x+ i∆y

∆x+ i∆y+ i

= lim∆x,∆y→0

∆x+ i∆y+ i

∆x+ i∆y

∆x+ i∆y+ i

∆x+ i∆y

Proof of “⇐.”

lim∆z→0

f (z+∆z)− f (z)∆z

= lim∆x,∆y→0

f (x+∆x,y+∆y)− f (x,y)∆x+ i∆y

= lim∆x,∆y→0

f (x+∆x,y+∆y)− f (x,y+∆y)∆x+ i∆y

+f (x,y+∆y)− f (x,y)

∆x+ i∆y

= lim∆x,∆y→0

+ iv(x+∆x,y+∆y)− v(x,y+∆y)

∆x+ i∆y

∆x+ i∆y+ i

= lim∆x,∆y→0

∆x+ i∆y+ i

∆x+ i∆y

∆x+ i∆y+ i

∆x+ i∆y

Proof of “⇐.”

lim∆z→0

f (z+∆z)− f (z)∆z

= lim∆x,∆y→0

f (x+∆x,y+∆y)− f (x,y)∆x+ i∆y

= lim∆x,∆y→0

f (x+∆x,y+∆y)− f (x,y+∆y)∆x+ i∆y

+f (x,y+∆y)− f (x,y)

∆x+ i∆y

= lim∆x,∆y→0

+ iv(x+∆x,y+∆y)− v(x,y+∆y)

∆x+ i∆y

∆x+ i∆y+ i

= lim∆x,∆y→0

∆x+ i∆y+ i

∆x+ i∆y

∆x+ i∆y+ i

∆x+ i∆y

Proof of “⇐.”

lim∆z→0

f (z+∆z)− f (z)∆z

= lim∆x,∆y→0

f (x+∆x,y+∆y)− f (x,y)∆x+ i∆y

= lim∆x,∆y→0

f (x+∆x,y+∆y)− f (x,y+∆y)∆x+ i∆y

+f (x,y+∆y)− f (x,y)

∆x+ i∆y

= lim∆x,∆y→0

+ iv(x+∆x,y+∆y)− v(x,y+∆y)

∆x+ i∆y

∆x+ i∆y+ i

= lim∆x,∆y→0

∆x+ i∆y+ i

∆x+ i∆y

∆x+ i∆y+ i

∆x+ i∆y

Proof of “⇐” (concl.)

lim∆x,∆y→0

∆x+ i∆y+ i

∆x+ i∆y+

∂u∂y (x,y+ηu)∆y

∆x+ i∆y+ i

∆x+ i∆y

= lim∆x,∆y→0

∆x+ i∆y+ i

∆x+ i∆y

+− ∂v

∂x (x,y+ηu)∆y∆x+ i∆y

+ i∂u∂x (x,y+ηv)∆y

∆x+ i∆y

= lim∆x,∆y→0

∂u∂x (x+ξu,y+∆y)∆x+ ∂u

∂x (x,y+ηv)i∆y∆x+ i∆y

+i∂v∂x (x+ξv,y+∆y)∆x+ ∂v

∂x (x,y+ηu)i∆y∆x+ i∆y

=∂u∂x

(x,y)+ i∂v∂x

(Technical argument at end omitted.)

lim∆x,∆y→0

∆x+ i∆y+ i

∆x+ i∆y+

∆x+ i∆y+ i

∆x+ i∆y

= lim∆x,∆y→0

∆x+ i∆y+ i

∆x+ i∆y

+− ∂v

∆x+ i∆y

= lim∆x,∆y→0

=∂u∂x

(x,y)+ i∂v∂x

lim∆x,∆y→0

∆x+ i∆y+ i

∆x+ i∆y+

∆x+ i∆y+ i

∆x+ i∆y

= lim∆x,∆y→0

∆x+ i∆y+ i

∆x+ i∆y

+− ∂v

∆x+ i∆y

= lim∆x,∆y→0

=∂u∂x

(x,y)+ i∂v∂x

lim∆x,∆y→0

∆x+ i∆y+ i

∆x+ i∆y+

∆x+ i∆y+ i

∆x+ i∆y

= lim∆x,∆y→0

∆x+ i∆y+ i

∆x+ i∆y

+− ∂v

∆x+ i∆y

= lim∆x,∆y→0

=∂u∂x

(x,y)+ i∂v∂x

lim∆x,∆y→0

∆x+ i∆y+ i

∆x+ i∆y+

∆x+ i∆y+ i

∆x+ i∆y

= lim∆x,∆y→0

∆x+ i∆y+ i

∆x+ i∆y

+− ∂v

∆x+ i∆y

= lim∆x,∆y→0

=∂u∂x

(x,y)+ i∂v∂x

lim∆x,∆y→0

∆x+ i∆y+ i

∆x+ i∆y+

∆x+ i∆y+ i

∆x+ i∆y

= lim∆x,∆y→0

∆x+ i∆y+ i

∆x+ i∆y

+− ∂v

∆x+ i∆y

= lim∆x,∆y→0

=∂u∂x

(x,y)+ i∂v∂x

Example.

The function f (z) = z3 is differentiable on C.

f (z) = z3 = (x+ iy)3

= x3 +3ix2y−3xy2− iy3

= x3−3xy2 + i(3x2y− y3)

∂u∂x

= 3x2−3y2 =∂v∂y

∂u∂y

= −6xy =−∂v∂x

So by the Cauchy Riemann Equations, f is differentiable on C.

Example. The function f (z) = z3 is differentiable on C.

f (z) = z3 = (x+ iy)3

= x3 +3ix2y−3xy2− iy3

= x3−3xy2 + i(3x2y− y3)

∂u∂x

= 3x2−3y2 =∂v∂y

∂u∂y

= −6xy =−∂v∂x

f (z) = z3

= (x+ iy)3

= x3 +3ix2y−3xy2− iy3

= x3−3xy2 + i(3x2y− y3)

∂u∂x

= 3x2−3y2 =∂v∂y

∂u∂y

= −6xy =−∂v∂x

f (z) = z3 = (x+ iy)3

= x3 +3ix2y−3xy2− iy3

= x3−3xy2 + i(3x2y− y3)

∂u∂x

= 3x2−3y2 =∂v∂y

∂u∂y

= −6xy =−∂v∂x

f (z) = z3 = (x+ iy)3

= x3 +3ix2y−3xy2− iy3

= x3−3xy2 + i(3x2y− y3)

∂u∂x

= 3x2−3y2 =∂v∂y

∂u∂y

= −6xy =−∂v∂x

f (z) = z3 = (x+ iy)3

= x3 +3ix2y−3xy2− iy3

= x3−3xy2 + i(3x2y− y3)

∂u∂x

= 3x2−3y2 =∂v∂y

∂u∂y

= −6xy =−∂v∂x

f (z) = z3 = (x+ iy)3

= x3 +3ix2y−3xy2− iy3

= x3−3xy2 + i(3x2y− y3)

∂u∂x

= 3x2−3y2 =∂v∂y

∂u∂y

= −6xy =−∂v∂x

f (z) = z3 = (x+ iy)3

= x3 +3ix2y−3xy2− iy3

= x3−3xy2 + i(3x2y− y3)

∂u∂x

= 3x2−3y2

=∂v∂y

∂u∂y

= −6xy =−∂v∂x

f (z) = z3 = (x+ iy)3

= x3 +3ix2y−3xy2− iy3

= x3−3xy2 + i(3x2y− y3)

∂u∂x

= 3x2−3y2 =∂v∂y

∂u∂y

= −6xy =−∂v∂x

f (z) = z3 = (x+ iy)3

= x3 +3ix2y−3xy2− iy3

= x3−3xy2 + i(3x2y− y3)

∂u∂x

= 3x2−3y2 =∂v∂y

∂u∂y

= −6xy =−∂v∂x

f (z) = z3 = (x+ iy)3

= x3 +3ix2y−3xy2− iy3

= x3−3xy2 + i(3x2y− y3)

∂u∂x

= 3x2−3y2 =∂v∂y

∂u∂y

= −6xy

=−∂v∂x

f (z) = z3 = (x+ iy)3

= x3 +3ix2y−3xy2− iy3

= x3−3xy2 + i(3x2y− y3)

∂u∂x

= 3x2−3y2 =∂v∂y

∂u∂y

= −6xy =−∂v∂x

f (z) = z3 = (x+ iy)3

= x3 +3ix2y−3xy2− iy3

= x3−3xy2 + i(3x2y− y3)

∂u∂x

= 3x2−3y2 =∂v∂y

∂u∂y

= −6xy =−∂v∂x

Example.

