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PMTH333 COMPLEX ANALYSIS

The tutorial questions contained in this booklet have mostly been selected from

Complex Variables and Applications, 6th ed. by J.W. Brown and R.V. Churchill,

McGraw-Hill.

References to relevant sections from this text are included with each tutorial. If

you have difficulty with particular questions, you should review these sections and

attempt similar exercises from the text.

Sample worked solutions for the tutorial questions will appear at the course

website. Resist the temptation to refer to the solution as soon as you encounter any

difficulty. You will learn and remember more from perseverance, successful or not,

than from simply following through a worked solution.

Printed at the University of New England

November, 2005

1

TUTORIAL 1

Complex Variables and Applications, Sect. 1-8

1. Let z be any complex number. Prove that

(a) Im(i z) = Re z, (b) Re(i z) = − Im z.

2. Solve the equation z2 + z + 1 = 0 by substituting z = x+ i y.

3. Verify that

(a) |z| = |z|, (b) z1z2 = z1z2, (c)

(z1

z2

)=z1

z2

4. Sketch (or describe) the set of points determined by

(a) |z − 1 + i | = 1, (b) |2z − i | ≤ 4, (c) Re(z − i) = 2

5. Find the four roots of the equation z4 + 4 = 0.

6. Sketch (or describe) the closure of the sets

(a)− π < Arg z < π (z 6= 0), (b) |Re z| < |z|.

2

TUTORIAL 2


1. Describe the natural domain of the function f(z) = zz+z

.

2. Let f(z) = x2 − y2 − 2y + i(2x − 2xy), where z = x + i y. Using x = z+z2

,

y = z−z2 i

or otherwise, express f(z) in terms of z.

3. Sketch the region onto which the sector r ≤ 1, 0 ≤ θ ≤ π4

is mapped by the

transformations

(a)w = z2, (b) w = z3, (c) w = z4.

4. Show that

(a) limz→∞

4z2

(z − 1)2= 4, (b) lim

z→1

1

(z − 1)3= ∞.

5. Use the definitions to show that

(a) limz→0

1

z= ∞, (b) lim

z→∞

1

z= 0.

TUTORIAL 3


1. Find f ′(z) where f(z) = z−12z+1

, (z 6= −12).

2. Show that f ′(z) does not exist at any point if

(a) f(z) = 2x+ i y2, (b) f(z) = ex e− i y .

3. Determine where f ′(z) exists and find its value when

(a) f(z) = x2 + i y2,

(b) f(z) = z Im z,

(c) f(z) = e−θ cos(ln r) + i e−θ sin(ln r).

4. Show that, for f(z) = u(r, θ) + i v(r, θ) where z = r ei θ, if f ′(z0) exists and

z0 6= 0, then

f ′(z0) =− i

z0

(uθ + i vθ).

3

TUTORIAL 4


1. Show that an analytic function f(z) in a domain D which takes only real

values for all z ∈ D must be a constant in D.

2. Find all values of z such that ez =√

3 i.

3. Show that ez = ez and ez is not analytic anywhere.

4. Show that sin(z) is not analytic anywhere.

5. Find all roots of the equation cosh z = 12.

6. Harmonic Functions

(a) Which of the following functions are harmonic on some domain in C?

Specify the domain.

f(x, y) =y

x2 + y2

f(x, y) = ex2−y2

(b) Using Laplace’s equation, show directly that Re 1z

is harmonic in a neigh-

bourhood of any point z0 except the origin.

(c) Let

φ(x, y) = 6x2y2 − x4 − y4 + y − x+ 1.

If φ is the real part of an analytic function f(z), what is the imaginary part?

Conversely, if φ is the imaginary part of f(z), what is the real part? Are the

answers equal?

(d) Find the harmonic conjugate of

φ(x, y) = arctanx

y

where −π2< arctan x

y≤ π

2.

4

TUTORIAL 5


1. Show that

(a) Log(1− i) =1

2ln 2− π

4i

(b) Log(1 + i)2 = 2 Log(1 + i), but Log(−1 + i)2 6= 2 Log(−1 + i)

2. Find the principal value of (a) ii, (b) (1− i)4 i.

3. Calculate the following:

(a)

∫ π6

0

e2 i t dt, (b)

∫ ∞

0

e−zt dt.

4. Evaluate∫ π

0ex cosx dx and

∫ π

0ex sin x dx by using∫ π

0

ex cosx dx+ i

∫ π

0

ex sin x dx =

∫ π

0

e(1+i)x dx.

