solution

Complex Analysis II, Michaelmas 2012. Sheet 3: Harmonic Fns and Conformal Maps page 1
Q.1 Check that the following functions are harmonic.
(i) cos x cosh y , (ii) 3x 2 y − y 3 − 2(x 2 − y 2 ) ,
(iii) log(x 2 + y 2 ) , (iv) y − y/(x 2 + y 2 ) .
S.1 See solution on separate file.
Q.2 Draw the region R and find real constants a and b such that:
(i) u(x, y) = ax+b solves the Dirichlet problem on the region R = {x+iy : 0 ≤ x ≤ 6} with u(x, y) = 10
on x = 0 and u(x, y) = 40 on x = 6.
(ii) u(x, y) = ay+b solves the Dirichlet problem on the region R = {x+iy : 1 ≤ y ≤ 8} with u(x, y) = 2
on y = 1 and u(x, y) = 100 on y = 8.
(iii) u(x, y) = axy + b solves the Dirichlet problem on the region R = {x + iy : x ≥ 0, y ≥ 0, xy ≤ 4}
with u(x, y) = 1 on the bounding half-lines and u(x, y) = 21 on the bounding hyperbola.
(iv) u(x, y) = axy+b solves the Dirichlet problem on the region R = {x+iy : x ≥ 0, y ≥ 0, 1 ≤ xy ≤ 2}
with u(x, y) = 0 on the left-lower bounding hyperbola and u(x, y) = 3 on the right-higher bounding
hyperbola.
(v) u(x, y) = a log(x 2 + y 2 ) + b solves the Dirichlet problem on the region R = {z : 1 ≤ |z| ≤ 5} with
u(x, y) = 0 on |z| = 1 and u(x, y) = 100 on |z| = 5.
(vi) u(x, y) = a log(x 2 + y 2 ) + b solves the Dirichlet problem on the region R = {z : 2 ≤ |z| ≤ e} with
u(x, y) = 10 on |z| = 2 and u(x, y) = 70 on |z| = e.
S.2 See solution on separate file
Q.3 Check that the functions in Question 1 are the real parts of the following holomorphic functions, and hence
find the corresponding harmonic conjugates of the harmonic functions in Question 1 (ie, write down the
corresponding imaginary parts).
(i) cos z (ii) −iz 3 − 2z 2 , (iii) 2 log z , (iv) −iz − (i/z) .
S.3 See solution on separate file
Q.4 Find the harmonic conjugates of the harmonic functions in Question 1 using the Cauchy-Riemann equations
as demonstrated in lectures.
S.4 See solution on separate file
Q.5 Find the harmonic conjugate of e x cos y and the corresponding holomorphic function.
S.5 Let u(x, y) = e x cos y; we look for v(x, y) satisfying the Cauchy-Riemann equations, i.e.
So we need
The first equation gives
∂u ∂v
=
∂x ∂y ,
∂v
∂y = ex cos y,
∂v
= −∂u
∂x ∂y
∂v
∂x = ex sin y
v = e x sin y + g(x)
∂v
∂x = ex sin y + g ′ (x)
and so the second equation gives e x sin y + g ′ (x) = e x sin y, hence g ′ (x) = 0 and therefore g(x) = C. So
v(x, y) = e x sin y + C.
And
e x cos y + ie x sin y = e x (cos y + i sin y) = e x e iy = e x+iy = e z
is the corresponding holomorphic function (or e z + iC).

Complex Analysis II, Michaelmas 2012. Sheet 3: Harmonic Fns and Conformal Maps page 2
Q.6 Find the harmonic conjugate of x 3 − 3xy 2 + 4x and the corresponding holomorphic function.
S.6 Let u(x, y) = x 3 − 3xy 2 + 4x; we look for v(x, y) satisfying the Cauchy-Riemann equations, i.e.
So we need
∂u ∂v
=
∂x ∂y ,
∂v
∂y = 3x2 − 3y 2 + 4,
∂v
= −∂u
∂x ∂y
∂v
= 6xy
∂x
The second equation gives v(x, y) = 3x 2 y + g(y) and then the first gives
so v(x, y) = 3x 2 y − y 3 + 4y + C.
