c M. K. Warby
20161117171541 MA3614 Complex variable methods and applications
1
Exercises involving analytic functions, harmonic functions and harmonic conjugates
Some of the questions have been taken from past May exams of MA3614 and MA3914.
The format of the past exams was answer 3 from 4 in 3 hours with each question worth
20 marks. Hence if a question given here was worth 10 marks then as a percentage this
was worth 16.7%. The connection between analytic functions and harmonic functions is
unlikely to be covered in the lectures before week 5.
Ex. 1 — Let f (z) = |z|2 = z z. Show that f (z) has a complex derivative at z = 0 but
at no other point. Is f analytic at z = 0?
Answer (Ex. 1) —
You can use the definition directly here or you can use the Cauchy Riemann equations.
If we use the Cauchy Riemann equations then f (z) = u(x, y) + i v(x, y) with
u(x, y) = x2 + y 2 ,
v(x, y) = 0.
Thus
∂u
∂v
∂v
∂u
= 2x,
= 2y,
=
= 0.
∂x
∂y
∂x
∂y
At x = y = 0 the Cauchy Riemann equations are satisfied (as all terms are 0) and thus
the complex derivative exists at 0. The Cauchy Riemann equations are not satisfied at
any other point and thus f is not analytic at z = 0.
Ex. 2 — Use L’Hôpital’s rule to determine the following limits. (The first case was a
question from an earlier question sheet.)
1.
z 2 + iz + 2
.
z→i z 2 − 5iz − 4
lim
2.
z3 + i
.
z→i z(z 2 + 1)
lim
Answer (Ex. 2) —
1. The numerator and denominator vanish as z = i but the derivative of the denominator does not and thus
z 2 + iz + 2
2z + i 3i
lim 2
=
=
= −1.
z→i z − 5iz − 4
2z − 5i
−3i
z=i
2. As in the other example the numerator and denominator vanish as z = i but the
derivative of the denominator does not.
z3 + i
1 z3 + i
3z 2 −3
3
=
= (−i)
lim
lim 2
= (−i)
= .
2
z→i z(z + 1)
z→i z + 1
z
2z
2i
2
z=i
z=i
c M. K. Warby
20161117171541 MA3614 Complex variable methods and applications
2
Ex. 3 — Let z1 , z2 , . . . , zn be points in the complex plane and let
pn (z) = (z − z1 )(z − z2 ) · · · (z − zn ).
Prove by induction on n that
p0n (z)
1
1
1
=
+
+ ··· +
.
pn (z)
z − z1 z − z2
z − zn
Answer (Ex. 3) —
To start the induction consider the case when n = 1 which gives p01 (z) = 1. Thus
p01 (z)
1
=
p1 (z)
z − z1
and the result is true.
For the induction hypothesis suppose that it is true with m terms with m ≥ 1. We now
consider the case with m + 1 terms and note that by the product rule
pm+1 (z) = (z − zm+1 )pm (z) and p0m+1 (z) = (z − zm+1 )p0m (z) + pm (z)
so that
p0m+1 (z)
p0 (z)
1
= m
+
.
pm+1 (z)
pm (z) z − zm+1
Now by the hypothesis we can replace the term p0m (z)/pm (z) by the sum of m terms and
hence
p0m+1 (z)
1
1
=
+ ··· +
.
pm+1 (z)
z − z1
z − zm+1
This shows that the result is also true for m + 1 terms and by induction it is true for all
m = 1, 2, . . ..
Ex. 4 — Verify that the following functions, denoted by u, of real variables x and y are
harmonic and find their harmonic conjugate. If f is given by f (z) = u(x, y) + iv(x, y) is
entire then find f (z) in each case.
1. u(x, y) = y 3 − 3x2 y.
2. u(x, y) = sinh x sin y.
c M. K. Warby
20161117171541 MA3614 Complex variable methods and applications
3
Answer (Ex. 4) —
1.
∂u
= −6xy,
∂x
∂ 2u
= −6y,
∂x2
∂u
= 3y 2 − 3x2 ,
∂y
∂ 2u
= 6y.