The function f (z) = ez is differentiable on C.

f (z) = ez = ex+iy = exeiy = ex cos(y)+ iex sin(y)∂u∂x

= ex cos(y) =∂v∂y

∂u∂y

= −ex sin(y) =−∂v∂x

Example. The function f (z) = ez is differentiable on C.

∂u∂y

= ez = ex+iy = exeiy = ex cos(y)+ iex sin(y)∂u∂x

∂u∂y

f (z) = ez

= ex+iy = exeiy = ex cos(y)+ iex sin(y)∂u∂x

∂u∂y

f (z) = ez = ex+iy

= exeiy = ex cos(y)+ iex sin(y)∂u∂x

∂u∂y

f (z) = ez = ex+iy = exeiy

= ex cos(y)+ iex sin(y)∂u∂x

∂u∂y

f (z) = ez = ex+iy = exeiy = ex cos(y)+ iex sin(y)

∂u∂x

∂u∂y

= ex cos(y)

=∂v∂y

∂u∂y

= −ex sin(y)

=−∂v∂x

∂u∂y

Example.

The function f (z) = |z| is not differentiable at any z ∈ C.

f (z) = |z|=√

x2 + y2 + i ·0∂u∂x

x2 + y26= 0 (x 6= 0)

∂u∂y

x2 + y26= 0 (y 6= 0)

So f is not differentiable for z 6= 0 and, as a square root function, it isnot differentiable at the origin.

Example. The function f (z) = |z| is not differentiable at any z ∈ C.

f (z) = |z|=√

x2 + y2 + i ·0∂u∂x

x2 + y26= 0 (x 6= 0)

∂u∂y

x2 + y26= 0 (y 6= 0)

= |z|=√

x2 + y2 + i ·0∂u∂x

x2 + y26= 0 (x 6= 0)

∂u∂y

x2 + y26= 0 (y 6= 0)

f (z) = |z|

x2 + y2 + i ·0∂u∂x

x2 + y26= 0 (x 6= 0)

∂u∂y

x2 + y26= 0 (y 6= 0)

f (z) = |z|=√

x2 + y2

+ i ·0∂u∂x

x2 + y26= 0 (x 6= 0)

∂u∂y

x2 + y26= 0 (y 6= 0)

f (z) = |z|=√

x2 + y2 + i ·0

∂u∂x

x2 + y26= 0 (x 6= 0)

∂u∂y

x2 + y26= 0 (y 6= 0)

f (z) = |z|=√

x2 + y2 + i ·0∂u∂x

x2 + y26= 0 (x 6= 0)

∂u∂y

x2 + y26= 0 (y 6= 0)

f (z) = |z|=√

x2 + y2 + i ·0∂u∂x

x2 + y2

6= 0 (x 6= 0)

∂u∂y

x2 + y26= 0 (y 6= 0)

f (z) = |z|=√

x2 + y2 + i ·0∂u∂x

x2 + y26= 0

(x 6= 0)

∂u∂y

x2 + y26= 0 (y 6= 0)

f (z) = |z|=√

x2 + y2 + i ·0∂u∂x

x2 + y26= 0 (x 6= 0)

∂u∂y

x2 + y26= 0 (y 6= 0)

f (z) = |z|=√

x2 + y2 + i ·0∂u∂x

x2 + y26= 0 (x 6= 0)

∂u∂y

x2 + y26= 0 (y 6= 0)

f (z) = |z|=√

x2 + y2 + i ·0∂u∂x

x2 + y26= 0 (x 6= 0)

∂u∂y

x2 + y2

6= 0 (y 6= 0)

f (z) = |z|=√

x2 + y2 + i ·0∂u∂x

x2 + y26= 0 (x 6= 0)

∂u∂y

x2 + y26= 0

(y 6= 0)

f (z) = |z|=√

x2 + y2 + i ·0∂u∂x

x2 + y26= 0 (x 6= 0)

∂u∂y

x2 + y26= 0 (y 6= 0)

f (z) = |z|=√

x2 + y2 + i ·0∂u∂x

x2 + y26= 0 (x 6= 0)

∂u∂y

x2 + y26= 0 (y 6= 0)

So f is not differentiable for z 6= 0

and, as a square root function, it isnot differentiable at the origin.

f (z) = |z|=√

x2 + y2 + i ·0∂u∂x

x2 + y26= 0 (x 6= 0)

∂u∂y

x2 + y26= 0 (y 6= 0)

Theorem.

Cauchy-Riemann Equations revisited. Letf (z) = f

(reiθ)

= u(r,θ)+ iv(r,θ) be a function on an open domainthat does not contain zero and with continuous partial derivatives inthe underlying real variables. Then f is differentiable at z = reiθ ifand only if

r∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

The latter equations are called the Cauchy-Riemann Equations inpolar form.The value of the derivative is

f ′(z) = e−iθ(

∂u∂ r

(r,θ)+ i∂v∂ r

(r,θ))

Theorem. Cauchy-Riemann Equations revisited.

Letf (z) = f

(reiθ)

r∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

f ′(z) = e−iθ(

∂u∂ r

(r,θ)+ i∂v∂ r

(r,θ))

Theorem. Cauchy-Riemann Equations revisited. Letf (z) = f

(reiθ)

= u(r,θ)+ iv(r,θ) be a function on an open domainthat does not contain zero and with continuous partial derivatives inthe underlying real variables.

Then f is differentiable at z = reiθ ifand only if

r∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

f ′(z) = e−iθ(

∂u∂ r

(r,θ)+ i∂v∂ r

(r,θ))

(reiθ)

= u(r,θ)+ iv(r,θ) be a function on an open domainthat does not contain zero and with continuous partial derivatives inthe underlying real variables. Then f is differentiable at z = reiθ

ifand only if

r∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

f ′(z) = e−iθ(

∂u∂ r

(r,θ)+ i∂v∂ r

(r,θ))

(reiθ)

r∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

f ′(z) = e−iθ(

∂u∂ r

(r,θ)+ i∂v∂ r

(r,θ))

(reiθ)

r∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

f ′(z) = e−iθ(

∂u∂ r

(r,θ)+ i∂v∂ r

(r,θ))

(reiθ)

r∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

The latter equations are called the Cauchy-Riemann Equations inpolar form.

The value of the derivative is

f ′(z) = e−iθ(

∂u∂ r

(r,θ)+ i∂v∂ r

(r,θ))

(reiθ)

r∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

f ′(z) = e−iθ(

∂u∂ r

(r,θ)+ i∂v∂ r

(r,θ))

Proof (preparatory steps).

Remember that differentiability meansthat the Cauchy-Riemann Equations hold.

∂ r∂x

=∂√

x2 + y2

x√x2 + y2

= cos(θ)

∂ r∂y

=∂√

x2 + y2

y√x2 + y2

= sin(θ)

∂ arctan( y

(− y

)=− y

x2 + y2 =−sin(θ)r

∂ arctan( y

x2 + y2 =cos(θ)

Proof (preparatory steps). Remember that differentiability meansthat the Cauchy-Riemann Equations hold.

∂ r∂x

=∂√

x2 + y2

x√x2 + y2

= cos(θ)

∂ r∂y

=∂√

x2 + y2

y√x2 + y2

= sin(θ)

∂ arctan( y

(− y

)=− y

∂ arctan( y

x2 + y2 =cos(θ)

∂ r∂x

=∂√

x2 + y2

x√x2 + y2

= cos(θ)

∂ r∂y

=∂√

x2 + y2

y√x2 + y2

= sin(θ)

∂ arctan( y

(− y

)=− y

∂ arctan( y

x2 + y2 =cos(θ)

∂ r∂x

=∂√

x2 + y2

x2 + y2= cos(θ)

∂ r∂y

=∂√

x2 + y2

y√x2 + y2

= sin(θ)

∂ arctan( y

(− y

)=− y

∂ arctan( y

x2 + y2 =cos(θ)

∂ r∂x

=∂√

x2 + y2

x√x2 + y2

= cos(θ)

∂ r∂y

=∂√

x2 + y2

y√x2 + y2

= sin(θ)

∂ arctan( y

(− y

)=− y

∂ arctan( y

x2 + y2 =cos(θ)

∂ r∂x

=∂√

x2 + y2

x√x2 + y2

= cos(θ)

∂ r∂y

=∂√

x2 + y2

y√x2 + y2

= sin(θ)

∂ arctan( y

(− y

)=− y

∂ arctan( y

x2 + y2 =cos(θ)

∂ r∂x

=∂√

x2 + y2

x√x2 + y2

= cos(θ)