TUTORIAL 6


1. Let z = z(t) be a smooth arc, and f(z) be analytic at z0 = z(t0). Show that

d

dtf(z(t)) = f ′(z(t))z′(t)

at t = t0.

2. Calculate∫

Cf(z) dz where C is the arc from z = −1− i to z = 1 + i along the

curve y = x3 and

f(z) =

{1 when y < 0

4y when y > 0.

3. Show that∫

Cdz

z−z0= 2π i where C is given by z = z0 + R ei θ, (R > 0 fixed, θ

changes from −π to π.)

4. Evaluate

(a)

∫ i2

i

eπz dz, (b)

∫ 3

i

(z − 2)3 dz.

5. Apply the Cauchy-Goursat Theorem to show that∫C

dz

z2 + 2z + 2= 0

where C is the circle |z| = 1 oriented counter-clockwise.

5

TUTORIAL 7


1. Find∫

Cg(z) dz where C is the circle |z − i | = 2 in the positive sense and

g(z) =1

(z2 + 4)2.

2. Let C be the circle |z| = 3 in the positive sense. Show that g(2) = 8π i and

g(4 i) = 0 where

g(w) =

∫C

2z2 − z − 2

z − wdz, (|w| 6= 3).

3. Let C be a simple closed contour oriented counter-clockwise. Define

g(w) =

∫C

z3 + 2z

(z − w)3dz.

Show that g(w) = 6πw when w is inside C and g(w) = 0 when C is outside

C.

4. Show that if f(z) is analytic whithin and on a simple closed contour C and z0

is not on C, then ∫C

f ′(z)

z − z0

dz =

∫C

f(z)

(z − z0)2dz.

6

TUTORIAL 8


1. Show that ez = e∑∞

n=0(z−1)n

n!(z ∈ C).

2. Find the MacLaurin series expansion of the function

f(z) =z

z4 + 9=z

9

[1

1 + z4

9

].

3. Find the MacLaurin series expansion of the function f(z) = sin z2 and use it

to show that f (4)(0) = f (2n+1)(0) = 0, n = 0, 1, 2, . . . .

4. Derive the Taylor series representation

1

1− z=

∞∑n=0

(z − i)n

(1− i)n+1, (|z − i | <

√2).

5. Show that when 0 < |z| < 4,

1

4z − z2=

1

4z+

∞∑n=0

zn

4n+2.

7

TUTORIAL 9


1. Represent f(z) = z+1z−1

by

(a) its MacLaurin series, and give the region of validity for the representation;

(b) its Laurent series for the domain 1 < |z| <∞.

2. Show that when 0 < |z − 1| < 2,

z

(z − 1)(z − 3)=

−1

2(z − 1− 3

∞∑n=0

(z − 1)n

2n+2.

3. Differentiate 11−z

=∑∞

n=0 zn, (|z| < 1), to obtain

1

(1− z)2=

∞∑n=0

(n+ 1)zn, (|z| < 1)

2

(1− z)3=

∞∑n=0

(n+ 1)(n+ 2)zn, (|z| < 1).

4. Prove that

f(z) =

cos z

z2−(π2 )

2 , when z 6= ±π2,

− 1π, when z = ±π

2

is entire.

5. Prove that if f(z) is analytic at z0 and

f(z0) = f ′(z0) = · · · = f (m)(z0) = 0,

the the function

g(z) =

f(z)

(z−z0)m+1 when z 6= z0,

f (m+1)(z0)(m+1)!

when z = z0,

is analytic at z0.

8

TUTORIAL 10

Complex Variables and Applications, Sect. 51, 53-55

1. Show that

(a)ez

z(z2 + 1)=

1

z+ 1− 1

2z − 5

6z2 + · · · , (0 < |z| < 1)

(b)1

ez −1=

1

z− 1

2+

1

12z − 1

720z3 + · · · , (0 < |z| < 1).

2. Find the residues at z = 0 of the functions

(a)1

z + z2, (b) z cos

1

z, (c)

z − sin z

z.

3. Evaluate the integral of f(z) around the positively oriented circle |z| = 2 when

f(z) is

(a)z5

1− z3, (b)

1

1 + z2, (c)

1

z.

TUTORIAL 11


1. Find∫∞

0x2

(x2+1)(x2+4)dx.

2. Find∫∞−∞

x sin axx4+4

dx where a ∈ R.