∂y 2
Hence
∂ 2u ∂ 2u
+
= 0.
∂x2 ∂y 2
To get the harmonic conjugate v we have by one of the Cauchy Riemann equations
that
∂v
∂u
=−
= −(3y 2 − 3x2 ) = 3x2 − 3y 2
∂x
∂y
and thus integrating with respect to x gives
∇2 u =
v = x3 − 3y 2 x + g(y),
where g(y) is any function of y. Using the other Cauchy Riemann equation gives
∂u
∂v
= −6yx + g 0 (y) =
= −6xy.
∂y
∂x
Hence g 0 (y) = 0 and g(y) = A where A is a constant. The function f is
f = u + i v = y 3 − 3x2 y + i(x3 − 3y 2 x) + iA.
2.
∂u
= cosh x sin y,
∂x
and
∂u
= sinh x cos y,
∂y
∂ 2u
= sinh x sin y = u
∂x2
∂ 2u
= − sinh x sin y = −u.
∂y 2
Hence
∂ 2u ∂ 2u
+
= 0.
∂x2 ∂y 2
To get the harmonic conjugate v we have by one of the Cauchy Riemann equations
that
∂v
∂u
=−
= − sinh x cos y
∂x
∂y
and thus integrating with respect to x gives
∇2 u =
v = − cosh x cos y + g(y),
where g(y) is any function of y. Using the other Cauchy Riemann equation gives
∂v
∂u
= cosh x sin y + g 0 (y) =
= cosh x sin y
∂y
∂x
which gives g 0 (y) = 0 and g(y) = A where A is a constant. The function f is
f = u + i v = sinh x sin y − i cosh x cos y + iA.
c M. K. Warby
20161117171541 MA3614 Complex variable methods and applications
4
Ex. 5 — This was in the class test in November 2013 and was worth 30 marks.
Let z = x + iy, x, y ∈ R, and let f (z) = u(x, y) + iv(x, y) be a function which is defined
on C with u and v denoting respectively the real and imaginary parts.
State the Cauchy Riemann equations.
Determine if the following functions are analytic in C and if a function is analytic express
it in terms of z alone.
1.
f (x + iy) = y.
2.
f (x + iy) = 2 + y 3 − 3x2 y + 2x + i(x3 − 3xy 2 + 2y).
Answer (Ex. 5) —
The Cauchy Riemann equations are
∂v
∂u
=
,
∂x
∂y
∂v
∂u
=− .
∂x
∂y
5 marks
1.
u = y, v = 0.
∂v
∂u
= 1,
= 0.
∂y
∂x
The Cauchy Riemann equation involving these two partial derivatives is not satisfied
and hence the function is not analytic.
2.
u = 2 + y 3 − 3x2 y + 2x and v = x3 − 3xy 2 + 2y
∂v
∂u
∂v
∂u
= −6xy + 2,
= −6xy + 2,
= 3y 2 − 3x2 and
= 3x2 − 3y 2 .
∂x
∂y
∂y
∂x
The Cauchy Riemann equations are satisfied at all points and thus the function is
analytic everywhere.
To express in terms of z note that f (0) = 2 and
∂f
= (−6xy + 2) + i(3x2 − 3y 2 ),
∂x
∂ 2f
00
= (−6y) + i(6x),
f (z) =
∂x2
∂ 3f
f 000 (z) =
= 6i.
∂x3
f 0 (z) =
Thus
f (z) = f (0) + f 0 (0)z +
f 00 (0) 2 f 000 (0) 3
z +
z = 2 + 2z + iz 3 .
2
6
25 marks
c M. K. Warby
20161117171541 MA3614 Complex variable methods and applications
5
Ex. 6 — This was in the class test in November 2013 and was worth 15 marks.
Show that the following function u is a harmonic function and determine the harmonic
conjugate v which satisfies v(0, 0) = 2.
u(x, y) = x + sinh x cos y,
x, y ∈ R,
(Here sinh x = (ex − e−x )/2 denotes the usual hyperbolic function.)