∂ r∂y

=∂√

x2 + y2

y√x2 + y2

= sin(θ)

∂ arctan( y

(− y

)=− y

∂ arctan( y

x2 + y2 =cos(θ)

∂ r∂x

=∂√

x2 + y2

x√x2 + y2

= cos(θ)

∂ r∂y

=∂√

x2 + y2

x2 + y2= sin(θ)

∂ arctan( y

(− y

)=− y

∂ arctan( y

x2 + y2 =cos(θ)

∂ r∂x

=∂√

x2 + y2

x√x2 + y2

= cos(θ)

∂ r∂y

=∂√

x2 + y2

y√x2 + y2

= sin(θ)

∂ arctan( y

(− y

)=− y

∂ arctan( y

x2 + y2 =cos(θ)

∂ r∂x

=∂√

x2 + y2

x√x2 + y2

= cos(θ)

∂ r∂y

=∂√

x2 + y2

y√x2 + y2

= sin(θ)

∂ arctan( y

(− y

)=− y

∂ arctan( y

x2 + y2 =cos(θ)

∂ r∂x

=∂√

x2 + y2

x√x2 + y2

= cos(θ)

∂ r∂y

=∂√

x2 + y2

y√x2 + y2

= sin(θ)

=∂ arctan

(− y

)=− y

∂ arctan( y

x2 + y2 =cos(θ)

∂ r∂x

=∂√

x2 + y2

x√x2 + y2

= cos(θ)

∂ r∂y

=∂√

x2 + y2

y√x2 + y2

= sin(θ)

∂ arctan( y

(− y

)=− y

∂ arctan( y

x2 + y2 =cos(θ)

∂ r∂x

=∂√

x2 + y2

x√x2 + y2

= cos(θ)

∂ r∂y

=∂√

x2 + y2

y√x2 + y2

= sin(θ)

∂ arctan( y

(− y

=− yx2 + y2 =−sin(θ)

∂ arctan( y

x2 + y2 =cos(θ)

∂ r∂x

=∂√

x2 + y2

x√x2 + y2

= cos(θ)

∂ r∂y

=∂√

x2 + y2

y√x2 + y2

= sin(θ)

∂ arctan( y

(− y

)=− y

x2 + y2

=−sin(θ)r

∂ arctan( y

x2 + y2 =cos(θ)

∂ r∂x

=∂√

x2 + y2

x√x2 + y2

= cos(θ)

∂ r∂y

=∂√

x2 + y2

y√x2 + y2

= sin(θ)

∂ arctan( y

(− y

)=− y

∂ arctan( y

x2 + y2 =cos(θ)

∂ r∂x

=∂√

x2 + y2

x√x2 + y2

= cos(θ)

∂ r∂y

=∂√

x2 + y2

y√x2 + y2

= sin(θ)

∂ arctan( y

(− y

)=− y

=∂ arctan

x2 + y2 =cos(θ)

∂ r∂x

=∂√

x2 + y2

x√x2 + y2

= cos(θ)

∂ r∂y

=∂√

x2 + y2

y√x2 + y2

= sin(θ)

∂ arctan( y

(− y

)=− y

∂ arctan( y

x2 + y2 =cos(θ)

∂ r∂x

=∂√

x2 + y2

x√x2 + y2

= cos(θ)

∂ r∂y

=∂√

x2 + y2

y√x2 + y2

= sin(θ)

∂ arctan( y

(− y

)=− y

∂ arctan( y

x2 + y2 =cos(θ)

∂ r∂x

=∂√

x2 + y2

x√x2 + y2

= cos(θ)

∂ r∂y

=∂√

x2 + y2

y√x2 + y2

= sin(θ)

∂ arctan( y

(− y

)=− y

∂ arctan( y

x2 + y2

=cos(θ)

∂ r∂x

=∂√

x2 + y2

x√x2 + y2

= cos(θ)

∂ r∂y

=∂√

x2 + y2

y√x2 + y2

= sin(θ)

∂ arctan( y

(− y

)=− y

∂ arctan( y

x2 + y2 =cos(θ)

Proof.

