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

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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


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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.



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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.






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






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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.




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functions

and a new way to look at certain real differentiablefunctions.




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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.



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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

(z).



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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

(z).



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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

(z).



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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

(z).



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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

(z).



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∂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

(z).



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∂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

(z).



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∂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

(z).



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Proof of “⇒.”

∂u∂x

(x,y) = limt→x

u(t,y)−u(x,y)t− x

= ℜ

(lim

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

(lim

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

t→y

v(x, t)− v(x,y)t− y

=∂v∂y

(x,y)

The other equation is proved similarly.



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Proof of “⇒.”∂u∂x

(x,y)

= limt→x


= ℜ

(lim

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

(lim

x+it→x+iyi


)= ℑ

(limt→y


)= lim

t→y


=∂v∂y

(x,y)




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(x,y) = limt→x


= ℜ

(lim

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

(lim

x+it→x+iyi


)= ℑ

(limt→y


)= lim

t→y


=∂v∂y

(x,y)




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(x,y) = limt→x


= ℜ

(lim

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

(lim

x+it→x+iyi


)= ℑ

(limt→y


)= lim

t→y


=∂v∂y

(x,y)




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(x,y) = limt→x


= ℜ

(lim

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

(lim

x+it→x+iyi


)= ℑ

(limt→y


)= lim

t→y


=∂v∂y

(x,y)




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(x,y) = limt→x


= ℜ

(lim

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

(lim

x+it→x+iyi


)= ℑ

(limt→y


)= lim

t→y


=∂v∂y

(x,y)




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(x,y) = limt→x


= ℜ

(lim

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

(lim

x+it→x+iyi


)= ℑ

(limt→y


)= lim

t→y


=∂v∂y

(x,y)




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(x,y) = limt→x


= ℜ

(lim

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

(lim

x+it→x+iyi


)

= ℑ

(limt→y


)= lim

t→y


=∂v∂y

(x,y)




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(x,y) = limt→x


= ℜ

(lim

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

(lim

x+it→x+iyi


)= ℑ

(limt→y


)

= limt→y


=∂v∂y

(x,y)




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(x,y) = limt→x


= ℜ

(lim

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

(lim

x+it→x+iyi


)= ℑ

(limt→y


)= lim

t→y


=∂v∂y

(x,y)




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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



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Proof of “⇐” (concl.)

lim∆x,∆y→0


∆x+ i∆y+ i


∆x+ i∆y+

∂u∂y (x,y+ηu)∆y

∆x+ i∆y+ i

∂v∂y (x,y+ηv)∆y

∆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

(x,y)

(Technical argument at end omitted.)



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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.



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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




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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




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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




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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




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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




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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




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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




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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




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



∂u∂y





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f (z)

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


∂u∂y





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f (z) = ez

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


∂u∂y





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f (z) = ez = ex+iy

= exeiy = ex cos(y)+ iex sin(y)∂u∂x


∂u∂y





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

= ex cos(y)+ iex sin(y)∂u∂x


∂u∂y





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

∂u∂x


∂u∂y





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∂u∂y





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= ex cos(y)

=∂v∂y

∂u∂y





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∂u∂y





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∂u∂y

= −ex sin(y)

=−∂v∂x




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∂u∂y





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Example.

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

f (z) = |z|=√

x2 + y2 + i ·0∂u∂x

=x√

x2 + y26= 0 (x 6= 0)

∂u∂y

=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.



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

f (z) = |z|=√

x2 + y2 + i ·0∂u∂x

=x√

x2 + y26= 0 (x 6= 0)

∂u∂y

=y√

x2 + y26= 0 (y 6= 0)




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f (z)

= |z|=√

x2 + y2 + i ·0∂u∂x

=x√

x2 + y26= 0 (x 6= 0)

∂u∂y

=y√

x2 + y26= 0 (y 6= 0)




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f (z) = |z|

=√

x2 + y2 + i ·0∂u∂x

=x√

x2 + y26= 0 (x 6= 0)

∂u∂y

=y√

x2 + y26= 0 (y 6= 0)