We check that
3x 2 − 3y 2 + 4 = ∂v
∂y = 3x2 + g ′ (y)
g ′ (y) = −3y 2 + 4
g(y) = −y 3 + 4y + C
x 3 − 3xy 2 + i 3x 2 y − y 3 = x 3 + 3ix 2 y − 3xy 2 − iy 3 = x 3 + 3x 2 iy + 3x(iy) 2 + (iy) 3 = (x + iy) 3 = z 3 .
Thus 3x 2 y − y 3 + 4y is the harmonic conjugate of x 3 − 3xy 2 + 4x, and z 3 + 4z is the corresponding
holomorphic function. (Or 3x 2 y − y 3 + 4y + C and z 3 + 4z + iC.)
Q.7 Find the harmonic conjugate of cosh x cos y.
S.7 Let u(x, y) = cosh x cos y; we look for v(x, y) satisfying the Cauchy-Riemann equations, i.e.
So we need
∂v
∂y
∂u ∂v
=
∂x ∂y ,
∂u
= = sinh x cos y,
∂x
∂v
= −∂u
∂x ∂y
∂v
= −∂u = cosh x sin y
∂x ∂y
The first equation gives v(x, y) = sinh x sin y + g(x) and then the second gives
cosh x sin y = ∂v
∂x = cosh x sin y + g′ (x)
g ′ (x) = 0
g(x) = C
So v(x, y) = sinh x sin y + C. You may care to check that cosh z (or cosh z + iC) is the corresponding
holomorphic function.
Q.8 (a) Find a harmonic function defined on the region of the (u, v)-plane between u = 2 and u = 3 taking
the value 3 on the left-hand boundary and 5 on the right-hand.
(b) Using part (a), find a harmonic function defined on the region of the (x, y)-plane between y = x−2 √ 2
and y = x − 3 √ 2 taking the value 3 on the upper boundary and 5 on the lower.
S.8 (a) Let ϕ(u, v) be such a function. On the boundary of the region, the value of ϕ is independent of v, so
we may expect that, if w = u + iv then
ϕ(u, v) = Au + B = Re(Aw + B),

Complex Analysis II, Michaelmas 2012. Sheet 3: Harmonic Fns and Conformal Maps page 3
where A, B ∈ R. Putting in the boundary conditions we get
giving A = 2, B = −1 and hence
2A + B = 3
3A + B = 5
ϕ(u, v) = 2u − 1 = Re(2w − 1).
(b) If you draw the two given lines in the (x, y)-plane (see left-hand picture below), you see that rotating
by π/4 could be a good thing to try. So, let w = f(z) = u + iv = e iπ/4 (x + iy). Then
so that
(x + iy) = e −iπ/4 (u + iv) = 1
√ 2 (1 − i)(u + iv),
x = 1
√ 2 (u + v) , y = 1
√ 2 (−u + v) .
So the first line in the (x, y)-plane is mapped to the line −u + v = u + v − 4, ie to the line u = 2,
and the second line in the (x, y)-plane is similarly mapped to the line u = 3. A theorem discussed in
the module now tells us that ˜ϕ(x, y) = ϕf(x, y) is harmonic on the (x, y)-plane, and the construction
shows that ˜ϕ(x, y) has the correct boundary conditions. Explicitly,
1
˜ϕ(x, y) = ϕf(x, y) = ϕ √2 (1 + i)(x + iy) = √ 2(x − y) − 1.
[You could check directly that this is harmonic on the (x, y)-plane and takes the correct values on the
bounding lines.]
z = x + iy
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√ 2
−→
y = x − 3 √ 2 w = u + iv
π
i Figure 1: w = ze 4
u = 2
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u = 3
Q.9 (a) Find a harmonic function defined on the region of the (u, v)-plane between u = −1 and u = −3 taking
the value −5 on the left-hand boundary and 1 on the right-hand.
(b) Using part (a), find a harmonic function defined on the region of the (x, y)-plane between y = √ 3x+2
and y = √ 3x + 6 taking the value 1 on the lower boundary and −5 on the upper.