∂u∂x

=∂u∂ r

∂ r∂x

+∂u∂θ

∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

)∂v∂y

=∂v∂ r

∂ r∂y

+∂v∂θ

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂y

=∂u∂ r

∂ r∂y

+∂u∂θ

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

∂v∂x

=∂v∂ r

∂ r∂x

+∂v∂θ

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

, then by direct substitution∂u∂x

=∂v∂y

and∂u∂y

=−∂v∂x

. Conversely, if∂u∂x

=∂v∂y

and∂u∂y

=−∂v∂x

, setting the

above equal to each other, we obtain

Proof.∂u∂x

=∂u∂ r

∂ r∂x

+∂u∂θ

∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

)∂v∂y

=∂v∂ r

∂ r∂y

+∂v∂θ

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂y

=∂u∂ r

∂ r∂y

+∂u∂θ

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

∂v∂x

=∂v∂ r

∂ r∂x

+∂v∂θ

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

=∂v∂y

and∂u∂y

=−∂v∂x

=∂v∂y

and∂u∂y

=−∂v∂x

, setting the

Proof.∂u∂x

=∂u∂ r

∂ r∂x

+∂u∂θ

=∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

)∂v∂y

=∂v∂ r

∂ r∂y

+∂v∂θ

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂y

=∂u∂ r

∂ r∂y

+∂u∂θ

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

∂v∂x

=∂v∂ r

∂ r∂x

+∂v∂θ

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

=∂v∂y

and∂u∂y

=−∂v∂x

=∂v∂y

and∂u∂y

=−∂v∂x

, setting the

Proof.∂u∂x

=∂u∂ r

∂ r∂x

+∂u∂θ

∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

∂v∂y

=∂v∂ r

∂ r∂y

+∂v∂θ

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂y

=∂u∂ r

∂ r∂y

+∂u∂θ

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

∂v∂x

=∂v∂ r

∂ r∂x

+∂v∂θ

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

=∂v∂y

and∂u∂y

=−∂v∂x

=∂v∂y

and∂u∂y

=−∂v∂x

, setting the

Proof.∂u∂x

=∂u∂ r

∂ r∂x

+∂u∂θ

∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

)∂v∂y

=∂v∂ r

∂ r∂y

+∂v∂θ

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂y

=∂u∂ r

∂ r∂y

+∂u∂θ

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

∂v∂x

=∂v∂ r

∂ r∂x

+∂v∂θ

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

=∂v∂y

and∂u∂y

=−∂v∂x

=∂v∂y

and∂u∂y

=−∂v∂x

, setting the

Proof.∂u∂x

=∂u∂ r

∂ r∂x

+∂u∂θ

∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

)∂v∂y

=∂v∂ r

∂ r∂y

+∂v∂θ

=∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂y

=∂u∂ r

∂ r∂y

+∂u∂θ

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

∂v∂x

=∂v∂ r

∂ r∂x

+∂v∂θ

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

=∂v∂y

and∂u∂y

=−∂v∂x

=∂v∂y

and∂u∂y

=−∂v∂x

, setting the

Proof.∂u∂x

=∂u∂ r

∂ r∂x

+∂u∂θ

∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

)∂v∂y

=∂v∂ r

∂ r∂y

+∂v∂θ

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂y

=∂u∂ r

∂ r∂y

+∂u∂θ

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

∂v∂x

=∂v∂ r

∂ r∂x

+∂v∂θ

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

=∂v∂y

and∂u∂y

=−∂v∂x

=∂v∂y

and∂u∂y

=−∂v∂x

, setting the

Proof.∂u∂x

=∂u∂ r

∂ r∂x

+∂u∂θ

∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

)∂v∂y

=∂v∂ r

∂ r∂y

+∂v∂θ

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂y

=∂u∂ r

∂ r∂y

+∂u∂θ

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

∂v∂x

=∂v∂ r

∂ r∂x

+∂v∂θ

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

=∂v∂y

and∂u∂y

=−∂v∂x

=∂v∂y

and∂u∂y

=−∂v∂x

, setting the

Proof.∂u∂x

=∂u∂ r

∂ r∂x

+∂u∂θ

∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

)∂v∂y

=∂v∂ r

∂ r∂y

+∂v∂θ

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂y

=∂u∂ r

∂ r∂y

+∂u∂θ

=∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

∂v∂x

=∂v∂ r

∂ r∂x

+∂v∂θ

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

=∂v∂y

and∂u∂y

=−∂v∂x

=∂v∂y

and∂u∂y

=−∂v∂x

, setting the

Proof.∂u∂x

=∂u∂ r

∂ r∂x

+∂u∂θ

∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

)∂v∂y

=∂v∂ r

∂ r∂y

+∂v∂θ

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂y

=∂u∂ r

∂ r∂y

+∂u∂θ

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

∂v∂x

=∂v∂ r

∂ r∂x

+∂v∂θ

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

=∂v∂y

and∂u∂y

=−∂v∂x

=∂v∂y

and∂u∂y

=−∂v∂x

, setting the

Proof.∂u∂x

=∂u∂ r

∂ r∂x

+∂u∂θ

∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

)∂v∂y

=∂v∂ r

∂ r∂y

+∂v∂θ

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂y

=∂u∂ r

∂ r∂y

+∂u∂θ

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

∂v∂x

=∂v∂ r

∂ r∂x

+∂v∂θ

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

=∂v∂y

and∂u∂y

=−∂v∂x

=∂v∂y

and∂u∂y

=−∂v∂x

, setting the

Proof.∂u∂x

=∂u∂ r

∂ r∂x

+∂u∂θ

∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

)∂v∂y

=∂v∂ r

∂ r∂y

+∂v∂θ

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂y

=∂u∂ r

∂ r∂y

+∂u∂θ

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

∂v∂x

=∂v∂ r

∂ r∂x

+∂v∂θ

=∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

=∂v∂y

and∂u∂y

=−∂v∂x

=∂v∂y

and∂u∂y

=−∂v∂x

, setting the

Proof.∂u∂x

=∂u∂ r

∂ r∂x

+∂u∂θ

∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

)∂v∂y

=∂v∂ r

∂ r∂y

+∂v∂θ

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂y

=∂u∂ r

∂ r∂y

+∂u∂θ

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

∂v∂x

=∂v∂ r

∂ r∂x

+∂v∂θ

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

If r∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

=∂v∂y

and∂u∂y

=−∂v∂x

=∂v∂y

and∂u∂y

=−∂v∂x

, setting the

Proof.∂u∂x

=∂u∂ r

∂ r∂x

+∂u∂θ

∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

)∂v∂y

=∂v∂ r

∂ r∂y

+∂v∂θ

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂y

=∂u∂ r

∂ r∂y

+∂u∂θ

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

∂v∂x

=∂v∂ r

∂ r∂x

+∂v∂θ

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

=∂v∂y

and∂u∂y

=−∂v∂x

=∂v∂y

and∂u∂y

=−∂v∂x

, setting the

Proof.∂u∂x

=∂u∂ r

∂ r∂x

+∂u∂θ

∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

)∂v∂y

=∂v∂ r

∂ r∂y

+∂v∂θ

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂y

=∂u∂ r

∂ r∂y

+∂u∂θ

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

∂v∂x

=∂v∂ r

∂ r∂x

+∂v∂θ

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

=∂v∂y

and∂u∂y

=−∂v∂x

=∂v∂y

and∂u∂y

=−∂v∂x

, setting the

Proof.∂u∂x

=∂u∂ r

∂ r∂x

+∂u∂θ

∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

)∂v∂y

=∂v∂ r

∂ r∂y

+∂v∂θ

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂y

=∂u∂ r

∂ r∂y

+∂u∂θ

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

∂v∂x

=∂v∂ r

∂ r∂x

+∂v∂θ

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

=∂v∂y

and∂u∂y

=−∂v∂x

Conversely, if∂u∂x

=∂v∂y

and∂u∂y

=−∂v∂x

, setting the

Proof.∂u∂x

=∂u∂ r

∂ r∂x

+∂u∂θ

∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

)∂v∂y

=∂v∂ r

∂ r∂y

+∂v∂θ

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂y

=∂u∂ r

∂ r∂y

+∂u∂θ

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

∂v∂x

=∂v∂ r

∂ r∂x

+∂v∂θ

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

=∂v∂y

and∂u∂y

=−∂v∂x

=∂v∂y

and∂u∂y

=−∂v∂x

, setting the

Proof.∂u∂x

=∂u∂ r

∂ r∂x

+∂u∂θ

∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

)∂v∂y

=∂v∂ r

∂ r∂y

+∂v∂θ

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂y

=∂u∂ r

∂ r∂y

+∂u∂θ

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

∂v∂x

=∂v∂ r

∂ r∂x

+∂v∂θ

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

=∂v∂θ

and∂u∂θ

=−r∂v∂ r

=∂v∂y

and∂u∂y

=−∂v∂x

=∂v∂y

and∂u∂y

=−∂v∂x

, setting the

Proof.

∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

= −∂v∂ r

cos(θ)+∂v∂θ

sin(θ)r

Multiplying the first with r cos(θ), the second with r sin(θ) and

adding gives r∂u∂ r

=∂v∂θ

. Multiplying the first with −r sin(θ), the

second with r cos(θ) and adding gives∂u∂θ

=−r∂v∂ r

.The above shows that f is differentiable if and only if theCauchy-Riemann Equations in polar form are satisfied. (We omittedthe technical proof that continuity of the partial derivatives in onecoordinate system gives their continuity in the other.)Now we turn to the formula for the derivative.

Proof.∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

=∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

= −∂v∂ r

cos(θ)+∂v∂θ

sin(θ)r

=∂v∂θ

=−r∂v∂ r

Proof.∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

= −∂v∂ r

cos(θ)+∂v∂θ

sin(θ)r

=∂v∂θ

=−r∂v∂ r

Proof.∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

= −∂v∂ r

cos(θ)+∂v∂θ

sin(θ)r

=∂v∂θ

=−r∂v∂ r

Proof.∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

= −∂v∂ r

cos(θ)+∂v∂θ

sin(θ)r

=∂v∂θ

=−r∂v∂ r

Proof.∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

= −∂v∂ r

cos(θ)+∂v∂θ

sin(θ)r

=∂v∂θ

Multiplying the first with −r sin(θ), the

=−r∂v∂ r

Proof.∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

= −∂v∂ r

cos(θ)+∂v∂θ

sin(θ)r

=∂v∂θ

=−r∂v∂ r

The above shows that f is differentiable if and only if theCauchy-Riemann Equations in polar form are satisfied. (We omittedthe technical proof that continuity of the partial derivatives in onecoordinate system gives their continuity in the other.)Now we turn to the formula for the derivative.

Proof.∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

= −∂v∂ r

cos(θ)+∂v∂θ

sin(θ)r

=∂v∂θ

=−r∂v∂ r

.The above shows that f is differentiable if and only if theCauchy-Riemann Equations in polar form are satisfied.

(We omittedthe technical proof that continuity of the partial derivatives in onecoordinate system gives their continuity in the other.)Now we turn to the formula for the derivative.

Proof.∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

= −∂v∂ r

cos(θ)+∂v∂θ

sin(θ)r

=∂v∂θ

=−r∂v∂ r

.The above shows that f is differentiable if and only if theCauchy-Riemann Equations in polar form are satisfied. (We omittedthe technical proof that continuity of the partial derivatives in onecoordinate system gives their continuity in the other.)

Now we turn to the formula for the derivative.

Proof.∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

∂v∂ r

sin(θ)+∂v∂θ

cos(θ)r

∂u∂ r

sin(θ)+∂u∂θ

cos(θ)r

= −∂v∂ r

cos(θ)+∂v∂θ

sin(θ)r

=∂v∂θ

=−r∂v∂ r

Proof.

f ′(z) =∂u∂x

+ i∂v∂x

=∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

cos(θ)+∂v∂ r

sin(θ)+ i(

∂v∂ r

cos(θ)+∂u∂ r

(−sin(θ)))

=∂u∂ r

cos(−θ)+∂u∂ r

isin(−θ)+ i∂v∂ r

cos(−θ)

= e−iθ(

∂u∂ r

(r,θ)+ i∂v∂ r

(r,θ))

Proof.

f ′(z)

=∂u∂x

+ i∂v∂x

=∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

cos(θ)+∂v∂ r

sin(θ)+ i(

∂v∂ r

cos(θ)+∂u∂ r

(−sin(θ)))

=∂u∂ r

cos(−θ)

= e−iθ(

∂u∂ r

(r,θ)+ i∂v∂ r

(r,θ))

Proof.

f ′(z) =∂u∂x

+ i∂v∂x

=∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

cos(θ)+∂v∂ r

sin(θ)+ i(

∂v∂ r

cos(θ)+∂u∂ r

(−sin(θ)))

=∂u∂ r

cos(−θ)

= e−iθ(

∂u∂ r

(r,θ)+ i∂v∂ r

(r,θ))

Proof.

f ′(z) =∂u∂x

+ i∂v∂x

=∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

cos(θ)+∂v∂ r

sin(θ)+ i(

∂v∂ r

cos(θ)+∂u∂ r

(−sin(θ)))