Answer (Ex. 6) —
The function u is harmonic if ∇2 u = 0. For the partial derivatives we have
∂u
= − sinh x sin y
∂y
∂u
= 1 + cosh x cos y,
∂x
and
∂ 2u
= sinh x cos y,
∂x2
∂ 2u
= − sinh x cos y.
∂y 2
This shows that ∇2 u = 0.
To determine v we use the Cauchy Riemann equations.
∂u
∂v
=−
= sinh x sin y
∂x
∂y
and hence partially integrating with respect to x gives
v(x, y) = cosh x sin y + g(y)
for some function g(y). Using the other Cauchy Riemann equation gives
∂v
∂u
= cosh x cos y + g 0 (y) =
= 1 + cosh x cos y.
∂y
∂x
Thus g 0 (y) = 1, which gives g(y) = y + A where A is a constant, and
v(x, y) = cosh x sin y + y + A.
The requirement that v(0, 0) = 2 gives A = 2.
Ex. 7 — This was question 1 of the 2016 MA3614 paper.
1. Let z = x + iy, with x, y ∈ R, and let f (z) = u(x, y) + iv(x, y) denote a function
defined in the complex plane C with u and v being real-valued functions which have
continuous partial derivatives of all orders.
State the Cauchy Riemann equations for the function f to be analytic. Give expressions for f 0 (z) in terms of only partial derivatives of u and also in terms of only
partial derivatives of v.
[3 marks]
c M. K. Warby
20161117171541 MA3614 Complex variable methods and applications
6
2. For each of the following functions determine if they are analytic in the complex
plane, giving reasons for your answer in each case.
(a)
f1 (z) = (2x2 − 2y 2 + 6xy + 4x + 6y) + i(−3x2 + 3y 2 + 4xy + 4y − 6x).
(b)
∂φ ∂φ
+
,
∂x ∂y
where φ(x, y) denotes any function which is harmonic at all points (x, y).
(c)
f2 (z) = i
f3 (z) = ex ((x cos(y) − y sin(y) + cos(y)) + i(x sin(y) + y cos(y) + sin(y))) .
(d)
f4 (z) = ex (cos(y) − i sin(y)) .
[11 marks]
3. Show that the function
u(x, y) = x3 y − xy 3
is harmonic and determine the harmonic conjugate v(x, y) satisfying v(0, 0) = 2.
Express u + iv in terms of z only.
[6 marks]
Answer (Ex. 7) —
1. The Cauchy Riemann equations are
∂u
∂v
=
,
∂x
∂y
∂u
∂v
=− .
∂y
∂x
By considering only the x direction the derivative f 0 (z) is given by
f 0 (z) =
∂u
∂v
+i .
∂x
∂x
By using the Cauchy Riemann equations we hence have
∂u
∂u
−i ,
∂x
∂y
∂v
∂v
=
+i
∂y
∂x
f 0 (z) =
which are relations only involving u or only involving v.
3 marks
c M. K. Warby
20161117171541 MA3614 Complex variable methods and applications
7
2. (a)
u = 2x2 − 2y 2 + 6xy + 4x + 6y
and v = −3x2 + 3y 2 + 4xy + 4y − 6x.
∂u
= 4x + 6y + 4,
∂x
and
∂u
= −4y + 6x + 6,
∂y
∂v
= 6y + 4x + 4
∂y
∂v
= −6x + 4y − 6.
∂x
Both Cauchy Riemann equations are satisfied everywhere and thus f1 (z) is analytic everywhere.
3 marks
(b) In this case
∂φ
∂φ
and v =
.
∂y
∂x
As mixed partial derivatives can be done in any order
u=
∂u
∂v
∂ 2φ
∂ 2φ
=
=
=
.
∂x
∂y
∂x∂y
∂y∂x
For the other Cauchy Riemann equation
∂u ∂v
∂ 2φ ∂ 2φ
+
= 2 + 2 =0
∂y ∂x
∂y
∂x
as φ is harmonic. Both Cauchy Riemann equations are satisfied and thus f2 (z)
is analytic everywhere.