3. Show that∫ 2π

0dθ

1+ 12

cos θ= 4π√

3.

9

COMPLEX ANALYSIS SOLUTIONSTUTORIAL PROBLEMS SET 1.

1. Let z = x + i y. Then i z = −y + ix and therefore, Im i z = x = Re z and

Re i z = −y = − Im z.

2. After the substitution we have

(x+ i y)2 + (x+ i y) + 1 = (x2 − y2 + x+ 1) + (2xy + y) i = 0

which gives two real equations

x2 − y2 + x+ 1 = 0

2xy + y = 0

From the second equation we find that y = 0 or x = −12. In the first case,

the first equation won’t have a real solution (x = Re z must be real!). In the

second case, the first equation becomes

1

4− y2 − 1

2+ 1 = 0.

Hence y2 = 34. This gives the two solutions z = −1

2± 3

4i.

3. (a) Let z = x+ i y. By definition

|z| =√x2 + (−y)2 =

√x2 + y2 = |z|.

(b)

(x1 + i y1)(x2 + i y2) = x1x2 − y1y2 + (x1y2 + x2y1) i

= x1x2 − y1y2 − (x1y2 + x2y1) i = (x1 − i y1)(x2 − i y2)

= z1z2

(c) This can be done analogously to (b) or as follows (by reduction to (b)):

The assertion is equivalent to (z1

z2

)z2 = z1.

Due to (b) the left hand side equals ( z1

z2)z2 = z1 which proves (c).

4. (a) a circle centred at 1− i of radius 1 (the distance of z to 1− i equals 1).

(b) The equations is equivalent to |z − i2| ≤ 2. This is a closed disc (disc

including the boundaring circle) centred at i2

of radius 2.

(c) This equations is equivalent to x = 2, where x = Re z. Thus the point set

is a vertical line that intersects the real axis at x = 2.

10

5. From z4 + 4 = we get z2 = ±2 i. This can be solved by passing to z = x+ i y.

We find

x2 − y2 = 0

2xy = ±2

Hence, x = ±y and x2 = 1. Therefore, the four solutions are 1+i, 1− i, −1+i,

−1− i.

6. (a) The given set is the complex plane with the real half line (−∞, 0] deleted.

The closure is the entire complex plane, since any point of the real half line is

limiting point for the given set.

(b) The given formula is equivalent to y2 > 0 which in turn is equivalent to

y 6= 0. This is the complex plane with deleted real axis. Again, the closure is

the entire complex plane.

TUTORIAL PROBLEMS SET 2.

1. The natural domain of this function consists of all complex points for which

the denominator z + z = 2x does not vanish. Thus the natural domain is the

complex plane with deleted imaginary axis.

2. f(z) = z2 + 2 i z

3. (a) The function w = z2 squares the absolute value and doubles the argument.

Therefore the image will consist of all point whose absolute value is betwee

0 and 1 and whose argument is between 0 and π2. This is the (closed) upper

right quadrant.

(b) This function cubes the absolute values and triples the argument. Thus,

the image is the sector r ≤ 1 and 0 ≤ θ ≤ 2π4

.

(c) Here the image is the closed upper half plane.

4. (a)According to (5) in Sect. 14 we have

limz→∞

4z2

(z − 1)2= lim

z→0

4/z2

(1/z − 1)2= lim

z→0

4

(1− z)2

Since the latter function is continuous at 0 the limit is the value of the function

at 0 which is 4.

(b) We show that for any (not matter how big) number R there is a circle

around 1 such that for all z within this circle the absolute value of∣∣∣ 1(z−1)3

∣∣∣ > R.

In fact, just take a circle of any radius δ which is less that 1/ 3√R. Then∣∣∣∣ 1

(z − 1)3

∣∣∣∣ =1

|z − 1|3<

1

(1/ 3√R)3

= R.

(You find how to choose δ by starting with the last inequality and going

backwards.)

11

5. (a) similarly to 4(b) we show that for any (not matter how big) number R

there is a circle around 0 such that for all z within this circle the absolute

value of∣∣1z

∣∣ > R. Here it suffices to take δ < 1/R.

(b) We have to show that for any (no matter how small) number ε there is a

circle of radius δ such that outside that circle the absolute value of∣∣1z

∣∣ < ε.

Take δ > 1/ε.


1. According to the quotient rule

f ′(z) =2z + 1− 2(z − 1)

(2z + 1)2=

3

(2z + 1)2

for z 6= −12.