=∂u∂ r

cos(−θ)

= e−iθ(

∂u∂ r

(r,θ)+ i∂v∂ r

(r,θ))

Proof.

f ′(z) =∂u∂x

+ i∂v∂x

=∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

=∂u∂ r

cos(θ)+∂v∂ r

sin(θ)+ i(

∂v∂ r

cos(θ)+∂u∂ r

(−sin(θ)))

=∂u∂ r

cos(−θ)

= e−iθ(

∂u∂ r

(r,θ)+ i∂v∂ r

(r,θ))

Proof.

f ′(z) =∂u∂x

+ i∂v∂x

=∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

cos(θ)+∂v∂ r

sin(θ)

∂v∂ r

cos(θ)+∂u∂ r

(−sin(θ)))

=∂u∂ r

cos(−θ)

= e−iθ(

∂u∂ r

(r,θ)+ i∂v∂ r

(r,θ))

Proof.

f ′(z) =∂u∂x

+ i∂v∂x

=∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

cos(θ)+∂v∂ r

sin(θ)+ i(

∂v∂ r

cos(θ)+∂u∂ r

(−sin(θ)))

=∂u∂ r

cos(−θ)

= e−iθ(

∂u∂ r

(r,θ)+ i∂v∂ r

(r,θ))

Proof.

f ′(z) =∂u∂x

+ i∂v∂x

=∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

cos(θ)+∂v∂ r

sin(θ)+ i(

∂v∂ r

cos(θ)+∂u∂ r

(−sin(θ)))

=∂u∂ r

cos(−θ)

= e−iθ(

∂u∂ r

(r,θ)+ i∂v∂ r

(r,θ))

Proof.

f ′(z) =∂u∂x

+ i∂v∂x

=∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

cos(θ)+∂v∂ r

sin(θ)+ i(

∂v∂ r

cos(θ)+∂u∂ r

(−sin(θ)))

=∂u∂ r

cos(−θ)

= e−iθ(

∂u∂ r

(r,θ)+ i∂v∂ r

(r,θ))

Proof.

f ′(z) =∂u∂x

+ i∂v∂x

=∂u∂ r

cos(θ)+∂u∂θ

(−sin(θ)

∂v∂ r

cos(θ)+∂v∂θ

(−sin(θ)

∂u∂ r

cos(θ)+∂v∂ r

sin(θ)+ i(

∂v∂ r

cos(θ)+∂u∂ r

(−sin(θ)))

=∂u∂ r

cos(−θ)

= e−iθ(

∂u∂ r

(r,θ)+ i∂v∂ r

(r,θ))

Example.

The function f (z) = f(reiθ)

= r12 ei θ

2 is differentiable onany set r > δ ≥ 0, α < θ < α +2π .

f (z) = r12 ei θ

= r12 cos

12 sin

∂u∂ r

r−12 cos

r12 cos

∂v∂θ

− r∂v∂ r

= −r12

r−12 sin

)= −1

12 sin

∂u∂θ

Example. The function f (z) = f(reiθ)

= r12 ei θ

f (z) = r12 ei θ

= r12 cos

12 sin

∂u∂ r

r−12 cos

r12 cos

∂v∂θ

− r∂v∂ r

= −r12

r−12 sin

)= −1

12 sin

∂u∂θ

= r12 ei θ

f (z) = r12 ei θ

= r12 cos

12 sin

∂u∂ r

r−12 cos

r12 cos

∂v∂θ

− r∂v∂ r

= −r12

r−12 sin

)= −1

12 sin

∂u∂θ

= r12 ei θ

f (z) = r12 ei θ

= r12 cos

12 sin

r∂u∂ r

r−12 cos

r12 cos

∂v∂θ

− r∂v∂ r

= −r12

r−12 sin

)= −1

12 sin

∂u∂θ

= r12 ei θ

f (z) = r12 ei θ

= r12 cos

12 sin

∂u∂ r

r−12 cos

r12 cos

∂v∂θ

− r∂v∂ r

= −r12

r−12 sin

)= −1

12 sin

∂u∂θ

= r12 ei θ

f (z) = r12 ei θ

= r12 cos

12 sin

∂u∂ r

r−12 cos

r12 cos

=∂v∂θ

− r∂v∂ r

= −r12

r−12 sin

)= −1

12 sin

∂u∂θ

= r12 ei θ

f (z) = r12 ei θ

= r12 cos

12 sin

∂u∂ r

r−12 cos

r12 cos

∂v∂θ

− r∂v∂ r

= −r12

r−12 sin

)= −1

12 sin

∂u∂θ

= r12 ei θ

f (z) = r12 ei θ

= r12 cos

12 sin

∂u∂ r

r−12 cos

r12 cos

∂v∂θ

− r∂v∂ r

= −r12

r−12 sin

= −12

r12 sin

∂u∂θ

= r12 ei θ

f (z) = r12 ei θ

= r12 cos

12 sin

∂u∂ r

r−12 cos

r12 cos

∂v∂θ

− r∂v∂ r

= −r12

r−12 sin

)= −1

12 sin

=∂u∂θ

= r12 ei θ

f (z) = r12 ei θ

= r12 cos

12 sin

∂u∂ r

r−12 cos

r12 cos

∂v∂θ

− r∂v∂ r

= −r12

r−12 sin

)= −1

12 sin

∂u∂θ

Example.

The function f (z) = f(

reiθ)

= ln(r)+ iθ is differentiableon any set r > δ ≥ 0, α < θ < α +2π .

f (z) = ln(r)+ iθ

r∂u∂ r

= 1 =∂v∂θ

−r∂v∂ r

= 0 =∂u∂θ

Example. The function f (z) = f(

reiθ)

f (z) = ln(r)+ iθ

r∂u∂ r

= 1 =∂v∂θ

−r∂v∂ r

= 0 =∂u∂θ

reiθ)

f (z) = ln(r)+ iθ

r∂u∂ r

= 1 =∂v∂θ

−r∂v∂ r

= 0 =∂u∂θ

reiθ)

f (z) = ln(r)+ iθ

r∂u∂ r

= 1 =∂v∂θ

−r∂v∂ r

= 0 =∂u∂θ

reiθ)

f (z) = ln(r)+ iθ

r∂u∂ r

= 1 =∂v∂θ

−r∂v∂ r

= 0 =∂u∂θ

reiθ)

f (z) = ln(r)+ iθ

r∂u∂ r

=∂v∂θ

−r∂v∂ r

= 0 =∂u∂θ

reiθ)

f (z) = ln(r)+ iθ

r∂u∂ r

= 1 =∂v∂θ

−r∂v∂ r

= 0 =∂u∂θ

reiθ)

f (z) = ln(r)+ iθ

r∂u∂ r

= 1 =∂v∂θ

−r∂v∂ r

= 0 =∂u∂θ

reiθ)

f (z) = ln(r)+ iθ

r∂u∂ r

= 1 =∂v∂θ

−r∂v∂ r

=∂u∂θ

reiθ)

f (z) = ln(r)+ iθ

r∂u∂ r

= 1 =∂v∂θ

−r∂v∂ r

= 0 =∂u∂θ

Theorem.

If f is analytic on a connected domain D and f ′(z) = 0 forevery z in the domain, then f must be constant.

Proof. Via the Fundamental Theorem of Calculus, we obtain

u(x1,y0)−u(x0,y0) =∫ x1

∂u∂x

(x,y0) dx

u(x0,y1)−u(x0,y0) =∫ y1

∂u∂y

(x0,y) dy

So, because∂u∂x

=∂u∂y

= 0, we conclude that u does not change along

horizontal or vertical lines. Similarly v does not change alonghorizontal or vertical lines. Hence f does not change along horizontalor vertical lines.

Theorem. If f is analytic on a connected domain D and f ′(z) = 0 forevery z in the domain, then f must be constant.

u(x1,y0)−u(x0,y0) =∫ x1

∂u∂x

(x,y0) dx

u(x0,y1)−u(x0,y0) =∫ y1

∂u∂y

(x0,y) dy

So, because∂u∂x

=∂u∂y

Proof.