3 marks
(c)
u = ex (x cos(y) − y sin(y) + cos(y)),
∂u
=
∂x
=
∂v
=
∂y
∂u
=
∂y
=
∂v
=
∂x
v = ex (x sin(y) + y cos(y) + sin(y)).
ex (x cos(y) − y sin(y) + cos(y) + cos(y))
ex (x cos(y) − y sin(y) + 2 cos(y)).
ex (x cos(y) − y sin(y) + cos(y) + cos(y)) =
∂u
.
∂x
ex (−x sin(y) − y cos(y) − sin(y) − sin(y))
ex (−x sin(y) − y cos(y) − 2 sin(y)).
ex (x sin(y) + y cos(y) + sin(y) + sin(y)) = −
∂v
.
∂x
Both Cauchy Riemann equations are satisfied and thus f2 (z) is analytic everywhere.
3 marks
c M. K. Warby
20161117171541 MA3614 Complex variable methods and applications
8
(d)
u = ex cos(y),
v = −ex sin(y).
∂v
= −ex cos(y).
∂y
∂u
= ex cos(y),
∂x
∂u
= −ex sin(y),
∂y
∂v
= −ex sin(y).
∂x
There is no value of y for which cos(y) = 0 and sin(y) = 0 and thus the Cauchy
Riemann equations are not satisfied at any point and hence f4 is not analytic at
any point.
2 marks
3.
∂ 2u
= 6xy
∂x2
∂u
= 3x2 y − y 3 ,
∂x
and
∂u
= x3 − 3xy 2 ,
∂y
∂ 2u
= −6xy.
∂y 2
Hence ∇2 u = 0. The harmonic conjugate v is such that
∂v
∂u
=−
= 3xy 2 − x3 .
∂x
∂y
Integrating partially with respect to x gives
2
x
x4
v=3
+ g(y)
y2 −
2
4
for any function g(y). Using the other Cauchy Riemann equation gives
∂v
∂u
= 3x2 y + g 0 (y) =
= 3x2 y − y 3 .
∂y
∂x
Thus g 0 (y) = −y 3 and
y4
g(y) = − + C
4
where C is a constant. As we require v(0, 0) = 2 this gives C = 2.
4 marks
3
3
f = u + iv = (x y − xy ) + i
6x2 y 2 − x4 − y 4
4
+2 .
c M. K. Warby
20161117171541 MA3614 Complex variable methods and applications
9
As by construction this is analytic it is a polynomial in z of degree 4. We use the
Maclaurin series to get the representation and get the derivatives by differentiating
in the x-direction.
f 0 (z)
f 00 (z)
f 000 (z)
f 0000 (z)
=
=
=
=
3x2 y − y 3 + i(3xy 2 − x3 ),
6xy + i(3y 2 − 3x2 ),
6y − i(6x),
−6i.
Thus
f 00 (0) 2 f 000 (0) 3 f 0000 (0) 4
f (z) = f (0) + f (0)z +
z +
z +
z =
2
6
24
0
−i
4
z 4 + 2i.
2 marks
Ex. 8 — This was question 1 of the 2015 MA3614 paper.
1. Let z = x + iy, with x, y ∈ R, and let f (z) = u(x, y) + iv(x, y) denote a function
defined in the complex plane C with u and v being real-valued functions which have
continuous partial derivatives of all orders. In the following h is a real number. Given
that the function f (z) is such that the following limits exist, express each limit in
terms of the partial derivatives of u and v as appropriate.
(a)
f (z + h) − f (z)
.
h→0
h
lim
(b)
f (z + hi) − f (z)
.
h→0
hi
lim
[6 marks]
State the Cauchy Riemann equations for an analytic function.
[1 mark]
2. Let z = x+iy with x, y ∈ R. For each of the following functions determine whether
or not it is analytic in the domain specified giving reasons for your answers in each
case.
(a)
f1 : C → C,
f1 (z) = x.
(b)
f2 : C → C,
f2 (z) = x4 − 6x2 y 2 + y 4 + 4ixy(x2 − y 2 ).
c M. K. Warby
20161117171541 MA3614 Complex variable methods and applications
10
(c)
f3 : C → C,
f3 (z) = cos(x) cosh(y) − i sin(x) sinh(y).
