5 Complex Numbers and Functions - University …...Before deﬁning division, we need the complex...

5 Complex Numbers and Functions

Consider the polynomial equation x2 +3x+2 = 0. Since x2 +3x+2 = (x+1)(x+2), the twosolutions are x = −1 and x = −2. Unfortunately not all such equations have (real number)solutions. For example, since x2 > 0 for all x ∈ R,

x2 + 1 > 0 + 1 = 1 > 0 for all x ∈ R ⇒ x2 6= −1 for all x ∈ R.

To get round this problem we introduce the symbol i whose defining property is that it satisfiesthe equation

i2 = −1.

It is clear from its definition that i /∈ R. [In engineering texts j is often used instead of i forthe square root of −1, to avoid conflict with the notation for electrical current.]

A complex number is any symbol of the form z = x + iy, or z = x + yi, where x and yare any two real numbers. The real part of z is the number x; the imaginary part of z is y— note in particular that both parts are real numbers. They are denoted

Re z = x, Im z = y.

The set of all complex numbers is denoted by C. Given z, w ∈ C, if we have z = x + iy andw = u + iv for some x, y, u, v ∈ R, then we define equality in C by saying

z = w ⇔ x = u and y = v ⇔ Re z = Re w and Im z = Im w.

That is, two complex numbers are equal if and only if their real and imaginary parts areequal. The set of real numbers can be thought of as a subset of the complex numbers byidentifying x ∈ R with x + 0i ∈ C. Indeed, we will write x in place of x + 0i in such cases,and similarly write iy instead of 0 + iy.

5.1 Arithmetic in C

Addition, subtraction and multiplication in C all have obvious definitions which run as follows,where again z = x + iy and w = u + iv:

z + w = (x + iy) + (u + iv) = (x + u) + i(y + v),

z − w = (x + iy) − (u + iv) = (x − u) + i(y − v),

zw = (x + iy)(u + iv) = xu + ixv + iyu + i2yv

= (xu − yv) + i(xv + yu).

In particular it follows from these definitions that z − w = z + (−1)w — subtraction isessentially defined in terms of the other two operations.

For example if z = 1 + 2i and w = 3 − 5i then

z + w = (1 + 3) + (2 − 5)i = 4 − 3i,

z − w = (1 − 3) +(

2 − (−5))

i = −2 + 7i,

zw = (1 + 2i)(3 − 5i) =(

1 × 3 − 2 × (−5))

+ i(

1 × (−5) + 2 × 3)

= 13 + i.

Before defining division, we need the complex conjugate of a complex number z = x + iywhich is z = x− iy, i.e. the number obtained by changing the sign of the imaginary part. Soif z = 1 + 2i and w = 3 − 5i as above then z = 1 − 2i and w = 3 + 5i. Note that in general

zz = (x − iy)(x + iy) = x2 + ixy − iyx − i2y2 = x2 + y2.

82

COMPLEX NUMBERS AND FUNCTIONS

So if z 6= 0, that is, if either x 6= 0 or y 6= 0, then x2 > 0 or y2 > 0 and so zz > 0, a positivereal number. If z = x + iy, and t ∈ R with t 6= 0, then the obvious definition of division inthis case is

z

t=

x

t+ i

y

t.

In general, if w = u + iv 6= 0 then ww = u2 + v2 > 0, and we want to define division so that

it is the same as multiplication by1

w, and so that

w

w= 1. But then

1

w=

1

w

w

w=

w

ww=

u − iv

u2 + v2=

1

u2 + v2× (u − iv)

and so division by w can be defined by dividing by the real number u2 + v2, and thenmultiplying by the conjugate u − iv of w. So if z = x + iy then we define

z

w=

z

w

w

w=

zw

ww=

xu + yv

u2 + v2+ i

yu − xv

u2 + v2.

For example if we take z = 1 + 2i and w = 3 − 5i once more then

z

w=

1 + 2i

3 − 5i=

(1 + 2i)

(3 − 5i)

(3 + 5i)

(3 + 5i)=

3 + 5i + 6i − 10

9 + 15i − 15i + 25=

−7

34+

11

34i.

The modulus of the complex number z = x + iy is defined to be

|z| =√

x2 + y2 = (zz)1/2.

Note that z = 0 precisely when |z| = 0.Having introduced all of these definitions it is then not too hard to show that the following

rules apply to arithmetic in C:

z1 + z2 = z2 + z1; z1z2 = z2z1; (z1 + z2) + z3 = z1 + (z2 + z3);

(z1z2)z3 = z1(z2z3); (z1 + z2)z3 = z1z3 + z2z3; z1 + z2 = z1 + z2

z1z2 = z1 z2;( z

w

)

=z

w; |z| = |z|; |z1z2| = |z1||z2|;

∣

∣

∣

∣

z1

z2

∣

∣

∣

∣

=|z1||z2|

;

Re z = 12 (z + z); Im z = 1

2i(z − z).

We motivated the introduction of complex numbers to allow us to solve polynomial equa-tions which do not have real number solutions. That the above is the right way to do this isconfirmed by the following:

The Fundamental Theorem of Algebra. Any polynomial of degree n has n roots in C,

if we allow for repeated roots. That is, given any numbers an, an−1, . . . , a2, a1, a0 ∈ C with

an 6= 0, there are further numbers α1, α2, . . . , αn ∈ C such that

anzn + an−1zn−1 + · · · + a2z

2 + a1z + a0 = an(z − α1)(z − α2) · · · (z − αn).

Exercise 5.1. Derive the identities z1 + z2 = z1 + z2 and Im z = 12i(z − z)

Solution.