Via the Fundamental Theorem of Calculus, we obtain

u(x1,y0)−u(x0,y0) =∫ x1

∂u∂x

(x,y0) dx

u(x0,y1)−u(x0,y0) =∫ y1

∂u∂y

(x0,y) dy

So, because∂u∂x

=∂u∂y

u(x1,y0)−u(x0,y0) =∫ x1

∂u∂x

(x,y0) dx

u(x0,y1)−u(x0,y0) =∫ y1

∂u∂y

(x0,y) dy

So, because∂u∂x

=∂u∂y

u(x1,y0)−u(x0,y0) =∫ x1

∂u∂x

(x,y0) dx

u(x0,y1)−u(x0,y0) =∫ y1

∂u∂y

(x0,y) dy

So, because∂u∂x

=∂u∂y

u(x1,y0)−u(x0,y0) =∫ x1

∂u∂x

(x,y0) dx

u(x0,y1)−u(x0,y0) =∫ y1

∂u∂y

(x0,y) dy

So, because∂u∂x

=∂u∂y

u(x1,y0)−u(x0,y0) =∫ x1

∂u∂x

(x,y0) dx

u(x0,y1)−u(x0,y0) =∫ y1

∂u∂y

(x0,y) dy

So, because∂u∂x

=∂u∂y

, we conclude that u does not change along

u(x1,y0)−u(x0,y0) =∫ x1

∂u∂x

(x,y0) dx

u(x0,y1)−u(x0,y0) =∫ y1

∂u∂y

(x0,y) dy

So, because∂u∂x

=∂u∂y

horizontal or vertical lines.

Similarly v does not change alonghorizontal or vertical lines. Hence f does not change along horizontalor vertical lines.

u(x1,y0)−u(x0,y0) =∫ x1

∂u∂x

(x,y0) dx

u(x0,y1)−u(x0,y0) =∫ y1

∂u∂y

(x0,y) dy

So, because∂u∂x

=∂u∂y

horizontal or vertical lines. Similarly v does not change alonghorizontal or vertical lines.

Hence f does not change along horizontalor vertical lines.

u(x1,y0)−u(x0,y0) =∫ x1

∂u∂x

(x,y0) dx

u(x0,y1)−u(x0,y0) =∫ y1

∂u∂y

(x0,y) dy

So, because∂u∂x

=∂u∂y

Proof.

Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines, f is constant.

Proof.

Because any two points in the domain can be connected via a pathconsisting of horizontal and vertical lines

, f is constant.

Proof.

Example.

Determine where f (x) =3z2 +5

(z3−2)2 (z4 +16)3 is analytic.

Note that there is no need to compute the derivative or to use theCauchy-Riemann Equations. (Just because we have new tools doesnot mean that the old ideas are discarded.) By product rule, quotientrule and chain rule, the function is analytic on its whole domain. So itis analytic except at z = 3

√2, 3√

2ei 2π

3 , 3√

2ei 4π

3 andz = 2ei π

4 ,2ei 3π

4 ,2ei 5π

4 ,2ei 7π

Example. Determine where f (x) =3z2 +5

√2, 3√

2ei 2π

3 , 3√

2ei 4π

3 andz = 2ei π

4 ,2ei 3π

4 ,2ei 5π

4 ,2ei 7π

Note that there is no need to compute the derivative

or to use theCauchy-Riemann Equations. (Just because we have new tools doesnot mean that the old ideas are discarded.) By product rule, quotientrule and chain rule, the function is analytic on its whole domain. So itis analytic except at z = 3

√2, 3√

2ei 2π

3 , 3√

2ei 4π

3 andz = 2ei π

4 ,2ei 3π

4 ,2ei 5π

4 ,2ei 7π

Note that there is no need to compute the derivative or to use theCauchy-Riemann Equations.

(Just because we have new tools doesnot mean that the old ideas are discarded.) By product rule, quotientrule and chain rule, the function is analytic on its whole domain. So itis analytic except at z = 3

√2, 3√

2ei 2π

3 , 3√

2ei 4π

3 andz = 2ei π

4 ,2ei 3π

4 ,2ei 5π

4 ,2ei 7π

Note that there is no need to compute the derivative or to use theCauchy-Riemann Equations. (Just because we have new tools doesnot mean that the old ideas are discarded.)

By product rule, quotientrule and chain rule, the function is analytic on its whole domain. So itis analytic except at z = 3

√2, 3√

2ei 2π

3 , 3√

2ei 4π

3 andz = 2ei π

4 ,2ei 3π

4 ,2ei 5π

4 ,2ei 7π

Note that there is no need to compute the derivative or to use theCauchy-Riemann Equations. (Just because we have new tools doesnot mean that the old ideas are discarded.) By product rule, quotientrule and chain rule, the function is analytic on its whole domain.

So itis analytic except at z = 3

√2, 3√

2ei 2π

3 , 3√

2ei 4π

3 andz = 2ei π

4 ,2ei 3π

4 ,2ei 5π

4 ,2ei 7π

Note that there is no need to compute the derivative or to use theCauchy-Riemann Equations. (Just because we have new tools doesnot mean that the old ideas are discarded.) By product rule, quotientrule and chain rule, the function is analytic on its whole domain. So itis analytic except at z =

2, 3√

2ei 2π

3 , 3√

2ei 4π

3 andz = 2ei π

4 ,2ei 3π

4 ,2ei 5π

4 ,2ei 7π

, 3√

2ei 2π

3 , 3√

2ei 4π

3 andz = 2ei π

4 ,2ei 3π

4 ,2ei 5π

4 ,2ei 7π

√2, 3√

2ei 2π

, 3√

2ei 4π

3 andz = 2ei π

4 ,2ei 3π

4 ,2ei 5π

4 ,2ei 7π

√2, 3√

2ei 2π

3 , 3√

2ei 4π

andz = 2ei π

4 ,2ei 3π

4 ,2ei 5π

4 ,2ei 7π

√2, 3√

2ei 2π

3 , 3√

2ei 4π

3 andz =

2ei π

4 ,2ei 3π

4 ,2ei 5π

4 ,2ei 7π

√2, 3√

2ei 2π

3 , 3√

2ei 4π

3 andz = 2ei π

,2ei 3π

4 ,2ei 5π

4 ,2ei 7π

√2, 3√

2ei 2π

3 , 3√

2ei 4π

3 andz = 2ei π

4 ,2ei 3π

,2ei 5π

4 ,2ei 7π

√2, 3√

2ei 2π

3 , 3√

2ei 4π

3 andz = 2ei π

4 ,2ei 3π

4 ,2ei 5π

,2ei 7π

√2, 3√

2ei 2π

3 , 3√

2ei 4π

3 andz = 2ei π

4 ,2ei 3π

4 ,2ei 5π

4 ,2ei 7π

Proposition.

Let f be an analytic function on a connected domain sothat f is analytic, too. Then f is constant.

Proof. Applying the Cauchy Riemann Equations to f = u+ iv and

f = u− iv, we obtain∂u∂x

=∂v∂y

and∂u∂y

=−∂v∂x

as well as∂u∂x

=−∂v∂y

and∂u∂y

=∂v∂x

. But then all partial derivatives are zero. Hence f ′ is

zero and f is constant.

Proposition. Let f be an analytic function on a connected domain sothat f is analytic, too.

Then f is constant.

=∂v∂y

and∂u∂y

=−∂v∂x

as well as∂u∂x

=−∂v∂y

and∂u∂y

=∂v∂x

Proposition. Let f be an analytic function on a connected domain sothat f is analytic, too. Then f is constant.

=∂v∂y

and∂u∂y

=−∂v∂x

as well as∂u∂x

=−∂v∂y

and∂u∂y

=∂v∂x

Proof.

Applying the Cauchy Riemann Equations to f = u+ iv and

=∂v∂y

and∂u∂y

=−∂v∂x

as well as∂u∂x

=−∂v∂y

and∂u∂y

=∂v∂x

=∂v∂y

and∂u∂y

=−∂v∂x

as well as∂u∂x

=−∂v∂y

and∂u∂y

=∂v∂x

=∂v∂y

and∂u∂y

=−∂v∂x

as well as∂u∂x

=−∂v∂y

and∂u∂y

=∂v∂x

=∂v∂y

and∂u∂y

=−∂v∂x

as well as∂u∂x

=−∂v∂y

and∂u∂y

=∂v∂x

=∂v∂y

and∂u∂y

=−∂v∂x

as well as∂u∂x

=−∂v∂y

and∂u∂y

=∂v∂x

But then all partial derivatives are zero. Hence f ′ is

=∂v∂y

and∂u∂y

=−∂v∂x

as well as∂u∂x

=−∂v∂y

and∂u∂y

=∂v∂x

. But then all partial derivatives are zero.

Hence f ′ is

=∂v∂y

and∂u∂y

=−∂v∂x

as well as∂u∂x

=−∂v∂y

and∂u∂y

=∂v∂x

and f is constant.