[7 marks]
3. As in part (1) let z = x + iy, with x, y ∈ R, and let f (z) = u(x, y) + iv(x, y).
Explain why when f is analytic the function u is harmonic. In your answer you may
assume that partial derivatives of u and v of all orders exist and are continuous.
[2 marks]
Let
u(x, y) = −2x − y + x3 − 6x2 y − 3xy 2 + 2y 3 .
Given that the function u is harmonic and the function
g(z) =
∂u
∂u
−i
∂x
∂y
is analytic, express g(z) in terms of z only.
[4 marks]
c M. K. Warby
20161117171541 MA3614 Complex variable methods and applications
11
Answer (Ex. 8) —
1.
(a) As h is real we have
f (z + h) − f (z)
h
u(x + h, y) − u(x, y)
v(x + h, y) − v(x, y)
=
+i
h
h
∂v
∂u
+ i , as h → 0,
→
∂x
∂x
with the partial derivatives evaluated at (x, y).
3 marks
(b)
f (z + ih) − f (z)
ih
u(x, y + h) − u(x, y)
v(x, y + h) − v(x, y)
=
+i
ih
ih
1
∂v
∂u
∂u
∂v
→
+i
− i , as h → 0,
=
i
∂y
∂y
∂y
∂y
with the partial derivatives evaluated at (x, y).
3 marks
The Cauchy Riemann equations in Cartesians are
∂u
∂v
=
,
∂x
∂y
∂u
∂v
=− .
∂y
∂x
1 mark
2.
(a) For f1 take u = x and v = 0.
∂u
= 1,
∂x
∂v
= 0.
∂y
As these are not the same the Cauchy Riemann equations are not satisfied and
thus the function is not analytic.
(b) For f2 take u = x4 − 6x2 y 2 + y 4 and v = 4xy(x2 − y 2 ).
∂u
= 4x3 − 12xy 2 ,
∂x
and
∂v
= 4x3 − 12xy 2
∂y
∂u
∂v
= −12x2 y + 4y 3 ,
= 12x2 y − 4y 3 .
∂y
∂x
Both Cauchy Riemann equations are satisfied and thus the function is analytic
in the entire complex plane.
c M. K. Warby
20161117171541 MA3614 Complex variable methods and applications
12
(c) For f3 take u = cos(x) cosh(y) and v = − sin(x) sinh(y).
∂u
∂v
=
= − sin(x) cosh(y)
∂x
∂y
and
∂u
∂v
=−
= cos(x) sinh(y).
∂y
∂x
Both Cauchy Riemann equations are satisfied and thus the function is analytic
in the entire complex plane.
7 marks
3. As all the derivatives are continuous the mixed partial derivatives can be done in
either order, i.e. in the case of v we have
∂ 2v
∂ ∂v
∂ ∂v
=
=
.
∂x∂y
∂x ∂y
∂y ∂x
Using both Cauchy Riemann equations then gives
∂ ∂u
∂
∂u
=
−
∂x ∂x
∂y
∂y
and the result that u is harmonic follows.
2 marks
g(z) = (−2 + 3x2 − 12xy − 3y 2 ) + i(1 + 6x2 + 6xy − 6y 2 ).
As g(z) is analytic we can get the complex derivative by differentiating in any direction and thus using the x-direction we have
g 0 (z) = (6x − 12y) + i(12x + 6y),
g 00 (z) = 6 + 12i.
Thus we get the representation in terms of z only by using the Taylor polynomial.
g 00 (0) 2
g(z) = g(0) + g (0)z +
z = (−2 + i) + (3 + 6i)z 2 .
2
0
4 marks
c M. K. Warby
20161117171541 MA3614 Complex variable methods and applications
13
Ex. 9 — This was question 1 of the 2014 MA3614 paper.
1. Let z = x + iy, with x, y ∈ R and let f (z) = u(x, y) + iv(x, y) denote a function
defined in the complex plane C with u(x, y) and v(x, y) being real valued.
Define what it means for f (z) to be analytic at a point z0 and state the Cauchy
Riemann equations.