83

Geometrical interpretation: the complex plane

5.2 Geometrical interpretation: the complex plane

The fact that any complex number z = x + iy is specified by a pair of real numbers meansthat we can identify z with the point (x, y) in the plane. Note that with this interpretation,

the modulus |z| =√

x2 + y2 is nothing but the distance of this point (x, y) from the origin.Also, we can view addition in this context as an application once more of the parallelogramlaw for addition of vectors, an immediate consequence of which is the triangle inequality:

|z + w| 6 |z| + |w| for all z, w ∈ C.

x

=

y z

z + w

z − w

−w

wz = x + iy

real axis

imaginary axis

Since the modulus corresponds to a distance, in particular since |z − w| represents thedistance between z and w in the above diagram, we can use this function to describe geometricobjects in a succinct fashion. For example the equation

∣

∣z − (1 + i)∣

∣ = 2

is satisfied by all those complex numbers z that are a distance 2 from the point 1 + i. Thatis, all the points on the circle with centre 1 + i and with radius 2.

In a similar way the equation∣

∣z − (1 + 3i)∣

∣ =∣

∣z − (5 + i)∣

∣ is satisfied by all those pointsthat are equidistant from the points 1+3i and 5+i in the complex plane. Thus z satisfies thisequation precisely if it lies on the line that is the perpendicular bisector of the line segmentjoining 1 + 3i to 5 + i. Along this line segment the change in the real part is 5 − 1 = 4, andthe corresponding change in the imaginary part is 1 − 3 = −2, so the midpoint of the linesegment is

1 + 3i + 12

(

(5 + i) − (1 + 3i))

= (1 + 3i) + (2 − i) = 3 + 2i = 12

(

(1 + 3i) + (5 + i))

.

Moreover the slope of the line segment connecting the points has gradient −24 = − 1

2 , so theslope of the perpendicular bisector is − 1

−1/2 = 2, and hence this line has equation

y − 2 = 2 × (x − 3) ⇔ y = 2x − 4,

in terms of the real and imaginary parts of z.

1 + i

2

1 + 3i

3 + 2i

5 + i

y = 2x − 4

84


Exercise 5.2. Use |z|2 = (Re z)2+(Im z)2 to derive the equation of the perpendicular bisectorof the line segment from 1 + 3i to 5 + i in the form y = mx + c.

Solution.

Exercise 5.3. Which region of the complex plane is described by the inequality |z|2 > z +z?

Solution.

5.2.1 Polar representation

If z 6= 0 then its position in the plane can be specified by its distance r from the origin andits argument, denoted arg z, which is the angle θ between the line connecting z to 0 and thepositive real axis, with θ measured in an anticlockwise direction. Note that r = |z|, which isuniquely determined by z, but that the argument θ is not uniquely determined since we cango round the origin in either direction any whole number of times, so that θ − 2π, θ + 2π,

85

Complex-valued functions

θ +4π etc. are all alternative values for the argument. If θ is chosen so that −π < θ 6 π thenthis value is known as the principal argument.

r

r cos θ

r sin θ

θ

If we are given r and θ then by standard trigonometrywe have Re z = r cos θ and Im z = r sin θ, so that

z = r cos θ + ir sin θ

On the other hand given z we know that r = |z|, and findthat

tan θ =r sin θ

r cos θ=

Im z

Re z.

However, we cannot conclude from this that θ = tan−1 Im z

Re z,

since the usual definition of the inverse tangent only takes

values between −π

2and

π

2, and so will give the wrong answers for numbers such as z = −1−i.

This z has argument −3π

4, not tan−1(1) =

π

4.

By writing complex numbers in polar form, we find the geometrical meaning of multipli-cation — the analogue to the parallelogram rule for addition. Indeed, if

z = r(cos θ + i sin θ) and w = s(cosϕ + i sinϕ)

then their product is

zw = rs[

(cos θ cosϕ − sin θ sin ϕ) + i(sin θ cosϕ + cos θ sinϕ)]

= rs(

cos(θ + ϕ) + i sin(θ + ϕ))

.

That is,|zw| = |z||w| and arg(zw) = arg z + argw.

So, if we are dealing with a multiplication by a fixed number w, then its effect is to scaleeverything by a factor of |w|, and change all arguments by adding argw, that is, rotatingeverything anticlockwise by this amount.

A similar calculation, assuming w 6= 0, shows that

∣

∣

∣

z

w

∣

∣

∣=

|z||w| and arg

( z

w

)

= arg z − arg w.

5.3 Complex-valued functions

We will consider two types of function that take values in the complex plane. One typedepends on a real argument, the other on a complex argument. That is, we consider functionsf : R → C and g : C → C.

Functions of the first type include f(t) = t2 + i(1− et), or f(t) = cos t+ i sin t. As we varyt we get a new point on the plane, and the collection of all points obtained from f producesa curve (if f is well-behaved).

For example if f(t) = cos t + i sin t then |f(t)| =√

cos2 t + sin2 t = 1 for all t, and so f(t)is a constant distance of 1 from the origin. Evaluating f at various values of t gives:

t = 0 π4

π2

3π4 π 3π

2 2π 9π4 · · ·

f(t) = 1 1√2(1 + i) i 1√

2(−1 + i) −1 −i 1 1√

2(1 + i) · · ·

That is, f(t) travels round the unit circle in an anticlockwise direction.

86


Similarly we can describe straight lines, by setting

z = (at + b) + i(ct + d) = (b + id) + t(a + ic),

for real numbers a, b, c, d ∈ R. This is the parametric equation of the line through the pointb + id in the direction specified by a + ic.

For any such function f : R → C we can take real and imaginary parts, and obtainfunctions u : R → R and v : R → R:

u(t) = Re f(t), v(t) = Im f(t).

If we take our first example, f(t) = t2 + i(1 − et), then u(t) = t2 and v(t) = 1 − et.A function f : R → C is continuous at a ∈ R if f(t) approaches the value f(a) as t → a.

To be more precise, this means that |f(t) − f(a)| → 0 as t → a, where this limit is oneinvolving real numbers. An equivalent definition is that f is continuous at t = a if both ofthe real-valued functions u(t) = Re f(t) and v(t) = Im f(t) are continuous at that point.

Similarly, f is differentiable at t = a, with derivative f ′(a), if

limh→0

f(a + h) − f(a)

h= f ′(a)

exists. Again, an alternative, equivalent definition is that f is differentiable at t = a if thefunctions u and v for real and imaginary parts are both differentiable there, in which case

f ′(t) = u′(t) + iv′(t).