2. (a) The matrix of partial derivatives of Re f and Im f with respect to Re z

and Im z is (2 0

y2 2xy

)Hence the Cauchy-Riemann equations imply y = 0 a then 2 = 0 which is

impossible.

(b) Here the matrix of partial derivatives is(ex cos y − ex sin x

−ex sin x − ex cosx

)The Cauchy-Riemann equations imply sinx = cos x = 0 which is impossible.

3. (a) The matrix of partial derivatives is(2x 0

0 2y

)The Cauchy-Riemann equations are equivalent to x = y. Thus f ′(z) exists at

the line x = y.

(b) The matrix of partial derivatives is(y x

0 2y

)The Cauchy-Riemann equations are equivalent to x = y = 0. Thus f ′(z) exists

only at the origin.

(c) Since Log z = ln r+i θ we have f(z) = ei Log z which has a complex derivative

for z 6= 0 and θ 6= π.

12

4. If f ′(z) exists then

f ′(z) =1

2

(∂f

∂x− i

∂f

∂y

)0 =

1

2

(∂f

∂x+ i

∂f

∂y

).

We have ∂r∂x

= xr, ∂r

∂y= y

r, ∂θ

∂x= − y

r2 ,∂θ∂y

= xr2 . According to the chain rule now

∂f

∂x=

1

2

(∂f

∂r

x

r+∂f

∂θ

−yr2

)∂f

∂y=

1

2

(∂f

∂r

y

r+∂f

∂θ

x

r2

).

Therefore,

∂f

∂z=

1

2

(∂f

∂r

z

r+∂f

∂θ

− i z

r2

)0 =

1

2

(∂f

∂r

z

r+∂f

∂θ

i z

r2

).

Subtracting z/z times the second equation from the first equation above yields

∂f

∂z=∂f

∂θ

− i z

r2=− i

z

∂f

∂θ.


1. If f = u + i v is analytic and takes only real values we have ∂v∂x

= ∂v∂y

= 0 and

therefore ∂u∂x

= ∂u∂y

= 0. But then u does not depend neither on x nor on y,

thus u and therefore f is a constant.

2.

ez = ex(cos y + i sin y) =√

3 i

is equivalent to ex =√

3 and sin y = 1. Hence z = ln 32

+ (π2

+ 2kπ) i where k

is any integer.

3.

ez = ex(cos y + i sin(−y)) = ex(cos y − i sin y) = ex(cos y + i sin y) = ez.

Since ez = ex e−Iy it is nowhere analytic due to 2 (b) of Tutorial 2.

4. The function f(z) = sin z = sin(x− i y) has a complex derivative where ∂f∂x

+

i ∂f∂y

= 0.

cos(x− i y) + i(− i) cos(x− i y) = 2 cos(x− i y) = 0

This happens for z = π2+kπ, i.e. at isolated points. Therefore f is not analytic

anywhere.

13

5. cosh z = 12(ez + e−z) = 1

2is equivalent to e2z − ez +1 = 0. Substitute ez = u

and solve u2 − u+ 1 = 0. It follows

ez =1

2±√

3

2i

z = log

(1

2±√

3

2i

)z = ±π

3

6. (a)

f(x, y) =y

x2 + y2= Im

z

|z|2= Im

1

z= − Im

1

z

Hence f is harmonic outside the origin.

For f(x, y) = ex2−y2we find ∆f = (4 + 8(x2 + y2)) ex2−y2

. Therefore f is not

analytic anywhere.

(b) f(z) = Re 1z

= 12

(1

x+i y+ 1

x−i y

). Therefore

∂2f

(∂x)2=

1

(x+ i y)3+

1

(x+ i y)3

and∂2f

(∂y)2= − 1

(x+ i y)3− 1

(x+ i y)3.

Hence ∆f = 0.

(c) From ∂φ∂x

= 12xy2−4x3−1 and ∂φ∂y

= 12x2y−4y3 +1 we get for the desired

imaginary part ψ:

∂ψ

∂x= −12x2y + 4y3 − 1,

∂ψ

∂y= 12xy2 − 4x3 − 1.

It follows ψ = −4x3y + 4y3x− x− g(y). Comparing the derivatives of ψ by y

gives

−4x3 + 12xy2 − g′(y) = 12xy2 − 4x3 − 1

and hence g′(y) = 1. Therefore g = y +C and the desired imaginary part has

the form

ψ = −4x3y + 4y3x− x− y − C.