S.9 (a) Let ϕ(u, v) be such a function. On the boundary of the region the value of ϕ is independent of v, so
we may expect that, if w = u + iv, then
ϕ(x, y) = Au + B = Re(Aw + B),

Complex Analysis II, Michaelmas 2012. Sheet 3: Harmonic Fns and Conformal Maps page 4
where A, B ∈ R. Putting in the boundary conditions we get
giving A = 3, B = 4 and hence
−3A + B = −5
−A + B = 1
ϕ(u, v) = 3u + 4 = Re(3w + 4).
(b) If you draw the two given lines in the (x, y)-plane (see left-hand picture below), you see that rotating
by π/6 could be a good thing to try. So, let w = f(z) = u + iv = e iπ/6 (x + iy). Then
so that
(x + iy) = e −iπ/6 (u + iv) = 1
2 (√ 3 − i)(u + iv),
x = 1
2 (√ 3u + v) , y = 1
2 (−u + √ 3v) .
So the first line in the (x, y)-plane is mapped to the line −u + √ 3v = √ 3( √ 3u + v) + 4, ie to the
line u − 1, and the second line in the (x, y)-plane is similarly mapped to the line u = −3. A theorem
discussed in the module now tells us that ˜ϕ(x, y) = ϕf(x, y) is harmonic on the (x, y)-plane, and the
construction shows that ˜ϕ(x, y) has the correct boundary conditions. Explicitly,
1
˜ϕ(x, y) = ϕf(x, y) = ϕ
2 (√
3 + i)(x + iy)
= 3
2 (√ 3x − y) + 4.
[You could check directly that this is harmonic on the (x, y)-plane and takes the correct values on the
bounding lines.]
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y = √ 3x + 2
y = √ . . . .
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3x + 6 u = −3 . . . .
. . . . u = −1
−→
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z = x + iy . . . .
. . . . w = u + iv
π
i Figure 2: w = ze 6
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Q.10 Let f(z) = 1 + i √ 3 z. Describe f as a rotation followed by a dilation. Hence draw the image under f of
the unit circle |z| = 1 and of the line x = y.
S.10 We have
f(z) =
1 + i √
√
1 3
3 z = 2 + i
2 2
z = 2 cos π
3
π
π
i
+ i sin z = 2e 3 z
3
Thus f is rotation about the origin through an angle π
3 followed by dilation by 2. The unit circle is ampped
to the
Q.11 Let f(z) = (1 − i) z. Describe f as a rotation followed by a dilation. Hence draw the image under f of the
unit circle |z| = 1 and of the line x = y.

Complex Analysis II, Michaelmas 2012. Sheet 3: Harmonic Fns and Conformal Maps page 5
S.11 We have
f(z) = (1 − i) z = √
1
2 √2 − i 1
√ z =
2
√
2 cos
− π
4
+ sin
Thus f is rotation about the origin through an angle − π
4 followed by dilation by √ 2.
− π
4
z = √ π
−i
2e 4 z
Q.12 Let f(z) = √ 2 (1 − i) z + 2 − 4i. Describe f as a rotation followed by a dilation followed by a translation.
Hence draw the image under f of the unit circle |z| = 1 and of the line x = y.
S.12 We have
f(z) = √ 2 (1 − i) z + 2 − 4i
iπ
−
= 2e 4 z + 2 − 4i .
Thus f is rotation about the origin through an angle − π
4 followed by dilation by 2, followed by translation
through 2 − 4i.
Q.13 Let f(z) = 2 1 − i √ 3 z +6+5i. Describe f as a rotation followed by a dilation followed by a translation.
Hence draw the image under f of the unit circle |z| = 1 and of the line x = y.
S.13 Here, f(z) = 4e i5π
3 z + 6 + 5i. Thus f is rotation about the origin through an angle 5π
3
by 4, followed by translation through 6 + 5i.
followed by dilation
Q.14 Let f(z) = −1 − i √ 3 z+3−2i. Describe f as a rotation followed by a dilation followed by a translation.
Hence draw the image under f of the unit circle |z| = 1 and of the line x = y.