So, for example,

f(t) = t2 + i(1 − et) ⇒ f ′(t) = 2t − iet,

g(t) = cos t + i sin t ⇒ g′(t) = − sin t + i cos t.

The second type of function (the one that we are more interested in for this course) arefunctions from C back to C. Simple examples include

f(z) = 2z3 − 3iz + 5 − 7i, g(z) = z − |z|2, h(z) =z2 + 9

z2 + 4.

The first two make good sense for any choice of z. Since z2 + 4 = (z + 2i)(z − 2i), h(z) is notdefined when z = ±2i.

It is less easy to give a geometrical view of such functions than say for functions R → R,when we can picture the graph, or functions R2 → R, when we can picture the surface, orfunctions R → C, when we can picture the curve or path in the plane. Indeed, the analogueto writing y = f(x) for functions R → R is to write w = f(z), where z is the input variable,which is viewed as a point in one plane, and w = f(z), the value coming from the function,is a point in another plane.

z

w = f(z)

87


Exercise 5.4. Consider the mapping w = f(z) = (2 + i)z + 3. Find the image of the liney = 3x + 1 under this mapping, where z = x + iy.

Solution.

It can be difficult to picture overall what such a function is doing, but it is often possibleto determine the image in the w-plane of various figures or shapes in the z-plane.

For example, fix complex numbers a, b ∈ C with a 6= 0, and define w = az + b. Then wecan rearrange:

w = az + b ⇔ z =w − b

a.

Now any circle in the z-plane has equation |z − α| = r, where α ∈ C is the centre and r > 0is the radius. If w = az + b then the image of this circle is

∣

∣

∣

∣

w − b

a− α

∣

∣

∣

∣

= r ⇔∣

∣

∣

∣

w − b − aα

a

∣

∣

∣

∣

=|w − b − aα|

|a| = r ⇔ |w − (b + aα)| = r|a|.

That is, the image is a circle with centre aα + b, and radius r|a|.On the other hand any straight line in the z-plane has an equation of the form |z − α| =

|z − β|, where α ∈ C is a point off of the line, and β is its reflection in the given line. Again,under the mapping w = az + b this line is taken to

∣

∣

∣

∣

w − b

a− α

∣

∣

∣

∣

=

∣

∣

∣

∣

w − b

a− β

∣

∣

∣

∣

⇔ |w − b − aα||a| =

|w − b − aβ||a|

⇔ |w − (b + aα)| = |w − (b + aβ)|

88


That is, the image of the line that is the perpendicular bisector of the line segment betweenα and β is the perpendicular bisector of line segment between b + aα and b + aβ.

These two parts together show that w = az + b maps circles to circles and straight linesto straight lines.

We next want to consider the mapping w = z−1, which is defined for all z 6= 0. Beforethis, we return to the equation of a circle. If α = a1 + ia2, β = b1 + ib2 ∈ C, and t ∈ R suchthat 0 < t < 1 or t > 1, then some tedious algebra shows that

|z − α| = t|z − β|

⇔(

x − t2b1 − a1

t2 − 1

)2

+(

y − t2b2 − a2

t2 − 1

)2

=( t

t2 − 1

)2[

(a1 − b1)2 + (a2 − b2)

2]

.

That is, the equation |z − α| = t|z − β| corresponds to a circle with centre and radius

t2b1 − a1

t2 − 1+ i

t2b2 − a2

t2 − 1and

t

|t2 − 1|[

(a1 − b1)2 + (a2 − b2)

2]1/2

respectively.So now consider the image of the circle |z−α| = r under w = z−1. If α = 0, i.e. the centre

is the origin, then this becomes

∣

∣

∣

1

w

∣

∣

∣=

1

|w| = r ⇔ |w| =1

r,

another circle centred on the origin. Otherwise we have

∣

∣

∣

1

w− α

∣

∣

∣=

|1 − αw||w| = r ⇔ |α|

∣

∣

∣

1

α− w

∣

∣

∣= r|w|

which is a line if |α| = r, otherwise it is a circle by the above calculation, where t = r/|α|.On the other hand consider any straight line. This can always be written as |z−α| = |z−β|,

which is then transformed to

∣

∣

∣

1

w− α

∣

∣

∣=

∣

∣

∣

1

w− β

∣

∣

∣⇔ |1 − αw|

|w| =|1 − βw|

|w| ⇔ |1 − αw| = |1 − βw|.

But α and β can always be chosen so that neither is 0 (why?), hence the final equality isequivalent to

∣

∣

∣w − 1

α

∣

∣

∣=

|β||α|

∣

∣

∣w − 1

β

∣

∣

∣,

which is a line if |α| = |β|, and a circle otherwise.

89


The preceding calculations can be summed up in the following:

Theorem 5.5. Any transformation of the form w =az + b

cz + dfor complex numbers a, b, c, d ∈ C

such that ad − bc 6= 0 is called a Mobius transformation, and maps circles to circles or lines,

and lines to circles or lines.

Proof. First note that if c = 0, hence d 6= 0, then w = (a/d)z +b/d, which is a transformationwe have already considered. So if c 6= 0, then

f(z) =1

c× a(z + d/c) + (b − ad/c)

z + d/c=

a

c+

b − ad/c

cz + d,

Note that if we did not assume ad− bc 6= 0 then this would be a constant map, which wouldmap all points in the z-plane to a/c.

But now we can write f(z) = f3

(

f2

(

f1(z))

)

where

f1(z) = cz + d, f2(z) =1

zand f3(z) =

a

c+

(

b − ad

c

)

z,

Each of the maps f1, f2 and f3 maps lines or circles to lines or circles, so doing all three, oneafter the other, will have the same effect.

Exercise 5.6 (A03 7(a)). Find the image of the circle |z−1| = 1 under each of the mappings(i) w = 2z + 3 (ii) w = 1/z

Solution.

90


Exercise 5.7 (S03 7(a)). Find the image of the line Re z = 1 under the mappings (i) w = z2

(ii) w = 1/z

Solution.