=∂v∂y

and∂u∂y

=−∂v∂x

as well as∂u∂x

=−∂v∂y

and∂u∂y

=∂v∂x

=∂v∂y

and∂u∂y

=−∂v∂x

as well as∂u∂x

=−∂v∂y

and∂u∂y

=∂v∂x

Proposition.

Let f be an analytic function on a connected domain sothat |f | is constant. Then f is constant.

Proof. Because f f = |f |2 we have that f =|f |2

fis analytic. By the

preceding result, f must be constant.

Proposition. Let f be an analytic function on a connected domain sothat |f | is constant.

Then f is constant.

Proposition. Let f be an analytic function on a connected domain sothat |f | is constant. Then f is constant.

Proof.

Because f f = |f |2 we have that f =|f |2

Proof. Because f f = |f |2

we have that f =|f |2

fis analytic.

By the

Definition.

A real valued function u of two variables is calledharmonic if and only if it has continuous first and second order

partial derivatives and satisfies the equation∂ 2u∂x2 +

∂ 2u∂y2 = 0. The

above equation is called Laplace’s equation.

Laplace’s equation describes steady state heat transfer as well as wavemotion, because the (two-dimensional) wave equation is

∂ 2u∂x2 +

∂ 2u∂y2 = k

∂ 2u∂ t2

and the (two-dimensional) heat equation is

∂ 2u∂x2 +

∂ 2u∂y2 = k

∂u∂ t

Definition. A real valued function u of two variables is calledharmonic

if and only if it has continuous first and second order

∂ 2u∂y2 = 0. The

∂ 2u∂x2 +

∂ 2u∂y2 = k

∂ 2u∂ t2

∂ 2u∂x2 +

∂ 2u∂y2 = k

∂u∂ t

Definition. A real valued function u of two variables is calledharmonic if and only if it has continuous first and second order

partial derivatives

and satisfies the equation∂ 2u∂x2 +

∂ 2u∂y2 = 0. The

∂ 2u∂x2 +

∂ 2u∂y2 = k

∂ 2u∂ t2

∂ 2u∂x2 +

∂ 2u∂y2 = k

∂u∂ t

∂ 2u∂y2 = 0.

∂ 2u∂x2 +

∂ 2u∂y2 = k

∂ 2u∂ t2

∂ 2u∂x2 +

∂ 2u∂y2 = k

∂u∂ t

∂ 2u∂y2 = 0. The

∂ 2u∂x2 +

∂ 2u∂y2 = k

∂ 2u∂ t2

∂ 2u∂x2 +

∂ 2u∂y2 = k

∂u∂ t

∂ 2u∂y2 = 0. The

Laplace’s equation describes steady state heat transfer as well as wavemotion

, because the (two-dimensional) wave equation is

∂ 2u∂x2 +

∂ 2u∂y2 = k

∂ 2u∂ t2

∂ 2u∂x2 +

∂ 2u∂y2 = k

∂u∂ t

∂ 2u∂y2 = 0. The

∂ 2u∂x2 +

∂ 2u∂y2 = k

∂ 2u∂ t2

∂ 2u∂x2 +

∂ 2u∂y2 = k

∂u∂ t

∂ 2u∂y2 = 0. The

∂ 2u∂x2 +

∂ 2u∂y2 = k

∂ 2u∂ t2

∂ 2u∂x2 +

∂ 2u∂y2 = k

∂u∂ t

Example.

Show that the function u(x,y) = e−y sin(x) is harmonic for0 < x < π and y > 0.

∂ 2u∂x2 +

∂ 2u∂y2 = e−y(− sin(x)

)+ e−y sin(x)

Example. Show that the function u(x,y) = e−y sin(x) is harmonic for0 < x < π and y > 0.

∂ 2u∂x2 +

∂ 2u∂y2 = e−y(− sin(x)

)+ e−y sin(x)

∂ 2u∂x2 +

∂ 2u∂y2

= e−y(− sin(x))+ e−y sin(x)

∂ 2u∂x2 +

∂ 2u∂y2 = e−y(− sin(x)

+ e−y sin(x)

∂ 2u∂x2 +

∂ 2u∂y2 = e−y(− sin(x)

)+ e−y sin(x)

∂ 2u∂x2 +

∂ 2u∂y2 = e−y(− sin(x)

)+ e−y sin(x)

Theorem.

If the complex function f (z) = u(x,y)+ iv(x,y) is analyticon its domain D, then its real and imaginary parts are harmonic on D.

Proof. (Existence of the second derivatives will follow from latertheorems.)

∂ 2u∂x2 +

∂ 2u∂y2 =

∂x∂u∂x

∂y∂u∂y

∂x∂v∂y

(−∂v

∂ 2v∂x∂y

− ∂ 2v∂y∂x

Harmonicity of v is proved similarly.

Theorem. If the complex function f (z) = u(x,y)+ iv(x,y) is analyticon its domain D, then its real and imaginary parts are harmonic on D.

∂ 2u∂x2 +

∂ 2u∂y2 =

∂x∂u∂x

∂y∂u∂y

∂x∂v∂y

(−∂v

∂ 2v∂x∂y

− ∂ 2v∂y∂x

Proof.

(Existence of the second derivatives will follow from latertheorems.)

∂ 2u∂x2 +

∂ 2u∂y2 =

∂x∂u∂x

∂y∂u∂y

∂x∂v∂y

(−∂v

∂ 2v∂x∂y

− ∂ 2v∂y∂x

∂ 2u∂x2 +

∂ 2u∂y2 =

∂x∂u∂x

∂y∂u∂y

∂x∂v∂y

(−∂v

∂ 2v∂x∂y

− ∂ 2v∂y∂x

∂ 2u∂x2 +

∂ 2u∂y2

∂x∂u∂x

∂y∂u∂y

∂x∂v∂y

(−∂v

∂ 2v∂x∂y

− ∂ 2v∂y∂x

∂ 2u∂x2 +

∂ 2u∂y2 =

∂x∂u∂x

∂y∂u∂y

∂x∂v∂y

(−∂v

∂ 2v∂x∂y

− ∂ 2v∂y∂x

∂ 2u∂x2 +

∂ 2u∂y2 =

∂x∂u∂x

∂y∂u∂y

∂x∂v∂y

(−∂v

=∂ 2v

∂x∂y− ∂ 2v

∂y∂x= 0

∂ 2u∂x2 +

∂ 2u∂y2 =

∂x∂u∂x

∂y∂u∂y

∂x∂v∂y

(−∂v

∂ 2v∂x∂y

− ∂ 2v∂y∂x

∂ 2u∂x2 +

∂ 2u∂y2 =

∂x∂u∂x

∂y∂u∂y

∂x∂v∂y

(−∂v

∂ 2v∂x∂y

− ∂ 2v∂y∂x

∂ 2u∂x2 +

∂ 2u∂y2 =

∂x∂u∂x

∂y∂u∂y

∂x∂v∂y

(−∂v

∂ 2v∂x∂y

− ∂ 2v∂y∂x

∂ 2u∂x2 +

∂ 2u∂y2 =

∂x∂u∂x

∂y∂u∂y

∂x∂v∂y

(−∂v

∂ 2v∂x∂y

− ∂ 2v∂y∂x

Example.

The real and imaginary parts of f (z) =1z3 are harmonic

for (x,y) 6= (0,0).

Proof. f is analytic away from the origin!

Example. The real and imaginary parts of f (z) =1z3 are harmonic

for (x,y) 6= (0,0).

Proof.

f is analytic away from the origin!

for (x,y) 6= (0,0).

Definition.

Let u and v be harmonic on the domain D. The function vis the called a harmonic conjugate of u if and only if

∂u∂x

=∂v∂y

and∂u∂y

=−∂v∂x

Theorem. The complex function f (z) = u(x,y)+ iv(x,y) is analytic onits domain D if and only if u and v are harmonic and v is a harmonicconjugate of u.

Proof. The proof that analyticity of f implies that u and v areharmonic and harmonic conjugates follows from the precedingtheorem and from the Cauchy-Riemann Equations. The proof of theconverse follows from the fact that the Cauchy-Riemann Equationsand continuous partial derivatives implied analyticity.

Definition. Let u and v be harmonic on the domain D. The function vis the called a harmonic conjugate of u if and only if

∂u∂x

=∂v∂y

and∂u∂y

=−∂v∂x

∂u∂x

=∂v∂y

and∂u∂y

=−∂v∂x

∂u∂x

=∂v∂y

and∂u∂y

=−∂v∂x

∂u∂x

=∂v∂y

and∂u∂y

=−∂v∂x

Theorem.