If φ was the imaginary part of an analytic function then the corresponding

real part would have the form ψ = 4x3y− 4y3x+ x+ y+C. The two answers

differ by sign.

(d) We know that Log z = ln |z|+ i Arg z = ln |z|+ i arctan yx. Hence

Log(i z) = ln |z|+ i arctan

(x

y

)and

−Log(− i z) = − ln |z|+ i arctan

(x

y

).

It follows that − ln |z| is a harmonic conjugate.

14


1. (a) Since |1− i | =√

2 and Arg(1− i) = −π4

it follows Log(1− i) = 12log 2− π

4i.

(b)

Log(1 + i)2 = log 2 +π

2i = 2 Log(1 + i)

but

Log(−1 + i)2 = log 2− π

2i 6= log 2− 3π

2= 2 Log(−1 + i)

2. (a) ii = ei Log i = ei · i π2 = e

π2

(b) (1− i)4 i = e4 i Log(1−i) = e2 i log 2+π = eπ(cos(2 log 2) + i sin(2 log 2))

3. (a)∫ π

6

0e2 i t dt =

[12 i

e2 i t]π

6

0= e

π3 i−12 i

=√

34

+ 14i

(b)∫∞

0e−zt dt =

[e−zt

−z

]∞0

= 1z

if Re z > 0. The improper integral diverges if

Re z ≤ 0.

4.∫ π

0e(1+i)x dx =

[e(1+i)x

1+i

]π0

= − eπ +12

+ eπ +12

i. Hence∫ π

0

ex cosx dx = Re

∫ π

0

e(1+i)x dx = −eπ +1

2∫ π

0

ex sin x dx = Im

∫ π

0

e(1+i)x dx =eπ +1

2.


1. Let f = u+ i v and z = x+ i y. According to the chain rule

du(z(t))

dt=∂u

∂x

dx

dt+∂u

∂y

dy

dt

dv(z(t))

dt=∂v

∂x

dx

dt+∂v

∂y

dy

dt

Now, since f(z) is analytic, we have f ′(z) = ∂u∂x− i ∂u

∂y= ∂v

∂y+ i ∂v

∂x. Hence

du(z(t))

dt= Re f ′(z)

d

dt(x(t) + i y(t))

dv(z(t))

dt= Im f ′(z)

d

dt(x(t) + i y(t))

and df(z(t))dt

= f ′(z(t))dz(t)dt

.

2. The arc can be parametrised by x = t, y = t3 for t ∈ [−1, 1]. Then z = t+ i t3

and ∫C

f(z) dz =

∫ 0

−1

1 · (1 + 3 i t2)dt+

∫ 1

0

4t3 · (1 + 3 i t2)dt

=[t+ i t3

]0−1

+[t4 + 2 i t6

]10

= 2 + 3 i .

15

3. For z = z0 +R ei θ we get dz = iR ei θ dθ. Therefore∫C

dz

z − z0

=

∫ π

−π

iR ei θ dθ

R ei θ= [i θ]π−π = 2π i .

4. (a)∫ i

2

ieπz dz =

[eπz

π

] i2

i= 1−i

π.

(b)∫ 3

i(z − 2)3dz =

[(z−2)4

4

]3i

= 2 + 6 i.

5. The integrand is singular for z2 + 2z + 2 = 0, thus for z = 1± i. These points

are outside the circle C. According to the Cauchy-Goursat theorem then the

integral vanishes.


1. The only singularity within the circle is at z = 2 i. Therefore, the integral

equals 2π i res2 i g(z). Since the singularity is a pole of second order

res2 i g(z) = [(z − 2 i)g(z)]′(2 i) =1

32 i

Hence the integral equals π16

.

2. For w = 2 we have exactly one singularity inside the circle. Therefore the

integral g(2) = 2π i res22z2−z−2

z−2= 2π i(8− 2− 2) = 8π i.

For w = 4 i the integrad has no singularities inside the circle and the integral

vanishes.

3. When w is outside C the integrand has no singularities inside C and therefore

the integral vanishes. If w is inside C then w is the only singularity and the

integral equals 2π i reswz3+2z(z−w)3

. Since w is a 3rd order pole we find

reswz3 + 2z

(z − w)3=

1

2!(z3 + 2z)′′|z=w = 3w.

This implies g(w) = 6π iw if w is inside C.