Exercise 5.8. Find the image of the sets {z : Re z = 0} and {z : Re z > 0 and Im z > 0}under the map w = z2.

Solution.

91


Example 5.9 (S05 5(a)). Sketch the following sets of points in the complex plane:

A = {z : |z − 3| = |z − 1|}, B = {z : 1 < |z| < 2, π2 < arg z < 2π

3 }

Find the image of A under the transformation w =1

z − 1and the image of B under the

transformation w = z3. Sketch these images.

Solution. {z : |z − 3| = |z − 1|} consists of points equidistant from 3 and 1, i.e. the lineRe z = 2.

{z : 1 < |z| < 2, π2 < arg z < 2π

3 } is the intersection of the sector {z : π2 < arg z < 2π

3 }with the annulus {z : 1 < |z| < 2}.

92


2

2

1

1

A B

3

θ

θ =2π3

Now w =1

z − 1⇔ (z − 1)w = 1 ⇔ z =

1 + w

w=

1

w+ 1. Thus the image of A is

∣

∣

∣

∣

1 +1

w− 3

∣

∣

∣

∣

=

∣

∣

∣

∣

1 +1

w− 1

∣

∣

∣

∣

⇔ |1 − 2w||w| =

1

|w| ⇔ 2

∣

∣

∣

∣

1

2− w

∣

∣

∣

∣

= 1 ⇔∣

∣

∣

∣

w − 1

2

∣

∣

∣

∣

=1

2

which is the circle with centre 12 and radius 1

2 .For the image of B note that if w = z3 then |w| = |z3| = |z|3 and argw = arg z3 = 3 arg z.

Hence

1 < |z| < 2 ⇒ 13 < |w| < 23, i.e. 1 < |w| < 8, and

π

2< arg z <

2π

3⇒ 3π

2< arg w < 2π.

81 112

Image of A Image of B

Example 5.10. In each case determine geometrically the set of points that satisfy the givenconditions:

(i) |2z − 6 + 2i| = 8 (ii) |z + i| < |z − 1|

Find the image of the set in (i) under the map w = iz + 5.

Solution. (i) |2z−6+2i| = |2(z−3+ i)| = 2|z−3+ i| = 8 ⇒ |z− (3− i)| = 4, so z satisfiesthis condition if and only if it is on the circle with centre 3 − i and radius 4.

(ii) |z + i| < |z − 1| if z is closer to −i than to 1.Note: |z + i| = |z − 1| if z is equidistant from −i and 1, i.e. on the line y = −x, so

|z + i| < |z − 1| if z is below the line y = −x.

If w = iz + 5 then z =w − 5

i= −iw + 5i, and so if |z − 3 + i| = 4 then

4 = | − iw + 5i − 3 + i| = | − i(w − 6 − 3i)| ⇒ |w − (6 + 3i)| = 4

Thus the circle from (i) is mapped to the circle with centre 6 + 3i and radius 4.

93

Exercises

1

−i

|z + i| = |z − 1|

|z + i| < |z − 1|

5.4 Exercises

1. Evaluate each of the following expressions, giving your answer in the form x + iy for realnumbers x and y:

(i) (2 + 4i)(6 − 31) (ii)(−4 − 5i)(8 − 4i)

6 + 2i(iii)

1

i(iv)

(

3 + 2i

−1 + i

)2

(v) (3 − 8i)(2i)(3 + 2i) (vi) i3 − 4i2 + 2 (vii) (2 − i)3 (viii) i7 +1

i7

2. (a) In each case determine the modulus and argument of the given complex number:

(i) 1 + 4i (ii) − 3 − 6i (iii) − 14i (iv) 3 + 9i

(b) In each case write the number z in the form x + iy:

(i) |z| = 3, arg z =π

4(ii) |z| = 14, arg z =

7π

6(iii) |z| = 7, arg z =

8π

3

3. Let z = r(cos θ + i sin θ) and w = s(cosϕ+ i sinϕ) be two complex numbers given in polarform. Use the fact that 1/w = w/(ww) to show the following:

1

w=

1

s(cosϕ − i sinϕ),

∣

∣

∣

z

w

∣

∣

∣=

r

sand arg

( z

w

)

= θ − ϕ

4. In each case determine geometrically the set of points that satisfy given conditions:

(i) |z − 8 + 4i| = 9 (ii) |z| = |z − i| (iii) |z|2 + Im z = 16

(iv) |z − i| < |z − 1| (v) z = |z|eiθ, π/4 < θ ≤ 3π/4 (vi) |z − 2| > 3 and |z| < 2

Find the image of (i) and (ii) under the map w = (−1 + i)z + (3 + 2i), and the image of (ii)under the maps w = iz2, w = 1/z and w = (z − i)/(z + 2).

5.5 Continuity; sequences and series

The definition for continuity of functions f : C → C, and for convergence of sequences andseries made up of complex numbers look entirely analogous to those for their real-valuedcounterparts. However, some care needs to be taken when applying these definitions. Forexample, a function f : C → C is continuous at z = a if lim

z→af(z) = f(a), that is, if the value

of f(z) gets close to f(a) as z gets close to a. Since we measure the distance between points zand w in the plane in terms of the number |z−w|, this means that we need |f(z)− f(a)| → 0

94


as |z − a| → 0. These limits here are concerned with real-valued quantities, so we can makeuse of earlier (Section A) results to help calculate them.

The subtlety in the complex case is that there are many different ways for us to makez → a. For example, writing a = α + iβ, we could take z = α + h + iβ for h > 0, and leth → 0, or we could take z = α + i(β + h) for h < 0, and let h → 0, or . . . The importantthing to note is that the definition allows z to approach a from all directions in the plane.

For example, consider the function f(z) = z2. Then at any point z = a we have

|f(z) − f(a)| = |z2 − a2| = |(z − a)(z + a)| = |z − a| × |z + a|.