[3 marks]
2. Let z = x+iy with x, y ∈ R. For each of the following functions determine whether
or not it is analytic in the domain specified giving reasons for your answers in each
case.
(a)
f1 : C → C,
f1 (z) = |z|2 .
(b)
f2 : C → C,
f2 (z) = cos(3x) cosh(2y) − i sin(3x) sinh(2y).
(c)
f3 : C → C,
f3 (z) =
∂φ
∂φ
−i ,
∂x
∂y
where φ denotes any harmonic function.
(d)
f4 : C → C,
1
f4 (z) = (x2 − y 2 ) + ixy.
2
[9 marks]
3. Show that the function
u(x, y) = 2x2 − 2y 2 − 2xy + 3x − y + 1
is a harmonic function and determine the harmonic conjugate v(x, y) which satisfies
v(0, 0) = 0. For this function v(x, y) express u + iv in terms of z only.
[5 marks]
4. Let f : C → C be an analytic function with f (z) = u(x, y) + iv(x, y) with, as
usual, x, y, u, v being real. Show that the function
g(z) = u(x, −y) − iv(x, −y)
is also analytic in the complex plane.
[3 marks]
c M. K. Warby
20161117171541 MA3614 Complex variable methods and applications
14
Answer (Ex. 9) —
1. f is complex differentiable at z0 if the following limit exists.
f (z0 + h) − f (z0 )
.
h→0
h
f is analytic at z0 if f is differentiable at all points in some neighbourhood of z0 .
The Cauchy Riemann equations are as follows.
f 0 (z0 ) = lim
∂u
∂v
=
∂x
∂y
and
∂u
∂v
=− .
∂y
∂x
3 marks
2.
(a)
f1 (z) = |z|2 = x2 + y 2 .
f1 = u + iv, with u = x2 + y 2 , v = 0.
∂u
∂v
∂v
∂u
= 2x,
= 2y,
=
= 0.
∂x
∂y
∂x
∂y
The Cauchy Riemann equations are only true at x = y = 0. f is not analytic at
z = 0 as the equations do not hold in a neighbourhood of 0.
f1 is not analytic at any other point as well.
(b)
f2 = u + iv, u = cos(3x) cosh(2y), v = − sin(3x) sinh(2y).
∂u
∂v
= −3 sin(3x) cosh(2y),
= −2 sin(3x) cosh(2y).
∂x
∂y
∂v
∂u
= 2 cos(3x) sinh(2y), −
= 3 cos(3x) sinh(2y).
∂y
∂x
The first Cauchy Riemann equation is only satisfied when sin(3x) = 0 and the
second Cauchy Riemann equation is only satisfied when cos(3x) = 0. It is not
possible for both to be true and thus the function f2 is not analytic anywhere.
(c)
∂φ
∂φ
f3 = u + iv, u =
, v=− ,
∂x
∂y
∂u ∂v
∂ 2φ ∂ 2φ
−
=
+
=0
∂x ∂y
∂x2 ∂y 2
because φ is harmonic.
∂u ∂v
∂ ∂φ
∂ ∂φ
+
=
−
=0
∂y ∂x
∂y ∂x ∂x ∂y
as a property of the partial differentiation operation.
The Cauchy Riemann equations are satisfied and f3 is analytic in C.
c M. K. Warby
20161117171541 MA3614 Complex variable methods and applications
15
(d)
1
u = (x2 − y 2 ), v = xy.
2
∂u
∂v
∂u
∂v
=
= x,
=−
= y.
∂x
∂y
∂y
∂x
The Cauchy Riemann equations are satisfied and f4 is analytic in C.
f4 = u + iv,
9 marks
3.
∂u
= 4x − 2y + 3,
∂x
∂u
= −4y − 2x − 1,
∂y
and
∂ 2u
∂ 2u
and hence ∇2 u = 0.
2 = 4,
2 = −4,
∂x
∂y
We use the Cauchy Riemann equations to get the harmonic conjugate v.
∂u
∂v
=−
= 4y + 2x + 1
∂x
∂y
and partially integrating with respect to x gives
v = 4xy + x2 + x + g(y)
where g(y) is any function. To satisfy the other Cauchy Riemann equation we need
∂u
∂v
= 4x + g 0 (y) =
= 4x − 2y + 3.