The complex function f (z) = u(x,y)+ iv(x,y) is analytic onits domain D if and only if u and v are harmonic and v is a harmonicconjugate of u.

∂u∂x

=∂v∂y

and∂u∂y

=−∂v∂x

∂u∂x

=∂v∂y

and∂u∂y

=−∂v∂x

Proof.

The proof that analyticity of f implies that u and v areharmonic and harmonic conjugates follows from the precedingtheorem and from the Cauchy-Riemann Equations. The proof of theconverse follows from the fact that the Cauchy-Riemann Equationsand continuous partial derivatives implied analyticity.

∂u∂x

=∂v∂y

and∂u∂y

=−∂v∂x

Proof. The proof that analyticity of f implies that u and v areharmonic and harmonic conjugates follows from the precedingtheorem and from the Cauchy-Riemann Equations.

The proof of theconverse follows from the fact that the Cauchy-Riemann Equationsand continuous partial derivatives implied analyticity.

∂u∂x

=∂v∂y

and∂u∂y

=−∂v∂x

∂u∂x

=∂v∂y

and∂u∂y

=−∂v∂x

Example.

Find a harmonic conjugate of the functionu(x,y) = e−y sin(x) on the domain given by 0 < x < π and y > 0.

∂v∂y

=∂u∂x

= e−y cos(x)

v =∫

e−y cos(x) dy =−e−y cos(x)+ cy

∂v∂x

= −∂u∂y

= e−y sin(x)

v =∫

e−y sin(x) dx =−e−y cos(x)+ cx

So v(x,y) =−e−y cos(x).

Example. Find a harmonic conjugate of the functionu(x,y) = e−y sin(x) on the domain given by 0 < x < π and y > 0.

∂v∂y

=∂u∂x

= e−y cos(x)

v =∫

∂v∂x

= −∂u∂y

= e−y sin(x)

v =∫

∂v∂y

=∂u∂x

= e−y cos(x)

v =∫

∂v∂x

= −∂u∂y

= e−y sin(x)

v =∫

∂v∂y

=∂u∂x

= e−y cos(x)

v =∫

∂v∂x

= −∂u∂y

= e−y sin(x)

v =∫

∂v∂y

=∂u∂x

= e−y cos(x)

v =∫

∂v∂x

= −∂u∂y

= e−y sin(x)

v =∫

∂v∂y

=∂u∂x

= e−y cos(x)

∂v∂x

= −∂u∂y

= e−y sin(x)

v =∫

∂v∂y

=∂u∂x

= e−y cos(x)

v =∫

e−y cos(x) dy

=−e−y cos(x)+ cy

∂v∂x

= −∂u∂y

= e−y sin(x)

v =∫

∂v∂y

=∂u∂x

= e−y cos(x)

v =∫

e−y cos(x) dy =−e−y cos(x)

∂v∂x

= −∂u∂y

= e−y sin(x)

v =∫

∂v∂y

=∂u∂x

= e−y cos(x)

v =∫

∂v∂x

= −∂u∂y

= e−y sin(x)

v =∫

∂v∂y

=∂u∂x

= e−y cos(x)

v =∫

∂v∂x

= −∂u∂y

= e−y sin(x)

v =∫

∂v∂y

=∂u∂x

= e−y cos(x)

v =∫

∂v∂x

= −∂u∂y

= e−y sin(x)

v =∫

∂v∂y

=∂u∂x

= e−y cos(x)

v =∫

∂v∂x

= −∂u∂y

= e−y sin(x)

v =∫

∂v∂y

=∂u∂x

= e−y cos(x)

v =∫

∂v∂x

= −∂u∂y

= e−y sin(x)

∂v∂y

=∂u∂x

= e−y cos(x)

v =∫

∂v∂x

= −∂u∂y

= e−y sin(x)

v =∫

e−y sin(x) dx

=−e−y cos(x)+ cx

∂v∂y

=∂u∂x

= e−y cos(x)

v =∫

∂v∂x

= −∂u∂y

= e−y sin(x)

v =∫

e−y sin(x) dx =−e−y cos(x)

∂v∂y

=∂u∂x

= e−y cos(x)

v =∫

∂v∂x

= −∂u∂y

= e−y sin(x)

v =∫

∂v∂y

=∂u∂x

= e−y cos(x)

v =∫

∂v∂x

= −∂u∂y

= e−y sin(x)

v =∫

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Transcript of Equivalent Conditions for Differentiability - USM · 2010. 8. 30. · Cauchy-Riemann EquationsPolar...

La propiedad de analiticidad induce ciertas relaciones ...gopar/TEACHING/web_slides_CV_3_2015.pdf · Ecuaciones de Cauchy-Riemann Comentario: Que se satisfagan las ecs. de Cauchy-Riemann

CAUCHY-RIEMANN EQUATIONS...to solve the Cauchy-Riemann equations on C and the tangential Cauchy-Riemann equations on a hypersurface in C. For example, Romanov [16] discovered a kernel

THE TANGENTIAL CAUCHY-RIEMANN COMPLEX …...1972] THE TANGENTIAL CAUCHY-RIEMANN COMPLEX ON SPHERES 85 boundary conditions if u\bM £ F(B'') and du\bM er(B ' '). A simple argument by

13.4 Cauchy-Riemann 방정식,라플라스방정식3 ※정리1 [ Cauchy-Riemann 방정식] f ( z ) = u ( x , y ) + i v ( x , y ) 가한점z = x + i y 의어떤근방에서 정의되고연속이며또z

Lezioni sulle Superﬁcie di Riemann 2017-2018 (II Semestre)nacinovi/files/suprim.pdf · 4 INDICE 8. Il teorema di approssimazione di Runge 73 9. L’equazione di Cauchy-Riemann non

Asymptotic Behavio orf Solution osf Difference Equations with …doktori.bibl.u-szeged.hu/4283/1/2000_peics_hajnalka.pdf · Cauchy, Abel Riemann, Weierstrass, Darboux, Hilbert, to,

Analytic functions; Cauchy-Riemann equations...Analytic functions; Cauchy-Riemann equations Monday, December 7, 2020 9:03 AM Course content Page 1 Course content Page 2 Course content

Complex Analysis revisited - Eindhoven University of ...€¦ · Volha Shchetnikava Complex Analysis revisited. Complex Numbers Complex Functions Introduction Di erentiability Cauchy-Riemann

An Invitation to Mathematical Physics and Its Historyjontalle.web.engr.illinois.edu/uploads/298.17/anInvitation.pdf4 CONTENTS 1.4.2 Lecture 24: (Week 9) Cauchy-Riemann conditions .

ANALYSIS OF CONTACT CAUCHY–RIEMANN MAPS I: A PRIORI C ...math.berkeley.edu/~ruiwang/pdf/CRI.pdf · Key words and phrases. Contact triad connection, contact Cauchy–Riemann map,

Cauchy riemann equations

Some Application of Cauchy- Riemann Equation to …...Page 1219 Some Application of Cauchy- Riemann Equation to Complex Analysis Naisan Khalaf Mosah Master of Science in Applied Mathematics

Topic 9: Analysis of Riemann-Hilbert Problems › ... › 651_Winter18 › Topic09-651-W18.pdfTopic 9: Analysis of Riemann-Hilbert Problems Cauchy integrals. De nition of arcs and

staffnew.uny.ac.idstaffnew.uny.ac.id/...kunci-persamaan-differensial.pdf · Give two examples of complex function which satisfy the Cauchy- Riemann condition but they do not have

pnautonomouscollege.inpnautonomouscollege.in/.../2017/08/Electronics.docx · Web viewComplex Variable, Complex Function, Continuity, Differentiability,Analyticity. Cauchy-Riemann

A Fast Cauchy-Riemann Solver · 2018. 11. 16. · A Fast Cauchy-Riemann Solver By Michael Ghil* and Ramesh Balgovind** ... v and the reduction to a discrete Poisson problem, and finally

Anakom Kelomok 6 Persamaan Cauchy Riemann

When is a Function that Satisfies the Cauchy …cms.dm.uba.ar/academico/materias/2docuat2010/analisis...WHEN IS A FUNCTION THAT SATISFIES THE CAUCHY-RIEMANN EQUATIONS ANALYTIC? J.

Turunan Fungsi Kompleks Dan Persamaan Cauchy Riemann

FUNGSI DARI VARIABEL KOMPLEK - ee.unud.ac.id · FUNGSI DARI VARIABEL KOMPLEK . PEMETAAN KOMPLEKS . INTEGRAL KOMPLEKS . INTEGRAL KOMPLEKS . DERET LAURENT . CAUCHY -RIEMANN . For a