Letting z → a is the same thing as letting |z − a| → 0. Also, by the triangle inequality,

|z| = |(z −w) + w| 6 |z − w|+ |w| and |w| 6 |w − z|+ |z| = |−(z − w)|+ |z| = |z − w|+ |z|

for all z, w ∈ C, from which we get∣

∣|z| − |w|∣

∣ 6 |z − w|. It follows that

0 6∣

∣|z + a| − |a + a|∣

∣ 6 |(z + a) − (a + a)| = |z − a| → 0,

so by the Squeeze Theorem |z + a| → |a + a|, and applying the limit laws we get

|z2 − a2| → 0 × |2a| = 0.

That is, z2 → a2, hence the function f(z) = z2 is continuous at a. Similar arguments, andsome induction show that all polynomials are continuous functions at all points in the plane.

Exercise 5.11. Consider the function f(z) =Re z Im z

|z|2 , defined for each z 6= 0. Show that

there is no α ∈ C such that defining f(0) = α will make the function continuous at the origin.

Solution.

We say that a sequence (an)∞n=1 of complex numbers is convergent to some limit L if an

gets closer and closer to L as n → ∞, which amounts to saying that |an −L| → 0 as n → ∞,or, equivalently, that

Re an → Re L and Im an → Im L as n → ∞.

If the sequence is not convergent then it is divergent.

95

Continuity; sequences and series

Complex series can be treated similarly: if an ∈ C for each integer n > 1 then the series∑∞

n=1 an is convergent if the sequence of partial sums SN =∑N

n=1 an is convergent. Again,this is equivalent to requiring that the real series

Re(

N∑

n=1

an

)

=N

∑

n=1

Re an and Im(

N∑

n=1

an

)

=N

∑

n=1

Im an

both be convergent. More importantly, a series∑∞

n=1 an is absolutely convergent if thesequence of

∑∞n=1 |an| of nonnegative real numbers is convergent, and this is a series to which

we can apply all the tests previously discussed (Comparison Test, Ratio Test, etc.). This isuseful since every absolutely convergent sequence is necessarily convergent.

Exercise 5.12. Determine whether or not the sequences( 2 + in

1 + 3n

)∞

n=1and (in)∞n=1 converge.

Solution.

Exercise 5.13. For which values of z does the series∞∑

n=0

(z + 1)n

2nconverge absolutely?

Solution.

96


Any a ∈ C and sequence (bn)∞n=1 of complex numbers determines a power series:

∞∑

n=1

bn(z − a)n.

Combining the fact that absolute convergence implies convergence of a series of complex num-

R

a

bers, together with an application of the Ratio Test, showsthat there is some number 0 6 R 6 ∞ such that the aboveseries converges for all z ∈ C that satisfy |z − a| < R, anddiverges if |z − a| > R. That is, the series is convergentinside the disc of radius R and centre a, and divergent out-side. On the boundary we have to apply further tests. Thefunction defined by the series is continuous within the disc,and moreover it is permissible to differentiate a power se-ries term-by-term within its radius of convergence (as soonas we have defined differentiation for such functions. . . )

5.6 Exponential and trigonometric functions

Recall that the exponential, sine and cosine functions can be written in terms of their Maclau-rin Series:

ex = exp x =

∞∑

n=0

xn

n!, sinx =

∞∑

n=0

(−1)nx2n+1

(2n + 1)!, cosx =

∞∑

n=0

(−1)nx2n

(2n)!,

where x is a real variable above. These were shown to converge for all x ∈ R by calculating theradius of convergence using the Ratio Test. In turns out that the calculation is equally valid forthe analogous complex power series. For example, consider the series

∑∞n=0 zn/n! =

∑∞n=0 an

where an = zn/n!. We have

|an+1||an|

=

∣

∣

∣

∣

zn+1

(n + 1)!

∣

∣

∣

∣

×∣

∣

∣

∣

n!

zn

∣

∣

∣

∣

=|z|n × |z| × n!

(n + 1) × n! × |z|n =|z|

n + 1→ 0 as n → ∞.

Thus, by the Ratio Test, the series is absolutely convergent and hence convergent for allz ∈ C. Consequently we can define ez, and similarly sin z and cos z, by setting:

ez = exp z =

∞∑

n=0

zn

n!, sin z =

∞∑

n=0

(−1)nz2n+1

(2n + 1)!, cos z =

∞∑

n=0

(−1)nz2n

(2n)!.

That is, we use the power series expansions to extend the domain of definition of thesefunctions to all of the complex plane C, noting that if z = x, i.e. if z has no imaginary part,then ez as defined above coincides with the usual value for ex, etc.

One can then check from these definitions of complex exponentials and trigonometricfunctions that the following familiar formulae hold:

ez+w = ezew, (i.e. exp(z + w) = exp z exp w)

sin(z + w) = sin z cosw + cos z sin w,

cos(z + w) = cos z cosw − sin z sin w.

In particular we see from this that

exp(z) exp(−z) = exp(z − z) = exp 0 = 1,

97

Exponential and trigonometric functions

and so exp z 6= 0 for all z; moreover exp(−z) = 1/ exp z. Also, sin(−z) = − sin z since theseries for sin z involves only odd powers of z, and cos(−z) = cos z since the series for cos zinvolves only even powers of z. For example

cos(−z) =

∞∑

n=0

(−1)n(−z)2n

(2n)!=

∞∑

n=0

(−1)n(−1)2nz2n

(2n)!=

∞∑

n=0

(−1)nz2n

(2n)!= cos z,

since (−1)2n = [(−1)2]n = 1n = 1.To see how these functions can be expressed in terms of well-known real-valued functions

of a real variable, first consider replacing z by iz in the definition of ez:

exp(iz) =∞∑

n=0

(iz)n

n!=

∞∑

n=0

inzn

n!

= 1 + iz − z2

2!− i

z3

3!+

z4

4!+ i

z5

5!− z6

6!− i

z7

7!+ · · ·

=[

1 − z2

2!+

z4

4!− z6

6!+ · · ·

]

+ i[

z − z3

3!+

z5

5!− z7

7!+ · · ·

]

= cos z + i sin z, (†)

since i2 = −1, i3 = −i, i4 = 1, i5 = i4.i = i, i6 = −1, etc. In particular setting z = θ ∈ R

above gives a useful representation of the polar form of z ∈ C:

z = r(cos θ + i sin θ) = reiθ .