∂y
∂x
and hence
g 0 (y) = −2y + 3,
g(y) = −y 2 + 3y + c,
c =constant.
As v(0, 0) = 0 this gives c = 0 and
f = u + iv = 2(x2 − y 2 ) − 2xy + 3x − y + 1 + i(4xy + x2 − y 2 + x + 3y).
As this function is analytic it is the polynomial in z of degree 2 given by
f 00 (0) 2
f (z) = f (0) + f (0)z +
z .
2
0
f (0) = 1,
f 0 (z) =
∂u
∂v
+i
= 4x − 2y + 3 + i(4y + 2x + 1),
∂x
∂x
and
f 00 (z) =
∂ 2u
∂ 2v
+
i
= 4 + 2i,
∂x2
∂x2
f 0 (0) = 3 + i,
f 00 (0)/2 = 2 + i.
c M. K. Warby
20161117171541 MA3614 Complex variable methods and applications
16
Hence
f (z) = 1 + (3 + i)z + (2 + i)z 2 .
5 marks
4.
g = u1 + iv1
with
u1 (x, y) = u(x, −y),
v1 (x, y) = −v(x, −y).
The partial derivatives with respect to x are related by
∂u
∂u1
=
,
∂x
∂x
∂v1
∂v
=− .
∂x
∂x
The partial derivatives with respect to x are related by
∂u1
∂u
=− ,
∂y
∂y
∂v1
∂v
=
.
∂y
∂y
The Cauchy Riemann equations hence hold for g = u1 + iv1 and thus the function is
also analytic in C.
3 marks
Ex. 10 — This was question 1 of the 2013 MA3914 paper.
1. Let z = x + iy, with x, y ∈ R and let f (z) = u(x, y) + iv(x, y) denote a function
defined in the complex plane C with u(x, y) and v(x, y) being real valued.
Define what it means for f (z) to be analytic at a point z0 , state the Cauchy Riemann
equations and show that when f (z) is analytic the Cauchy Riemann equations are
satisfied.
[6 marks]
2. Let z = x+iy with x, y ∈ R. For each of the following functions determine whether
or not it is analytic in the domain specified giving reasons for your answers in each
case.
(a)
f1 : C → C,
f1 (z) = z.
(b)
f2 : C → C,
f2 (z) = 3x + y + x2 − y 2 + 4xy + i(3y − x + 2xy − 2x2 + 2y 2 ).
(c)
f3 : C → C,
f3 (z) = cos x cosh y − i sin x sinh y.
c M. K. Warby
20161117171541 MA3614 Complex variable methods and applications
17
(d)
f4 : D → C,
f4 (z) =
x − iy
,
x2 + y 2
where D = {z : |z| > 1}.
[9 marks]
3. Show that the function
u(x, y) = x3 − 3xy 2
is a harmonic function and determine the harmonic conjugate v(x, y) which satisfies
v(1, 0) = 1. For this function v(x, y) express u + iv in terms of z only.
[5 marks]
Answer (Ex. 10) —
1. f is complex differentiable at z0 if the following limit exists.
f (z0 + h) − f (z0 )
.
h→0
h
f 0 (z0 ) = lim
f is analytic at z0 if f is differentiable at all points in some neighbourhood of z0 .
The Cauchy Riemann equations are as follows.
∂u
∂v
=
∂x
∂y
and
∂u
∂v
=− .
∂y
∂x
If h is real then
f (z0 + h) − f (z0 )
=
h
u(x0 + h, y0 ) − u(x0 , y0 )
v(x0 + h, y0 ) + iv(x0 , y0 )
+i
h
h
∂u
∂v
→
+i
(x0 , y0 ) as h → 0.
∂x
∂x
If h is pure imaginary then h = ik with k being real and
u(x0 , y0 + k) − u(x0 , y0 )
v(x0 , y0 + k) + iv(x0 , y0 )
f (z0 + h) − f (z0 )
=
+i
h
ik
ik
1 ∂u ∂v
→
+
(x0 , y0 ) as h → 0.
i ∂y ∂y
As the limit is the same in both cases equating the real and imaginary parts gives
the Cauchy Riemann equations.