4. If z0 outside C then both integrals vanish. If z0 is inside C then z0 is a simple

pole of f ′

(z−z0)and a second order pole of f

(z−z0)2. We have to prove

resz0

f ′

(z − z0)= resz0 fracf(z − z0)

2.

But both residues equal f ′(0).


16

1. From eu =∑∞

n=0un

n!we get for u = z − 1

ez−1 =ez

e=

∞∑n=0

(z − 1)n

n!.

Hence ez = e∑∞

n=0(z−1)n

n!.

2.

f(z) =z

9

[1

1 + z4

9

]=z

9

∞∑n=0

z4n

9n=

∞∑n=0

z4n+1

9n+1

for |z| <√

3.

3. From sinu =∑∞

n+0(−1)n u2n+1

(2n+1)!with u = z2 we get

sin z2 =∞∑

n+0

(−1)n z4n+2

(2n+ 1)!.

In the expansion the coefficients at all odd powers and at powers divisible by

4 vanish. Thus for the corresponding derivatives at 0 must vanish.

4.

1

z=

1

1− i−(z − i)=

1

1− i· 1

1− z−i1−i

=1

1− i

∞∑n=0

(z − i

1− i

)n

=∞∑

n=0

(z − i)n

(1− i)n+1

for |z − i | < |1− i | =√

2.

5.

1

4z − z2=

1

4z· 1

1− z4

=1

4z

∞∑n=0

(z4

)n

=1

4z+

1

4z

∞∑n=1

(z4

)n

=1

4z+

∞∑n=1

zn−1

4n+1

=1

4z+

∞∑n=0

zn

4n+2


1. (a)

z + 1

z − 1=−z − 1

1− z= −

∞∑n=0

(zn+1 + zn) = −1− 2∞∑

n=1

zn

This representation is valid for |z| < 1.

(b)

z + 1

z − 1=

1 + 1z

1− 1z

=∞∑

n=0

(1

zn+

1

zn+1

)=

1

z+ 2

∞∑n=1

1

zn

17

2.

z

(z − 1)(z − 3)=

z − 1 + 1

(z − 1)(z − 3)=

−1

2− (z − 1)− 1

z − 1· −1

2− (z − 1)

=−1

2(1− z−1

2

) − 1

z − 1· −1

2(1− z−1

2

)= −1

2

∞∑0

(z − 1)n

2n− 1

2

∞∑0

(z − 1)n−1

2n

= − 1

2(z − 1)− 3

∞∑0

(z − 1)n

2n+2

3. Differentiating both sides of the geometric series

1

1− z=

∞∑n=0

zn

yields

1

(1− z)2=

∞∑n=0

nzn−1 =∞∑

n=0

(n+ 1)zn,

and2

(1− z)3=

∞∑n=0

n(n+ 1)zn−1 =∞∑

n=0

(n+ 1)(n+ 2)zn.

4. We prove that f(z) is continuous at z = ±π2. Then, according to Riemann’s

theorem, f is analytic at those points and therefore f is entire. Due to

l’Hopital’s rule

limz→±π

2

cos z

z2 −(

π2

)2 = limz→±π

2

− sin z

2z= −±1

±π=−1

π.

5. We show that g is continuous at z0. Due to l’Hopital’s rule (applied m + 1

times)

limz→z0

f(z)

(z − z0)m+1= lim

z→z0

fm+1(z)

m+ 1!=fm+1(z0)

m+ 1!.


1. (a)

ez

z(z2 + 1)=

1

z(1 + z +

z2

2+z3

6+ · · · )(1− z2 + · · · )

=1

z(1 + z − z2

2− 5z3

6+ · · · ) =

1

z+ 1− z

2− 5z2

6+ · · ·

18

(b)

1

ez −1=

1

z· 1

1 + z2

+ z2

6+ z3

24+ z4

120+ · · ·

=1

z(1− z

2− z2

6− z3

24− z4

120+z2

4+ 2

z3

12+z4

36+ 2

z4

48−

z3

8− 3

z4

24+z4

16+ · · · )

=1

z− 1

2+

z

12− z3

720+ · · ·

2. (a) This function has a pole of order 1 at 0, thus

res01

z + z2= lim

z→0

z

z + z2= 1.

(b)

z cos1

z= z(1− 1

2z2+ · · · )

Therefore,

res0 z cos1

z= −1

2.