This in turn leads to quick proofs of how arguments and moduli change on multiplication anddivision, for example if z = reiθ and w = seiϕ then

z

w=

reiθ

seiϕ=

r

sei(θ−ϕ) ⇒

∣

∣

∣

z

w

∣

∣

∣=

|z||w| and arg

( z

w

)

= arg z − arg w.

Also note that the power law for the exponential function and (†) give

exp(x + iy) = exp x exp(iy) = ex(cos y + i sin y),

which is an alternative way to define ez, as used in the other texts. For similar formulae forcos z and sin z note that from (†) we have

eiz = cos z + i sin z ⇒ e−iz = ei(−z) = cos(−z) + i sin(−z) = cos z − i sin z

⇒ cos z = 12 (eiz + e−iz),

= 12 (e−y+ix + ey−ix)

= 12

(

e−y(cosx + i sinx) + ey(cosx − i sin x))

= 12 (ey + e−y) cosx − i

2 (ey − e−y) sin x

Thus cos z = cos(x + iy) can be expressed in terms of the usual trigonometric functions of x,together with the hyperbolic functions of y. A similar calculation for sin z yields:

cos z = cosx cosh y − i sinx sinh y, sin z = sinx cosh y + i cosx sinh y

where recall that cosh y = 12 (ey + e−y) and sinh y = 1

2 (ey − e−y). In particular, taking apurely imaginary number as the argument in the above gives

cos(iy) = cosh y, sin(iy) = i sinh y

98


Our earlier identities about cos(z + w) etc. made it look like complex trigonometric func-tions behave in essentially the same way as their real counterparts. However the aboveidentities show that this is not always the case. For example:

cos(iy) = cosh y = 12 (ey + e−y) → +∞ as y → ±∞.

In particular, it is not true that | cos z| 6 1. Moreover ez can take any value in the complexplane apart from 0, so is not restricted to just the positive real numbers as is ex, x ∈ R. Forexample

exp(

ln(√

2 ) + iπ4

)

= exp(

ln(√

2 ))(

cos π4 + i sin π

4

)

=√

2 ×(

1√2

+ i 1√2

)

= 1 + i.

Exercise 5.14 (A03 7(b)). Define ez and cos z for z = x + iy. Write cos(2− 3i) in the forma + ib for a, b real numbers.

Solution.

Exercise 5.15. Show that sin(z) = sin z, and hence that | sin z|2 = sin2 x + sinh2 y. Use thisto write e4+i/ sin(2 − 5i) in the form a + ib for a, b ∈ R.

Solution.

Example 5.16 (S05 5(b)). Write e2+3i and cos(4 − i) sin(3 + 5i) in the form x + iy for realnumbers x and y.

99

De Moivre’s Theorem; nth roots

Solution. We have

e2+3i = e2 cos 3 + ie2 sin 3

and

cos(4 − i) sin(3 + 5i) =(

cos 4 cosh 1 + i sin 4 sinh 1)(

sin 3 cosh 5 + i cos 3 sinh 5)

=(

cos 4 cosh 1 sin 3 cosh5 − sin 4 sinh 1 cos 3 sinh 5)

+ i(

cos 4 cosh 1 cos 3 sinh 5 + sin 4 sinh 1 sin 3 cosh5)

5.7 De Moivre’s Theorem; nth roots

Let z ∈ C, z 6= 0, hence we can write z = reiθ for some r > 0 and θ ∈ R. Then for any integern > 1 we have

zn = (reiθ)n = rn(eiθ)n = rneinθ.

Moreover, if m < 0 is an integer then we define zm = z−(−m) = (z−m)−1 = 1/z−m, and itfollows that the above equation holds for all integers m. So in particular if we take |z| = 1,i.e. r = 1, then the above becomes

(cos θ + i sin θ)m = cosmθ + i sinmθ.

This identity is known as De Moivre’s Theorem and can be used to prove various trigono-metric identities, as well as calculate powers and roots of complex numbers.

For example, given a complex number z = reiθ 6= 0 and an integer n > 2, which w ∈ C

satisfy wn = z? Any solution of this equation is an nth root of z. For the case n = 2 wewould be finding square roots. To solve this, let w = seiϕ, then

wn = z ⇔ (seiϕ)n = sneinϕ = reiθ.

For these to be equal we need the moduli to agree, i.e. sn = r, hence s = n

√r = r1/n, the

usual positive nth root of the positive number r, and the arguments must also agree — upto a multiple of 2π, since we could have nϕ = θ, or nϕ = θ + 2π, or nϕ = θ + 4π, or. . . If wedeal with the principle arguments of z and w then −π < θ 6 π and −π < ϕ 6 π, but for annth root this will still produce a total of n possibilities for ϕ, each equally spaced from thenext by 2π/n radians.

Such behaviour is already well-known for square roots of positive real numbers. Forexample 9 = (±3)2, where

1

ω

ω2

9 = 9ei×0, 3 = 3ei×0 and − 3 = 3ei×π = 3ei×2π/2.

Similarly, since 1 = 1ei×0, it follows that 1 has three cuberoots which are

ω = e2πi/3, ω2 = e4πi/3 = e−2πi/3 and 1 = ω3 = e6πi/3 = e2πi.

Exercise 5.17. Apply De Moivre’s Theorem with n = 3 to derive expressions for cos 3θ andsin 3θ in terms of cos θ and sin θ.

100


Solution.

Exercise 5.18 (A04 5(a)). By converting to polar form find:

(i) (−1 + i)6 (ii) 5√−i (iii) (2 + 2

√3i)5

Solution.