6 marks
c M. K. Warby
20161117171541 MA3614 Complex variable methods and applications
18
2.
(a) Let f1 = u + iv with u = x and v = −y.
∂u
= 1,
∂x
∂v
= −1.
∂y
As this Cauchy Riemann equation is not satisfied at any point it follows that
f1 (z) is not analytic at any point.
1 mark
(b) Let f2 = u + iv with
u = 3x + y + x2 − y 2 + 4xy,
v = 3y − x + 2xy − 2x2 + 2y 2 .
∂u
∂v
= 3 + 2x + 4y =
,
∂x
∂y
∂u
∂v
= 1 − 2y + 4x = − .
∂y
∂x
The Cauchy Riemann equations are satisfied at all points which implies that f2
is analytic in the domain specified.
3 marks
(c) Let f3 = u + iv with
u = cos x cosh y,
v = − sin x sinh y.
∂u
∂v
= − sin x cosh y =
,
∂x
∂y
∂u
∂v
= cos x sinh y = − .
∂y
∂x
The Cauchy Riemann equations are satisfied at all points which implies that f3
is analytic in the domain specified.
2 marks
(d) We note that
z
1
2 = .
z
|z|
For z 6= 0 this is a composition of the analytic function 1 and z and is analytic
with derivative
1
f4 (z) = − 2 .
z
in the domain specified.
3 marks
f4 (z) =
c M. K. Warby
20161117171541 MA3614 Complex variable methods and applications
3.
∂ 2u
= 6x,
∂x2
∂ 2u
= −6x
∂y 2
19
and thus ∇2 u = 0.
To get the harmonic conjugate v we have
∂v
∂u
=−
= 6xy,
∂x
∂y
giving v(x, y) = 3x2 y + g(y)
for some function g(y). Using the other Cauchy Riemann equation gives
∂v
∂u
= 3x2 + g 0 (y) =
= 3x2 − 3y 2 ,
∂y
∂x
giving g 0 (y) = −3y 2 ,
g(y) = −y 3 + C
where C is a constant. v(x, y) = 3x2 y − y 3 + C. Now v(1, 0) = C and hence C = 1.
f = u + iv = (x3 − 3xy 2 ) + i(3x2 y − y 3 ) + i = (x + iy)3 + i = z 3 + i.
5 marks
Ex. 11 — This was question 2b of the 2009 MA3914 paper and was worth 10 marks.
Show that the following function is harmonic in the whole complex plane except at the
origin
−cy
with c 6= 0 , c ∈ R.
u(x, y) = 2
x + y2
Write down the Cauchy-Riemann partial differential equations for the function f (z) =
u(x, y) + iv(x, y) and use them to determine the harmonic conjugate v(x, y) of u(x, y).
Can you write f (z) as a function of z alone? (hint: what is z z̄?) Is f (z) conformal on
the domain of its definition?
(Very little is done in MA3614 on conformal mapping at the moment but a function f has
this property if f (z) is analytic and f 0 (z) 6= 0.)
Answer (Ex. 11) —
The shortest way to answer this question is to note that
1
z
x − iy
= 2 = 2
z
|z|
x + y2
Thus u is the real part of
and
−i
−y − ix
= 2
,
z
x + y2
−ic
.
z
As −ic/z is analytic except at z = 0 it follows that u is a harmonic function.
The harmonic conjugate is
cx
v=− 2
+A
x + y2
c M. K. Warby
20161117171541 MA3614 Complex variable methods and applications
20
where A is a constant.
The question did ask for the Cauchy Riemann equations as well and these are
∂u
∂v
=
,
∂x
∂y
∂u
∂v
=− .
∂y
∂x
The expression for f in terms of z is
f (z) = −
ic
+ iA.
z
In this case
ic
.
z2
The derivative is non-zero on {z : z ∈ C, z 6= 0} and hence f (z) is conformal on this
domain.
f 0 (z) =