(c) This function has a pole of order 1 at 0, thus

res0z − sin z

z= lim

z→0(z − sin z) = 0.

3. (a) The singularities of the function f = z5

1−z3 are 1, ε = −12+√

32

, ε2 = −12−

√3

2.

They are all inside the circle C = {|z| = 2}. Therefore∫C

fdz = 2π i(res1 f + resε f + resε2 f).

We have

res1 f = limz→1

z5(z − 1)

1− z3= lim

z→1

−z5

1 + z + z2= −1

3

resε f = limz→ε

z5(z − ε)

1− z3=

−ε2

(ε− 1)(ε− ε2)

resε2 f = limz→ε2

z5(z − ε2)

1− z3=

−ε(ε2 − 1)(ε2 − ε)

This yields ∫C

fdz = 2π i

(−1

3− ε2

2ε2 − 1− ε− ε

2ε− 1− ε2

)We take into account 1 + ε+ ε2. Thus∫

C

fdz = 2π i

(−1

3− ε2

3ε2− ε

3ε

)= −2π i

19

(b) The singularities of the function f = 1z2 are ± i which are all inside C.

Thus ∫C

fdz = 2π i(resi f + res− i f).

resi f = limz→i

z − i

z2 + 1=

1

2 i

res− i f = limz→− i

z + i

z2 + 1= − 1

2 i

Hence∫

Cfdz = 0.

(c) f(z) = 1z

has a simple pole in C with residue 1. Thus∫

Cfdz = 2π i.


1. Since the integrand is an even function the integral equals 12

∫∞−∞ f(z) dz. The

integral along a half circle of (sufficiently big) radius R in the upper half plane

consists of the straight part which tends to the desired integral and the circle

part which tends to zero. On the other hand, this integral equals to the sum

of residues in the upper half plane multiplied with 2π i. Thus∫ ∞

0

x2

(x2 + 1)(x2 + 4)dx =

1

22π i (resi f + res2 i f)

=π i

(−1

2 i ·3+

−4

(−3) · 4 i

)=π

6

2.

I(a) =

∫ ∞

−∞

x sin ax

x4 + 4dx =

1

2 i

∫ ∞

−∞

x ei ax

x4 + 4dx− 1

2 i

∫ ∞

−∞

x e− i ax

x4 + 4dx

Since I(a) is an odd function of a we may restict ourselves to a > 0. Then

Jordan’s lemma applies for the first integral in the upper half plane and for

the second integral in the lower half plane. Thus the first integral equals 2π i

times the two residues at ±1 + i and the second integral equals −2π i times

the two residues at ±1− i.

I(a) =2π i(

res1+i1

2 i

∫ ∞

−∞

x ei ax

x4 + 4+ res−1+i

1

2 i

∫ ∞

−∞

x ei ax

x4 + 4−

res1−i1

2 i

∫ ∞

−∞

x e− i ax

x4 + 4− res−1−i

1

2 i

∫ ∞

−∞

x e− i ax

x4 + 4

)For the first integral we get

I1(a) =π

((1 + i) e(−1+i)a

2 · 2 i ·(2 + 2 i)+

(−1 + i) e(−1−i)a

(−2) · (−2 + 2 i) · 2 i

)=π

(e−a(cos a+ i sin a)

8 i− e−a(cos a− i sin a)

8 i

)=π e−a sin a

4

20

Analogously we get for the second integral

I2(a) =− π

((1− i) e−(1+i)a

(−2 i) · (2− 2 i) · 2+

(−1− i) e−(1−i)a

(−2− 2 i) · (−2 i) · (−2)

)=π

(e−a(cos a− i sin a)

8 i− e−a(cos a+ i sin a)

8 i

)=− π e−a sin a

4

(Note that the orientation of the contour of integration is clockwise!)

Altogether we find for positive a

I(a) = I1(a)− I2(a) =π e−a sin a

2.

Since I(a) is odd we find I(a) = −π ea sin(−a)2

= π ea sin a2

for negative a. Clearly,

I(0) = 0. Hence

I(a) =π sin a

2 e|a|.

3. The integral equals I =∫ 2π

0dθ

1+ 14(ei θ +e− i θ)

. We substitute z = ei θ. This yields

I =

∫|z|=1

dz

i z(1 + 14(z + 1

z))

=4

i

∫|z|=1

dz

z2 + 4z + 1=

2π i ·4ires−2+

√3

1

z2 + 4z + 1=

4π√3.

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