101

Exercises

5.8 Exercises

1. Which of the following complex sequences converge? [Hint: considering real and imaginaryparts of an, or considering |an| maybe be useful]

(i)

(

in

n

)∞

n=1

(ii)

(

(−1)nn

n + i

)∞

n=1

(iii)

(

n2 + in

n2 + i

)∞

n=1

(iv) (e(1+i)n)∞n=1

2. Find all of the solutions of the following equations:

(i) cos z = 0 (ii) sin z = cos z (iii) ez = i

3. Find the following powers and roots and plot them on an Argand diagram:

(i) (−1 +√

3i)7 (ii) (1 − i)4 (iii)√

i (iv) 3√−8i (v)

4

√

81 + 81√

3i

4. In each case find the radius of convergence of the given power series:

(i)

∞∑

n=0

(n2 + 1)(z + 1)n

n!(ii)

∞∑

n=0

(n − 1)7nzn

3n + 5(iii)

∞∑

n=0

i(z − i)n

in + 1

5.9 Differentiable and analytic functions

As with continuity, the definition of a differentiable function f : C → C is done by analogywith the real case. Indeed, the function f is differentiable at z0 ∈ C if

limw→0

f(z0 + w) − f(z0)

wexists,

in which case it is denoted f ′(z0). One immediate consequence of this definition is that theproduct, quotient and chain rules remain valid for derivatives of complex functions.

However, as with continuity, the subtlety here is that the limit is taken over all w ∈ C

approaching 0 from any possible direction. This is distinct from partial derivatives when wewere concerned with functions g : R2 → R, and looked at changes of the form g(x + h, y) −g(x, y) or g(x, y + h)− g(x, y) — that is, varying either one coordinate or the other, but onlymaking changes parallel to the coordinate axes.

There is a very important connection with partial derivatives, which comes by looking atthe real and imaginary parts of a differentiable function. Given any function f : C → C wecan define functions u : R2 → R and v : R2 → R by writing z = x + iy, and then setting

u(x, y) = Re f(z) = Re f(x + iy), v(x, y) = Im f(z) = Im f(x + iy).

If we also know that f is differentiable at the point z0 = a + ib then we can calculate thederivative in a number of different ways, the most obvious being to either vary the real partwhile leaving the imaginary part fixed, or vice versa. For the first case we are making thesmall change w = h + i0 = h with h ∈ R, h 6= 0, and so

f ′(z0) = limw→0

f(z0 + w) − f(z0)

w= lim

h→0

f(a + ib + h) − f(a + ib)

h

= limh→0

[u(a + h, b) + iv(a + h, b)] − [u(a, b) + iv(a, b)]

h

= limh→0

{

u(a + h, b) − u(a, b)

h+ i × v(a + h, b) − v(a, b)

h

}

=∂u

∂x(a, b) + i

∂v

∂x(a, b).

102


For the second case w = 0 + ik = ik for k ∈ R, k 6= 0, and so now

f ′(z0) = limw→0

f(z0 + w) − f(z0)

w= lim

k→0

f(a + ib + ik) − f(a + ib)

ik

= limk→0

[u(a, b + k) + iv(a, b + k)] − [u(a, b) + iv(a, b)]

ik

= limk→0

{

1

i× u(a, b + k) − u(a, b)

k+ i × 1

i× v(a, b + k) − v(a, b)

k

}

= −i∂u

∂y(a, b) +

∂v

∂y(a, b).

But both of these calculations must yield the same result, so equating the real and imaginaryparts shows that if f(z) is differentiable at z0 = a + ib then

∂u

∂x=

∂v

∂yand

∂u

∂y= −∂v

∂x.

The equations above are known as the Cauchy-Riemann equations, and are necessarily sat-isfied by any differentiable function f : C → C.

Example 5.19. Show that the real and imaginary parts of the function f(z) = zz2 satisfythe Cauchy-Riemann equations only at the origin.

Solution. f(z) = f(x+iy) = (x−iy)(x+iy)2 = (x−iy)(x2−y2+2ixy) = x3+xy2+i(y3+x2y)and so the real and imaginary parts of f are

u(x, y) = x3 + xy2 and v(x, y) = y3 + x2y

⇒ ∂u

∂x= 3x2 + y2,

∂u

∂y= 2xy,

∂v

∂x= 2xy and

∂v

∂y= 3y2 + x2.

So if∂u

∂y= −∂v

∂xthen 2xy = −2xy ⇒ xy = 0 ⇒ x = 0 or y = 0. On the other hand if

∂u

∂x=

∂v

∂ythen 3x2 + y2 = 3y2 + x2 ⇒ x2 = y2 ⇒ x = ±y. So if both of these equations

hold then both x and y must be zero, i.e. the Cauchy-Riemann equations hold only at theorigin.

A stronger condition on a function f : C → C than that of being differentiable at thepoint z0 is if it is analytic at z0, which means that not only is it differentiable at z0, but thatit is also differentiable at every point in some disc of some positive radius r > 0 with centrez0. That is, it also must be differentiable at all points close to z0 as well as at z0. A functionis called entire if it is differentiable at all points in the complex plane.

We have shown that the Cauchy-Riemann equations are a necessary condition for a func-tion to be differentiable, but they turn out not to be a sufficient condition — that is, thereare functions that satisfy these equations which are not differentiable. However if we add oneextra technical condition then everything works well:

Theorem 5.20. (i) If a function f : C → C satisfies the Cauchy-Riemann equations in the

disc D = {z : |z − z0| < r}, and if the four partial derivatives are continuous on D, then f is

analytic at every point in D.

(ii) If f : C → C is analytic in the disc D = {z : |z − z0| < r} then it is infinitely

differentiable at each point in D, and

f(z) =

∞∑

n=0

f (n)(z0)

n!(z − z0)

n, z ∈ D

103

Differentiable and analytic functions

where f (n) denotes the nth derivative of f .

(iii) If the power series∞∑

n=1

bn(z − z0)n.

has radius of convergence r, then f is analytic in the disc D = {z : |z − z0| < r}. Hence it is

infinitely differentiable in D, and

f ′(z) =

∞∑

n=1

nbn(z − z0)n−1.

In particular, differentiating m times and setting z = z0 gives f (m)(z0) = m! bm.

Since the exponential and trigonometric functions were defined in terms of power serieswith infinite radius of convergence, it follows that these functions are entire, and that we candifferentiate them term-by-term. It is then not hard to check that

d

dzez = ez,

d

dzsin z = cos z,

d

dzcos z = − sin z.