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f (z) = |z|=√

x2 + y2

+ i ·0∂u∂x

=x√

x2 + y26= 0 (x 6= 0)

∂u∂y

=y√

x2 + y26= 0 (y 6= 0)




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f (z) = |z|=√

x2 + y2 + i ·0

∂u∂x

=x√

x2 + y26= 0 (x 6= 0)

∂u∂y

=y√

x2 + y26= 0 (y 6= 0)




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f (z) = |z|=√

x2 + y2 + i ·0∂u∂x

=x√

x2 + y26= 0 (x 6= 0)

∂u∂y

=y√

x2 + y26= 0 (y 6= 0)




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f (z) = |z|=√

x2 + y2 + i ·0∂u∂x

=x√

x2 + y2

6= 0 (x 6= 0)

∂u∂y

=y√

x2 + y26= 0 (y 6= 0)




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f (z) = |z|=√

x2 + y2 + i ·0∂u∂x

=x√

x2 + y26= 0

(x 6= 0)

∂u∂y

=y√

x2 + y26= 0 (y 6= 0)




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f (z) = |z|=√

x2 + y2 + i ·0∂u∂x

=x√

x2 + y26= 0 (x 6= 0)

∂u∂y

=y√

x2 + y26= 0 (y 6= 0)




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f (z) = |z|=√

x2 + y2 + i ·0∂u∂x

=x√

x2 + y26= 0 (x 6= 0)

∂u∂y

=y√

x2 + y2

6= 0 (y 6= 0)




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f (z) = |z|=√

x2 + y2 + i ·0∂u∂x

=x√

x2 + y26= 0 (x 6= 0)

∂u∂y

=y√

x2 + y26= 0

(y 6= 0)




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f (z) = |z|=√

x2 + y2 + i ·0∂u∂x

=x√

x2 + y26= 0 (x 6= 0)

∂u∂y

=y√

x2 + y26= 0 (y 6= 0)




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f (z) = |z|=√

x2 + y2 + i ·0∂u∂x

=x√

x2 + y26= 0 (x 6= 0)

∂u∂y

=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.



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f (z) = |z|=√

x2 + y2 + i ·0∂u∂x

=x√

x2 + y26= 0 (x 6= 0)

∂u∂y

=y√

x2 + y26= 0 (y 6= 0)




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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))

.



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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))

.



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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))

.



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(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))

.



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(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))

.



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(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))

.



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(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))

.



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Proof (preparatory steps).

Remember that differentiability meansthat the Cauchy-Riemann Equations hold.

∂ r∂x

=∂√

x2 + y2

∂x=

x√x2 + y2

= cos(θ)

∂ r∂y

=∂√

x2 + y2

∂y=

y√x2 + y2

= sin(θ)

∂θ

∂x=

∂ arctan( y

x

)∂x

=1

1+( y

x

)2

(− y

x2

)=− y

x2 + y2 =−sin(θ)r

∂θ

∂y=

∂ arctan( y

x

)∂y

=1

1+( y

x

)21x

=x

x2 + y2 =cos(θ)

r



logo1


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

∂ r∂x

=∂√

x2 + y2

∂x=

x√x2 + y2

= cos(θ)

∂ r∂y

=∂√

x2 + y2

∂y=

y√x2 + y2

= sin(θ)

∂θ

∂x=

∂ arctan( y

x

)∂x

=1

1+( y

x

)2

(− y

x2

)=− y


∂θ

∂y=

∂ arctan( y

x

)∂y

=1

1+( y

x

)21x

=x

x2 + y2 =cos(θ)

r



logo1



∂ r∂x

=∂√

x2 + y2

∂x

=x√

x2 + y2= cos(θ)

∂ r∂y

=∂√

x2 + y2

∂y=

y√x2 + y2

= sin(θ)

∂θ

∂x=

∂ arctan( y

x

)∂x

=1

1+( y

x

)2

(− y

x2

)=− y


∂θ

∂y=

∂ arctan( y

x

)∂y

=1

1+( y

x

)21x

=x

x2 + y2 =cos(θ)

r