Exercise 5.21. Let u(x, y) = Re f(z) and v(x, y) = Im f(z) for f(z) = z2 + iz, wherez = x + iy. Show that u and v satisfy the Cauchy-Riemann equations everywhere.

Solution.

Exercise 5.22. Let f(z) = |z|2, and let u and v denote the real and imaginary parts of f .Show that u and v only satisfy the Cauchy-Riemann equations at the origin, and that thefunction f is differentiable there.

Solution.

104


Exercise 5.23 (S03 7(b)). Simplify u(x, y) = Re zez and v(x, y) = Im zez, for z = x + iy,

and show that u and v satisfy∂u

∂x=

∂v

∂yat all points (x, y).

Solution.

Exercise 5.24. Show that the function f(z) = z is not differentiable anywhere.

Solution.

5.10 Harmonic functions

As a final note, suppose that f : C → C is analytic in the disc D = {z : |z − z0| < r}.Its real and imaginary parts satisfy the Cauchy-Riemann equations, and since f is infinitelydifferentiable it follows that u and v have partial derivatives of all orders, and that the mixedpartial derivatives are equal. This in turn leads to the following observation:

∂u

∂x=

∂v

∂yand

∂u

∂y= −∂v

∂x

⇒ ∂2u

∂x2+

∂2u

∂y2=

∂

∂x

(∂v

∂y

)

+∂

∂y

(

−∂v

∂x

)

=∂2v

∂x∂y− ∂2v

∂y∂x

⇒ ∂2u

∂x2+

∂2u

∂y2= 0.

That is, u satisfies Laplace’s equation, as does the imaginary part v of f , as can be shownby a similar calculation. Alternatively, we say that u and v are harmonic functions.

Such functions play an extremely important role in applied mathematical problems, forexample within the study of fluid mechanics which is an area of mathematics that drawsheavily on complex analytic methods. Not only do harmonic functions appear, but, forexample, the calculations involved in determining the lift generated by air flowing round thewing of an aircraft can be greatly simplified by making an appropriate transformation of thecomplex plane, i.e. some map of the form w = f(z) to turn the wing profile into a circle.

105

Harmonic functions

When doing such transformations we usually further require them to be conformal at allpoints off of the wing’s surface: a map is conformal at a point z = z0 if it preserves anglesbetween lines passing through z0 when looking at their images in w-plane. This property isguaranteed by the simple condition that f ′(z0) 6= 0, which is true of the Mobius transformation

w =az + b

cz + dexcept when z = −d/c (assuming c 6= 0).

Exercise 5.25 (A03 7(c)). Simplify u(x, y) = Re(z2 − iz) and show that it satisfies the

equation∂2u

∂x2+

∂2u

∂y2= 0.

Solution.

Exercise 5.26 (A04 5(b)). Let u(x, y) = Re e−z2

and v(x, y) = Im e−z2

for z = x+ iy. Show

that∂u

∂x=

∂v

∂yand

∂2u

∂x2+

∂2u

∂y2= 0 at all points (x, y).

Solution.

106


Example 5.27 (S05 5(c)). Let z = x + iy and simplify u(x, y) = Re z sin z. Show that usatisfies the equation

∂2u

∂x2+

∂2u

∂y2= 0.

Solution. If z = x + iy then

z sin z = (x + iy)(sinx cosh y + i cosx sinh y)

= (x sin x cosh y − y cosx sinh y) + i(x cos x sinh y + y sin x cosh y)

and so u(x, y) = Re z sin z = x sin x cosh y − y cosx sinh y. Thus

∂u

∂x= sinx cosh y + x cosx cosh y + y sin x sinh y

∂2u

∂x2= 2 cosx cosh y − x sin x cosh y + y cosx sinh y

∂u

∂y= x sin x sinh y − cosx sinh y − y cosx cosh y

∂2u

∂y2= x sin x cosh y − 2 cosx cosh y − y cosx sinh y

and so∂2u

∂x2+

∂2u

∂y2= 0.

Example 5.28. Let u(x, y) and v(x, y) be the real-valued functions defined by

u(x, y) = Re(iz2 + cos z), v(x, y) = Im(iz2 + cos z)

where z = x + iy. Show that∂u

∂x=

∂v

∂yholds at all points (x, y), and that v satisfies the

equation∂2v

∂x2+

∂2v

∂y2= 0.

Solution. Now iz2 + cos z = i(x2 − y2 + 2ixy) + cosx cosh y − i sinx sinh y, and so

u = −2xy + cosx cosh y, v = x2 − y2 − sin x sinh y.

Thus∂u

∂x= −2y − sin x cosh y and

∂v

∂y= −2y − sin x cosh y, hence these partial derivatives

are equal.

Also,∂v

∂x= 2x − cosx sinh y ⇒ ∂2v

∂x2= 2 + sinx sinh y, while

∂2v

∂y2= −2 − sin x sinh y,

so that∂2v

∂x2+

∂2v

∂y2= 0.

107

Exercises

5.11 Exercises

1. In each case, write the function w = f(z) as w = u(x, y) + iv(x, y), where z = x + iy.

Then verify that∂u

∂x=

∂v

∂y, that

∂u

∂x= −∂v

∂y, and that

∂2u

∂x2+

∂2u

∂y2= 0:

(i) w = z3 (ii) w = sin z + e2z (iii) w = cos z2 (iv) w = zez

(v) w = iz2 − ez2

(vi) w = 3z2 − iz4 (vii) w = z + i + ez (viii) w = sin z cos z

2. In each case show that the given function is not analytic at any point in C by showingthat the Cauchy-Riemann equations do not hold at most points in C:

(i) f(z) = Re z (ii) f(z) = |z|2 (iii) f(z) = iz + |z|

By reverting to the definition f ′(z) = limh→0 h−1(

f(z + h) − f(z))

, determine where thesefunctions are differentiable.

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5 Complex Numbers and Functions - University …...Before deﬁning division, we need the complex...

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