S.14 Here, f(z) = 2e i4π
3 z + 3 − 2i. Thus f is rotation about the origin through an angle 4π
3
by 2, followed by translation through 3 − 2i.
followed by dilation
Q.15 By noting that az + b = a(z + (b/a)) show that if f(z) = az + b, with a = 0, then f may be decribed as
a translation followed by a rotation followed by a dilation. Describe f(z) = √ 2 (1 − i) z + 2 − 4i in this
way.
S.15 We have
where
Thus
iπ
−
2e 4 w = 2 − 4i
w = e iπ
4 (1 − 2i) =
f(z) = √ 2 (1 − i) z + 2 − 4i
iπ
iπ
− −
= 2e 4 z + 2 − 4i = 2e 4 (z + w)
1
√2 + i
√ (1 − 2i) =
2
1
3 − i
√ (1 + i) (1 − 2i) = √
2 2
iπ
−
f(z) = 2e 4 z +
3 − i
√
2
So f is translation through 3−i
√ 2 , followed by rotation about the origin through an angle − π
4
dilation by 2.
Q.16 In what subset of the complex plane is 2z 3 + 3z 2 conformal?
followed by
S.16 Let f(z) = 2z 3 + 3z 2 ; then f ′ (z) = 6z 2 + 6z = 6z(z + 1). Thus for z = 0, −1, f ′ (z) = 0 and so f is
conformal in C \ {0, −1} but not at 0 or −1.

Complex Analysis II, Michaelmas 2012. Sheet 3: Harmonic Fns and Conformal Maps page 6
Q.17 In what subset of the complex plane is 2z 3 − 6z 2 + 6z conformal?
S.17 Let f(z) = 2z 3 − 6z 2 + 6z; then f ′ (z) = 6z 2 − 12z + 6 = 6(z − 1) 2 . Thus for z = 1, f ′ (z) = 0 and so f
is conformal in C \ {1}, but not at z = 1 because f ′ (1) = 0.
Q.18 In what subset of the complex plane is 2z 3 − 3 (1 + i) z 2 + 6iz conformal?
S.18 Let f(z) = 2z 3 − 3 (1 + i) z 2 + 6iz; then
f ′ (z) = 6z 2 − 6 (1 + i) z + 6i
= 6 z 2 − (1 + i) z + i
= 6 (z − 1) (z − i) .
Thus for z = 1, i, we see that f ′ (z) = 0. Hence f is conformal in C \ {1, i}, but not for z ∈ {1, i} because
f ′ (z) = 0 there.
Q.19 In what subset of the complex plane is z 3 + 3
2 (1 − 2i) z2 − 6iz + 4 conformal?
S.19 Let f(z) = z 3 + 3
2 (1 − 2i) z2 − 6iz + 4; then
f ′ (z) = 3 z 2 + (1 − 2i)z − 2i
= 3 (z − 2i) (z + 1)
Thus for z = −1, 2i, we see that f ′ (z) = 0. Hence f is conformal in C \ {−1, 2i}, but not for z ∈ {−1, 2i}
because f ′ (z) = 0 there.
Q.20 In what subset of the complex plane is sinh z conformal?
S.20 Let f(z) = sinh z, so that f ′ (z) = cosh z. Certainly f(z) is holomorphic on C and
where n ∈ Z. Let
f ′ (z) = 0
⇐⇒ e z + e −z = 0
S =
Thus f(z) is conformal on C − S but not on S.
⇐⇒ e 2z = −1
⇐⇒ e 2z = e i(π+2nπ)
π
⇐⇒ z = i + nπ
2
π
i
2
+ nπ : n ∈ Z
Q.21 At which points in C are the following maps conformal?
(a) x + iy ↦−→ x − 2yi (b) z ↦−→ z 2 + 1
In both cases give an example of two curves through the origin such that the angle between them is not
preserved by f(z).
S.21 (a) f(z) = x − 2iy is not holomorphic anywhere, since x and −2y do not satisfy the Cauchy-Riemman
equations anywhere. Therefore it is not conformal at any point of C. (You can also write f(z) = 1
2 (−z +3¯z)
to show that it is not holomorphic.)
(b) f(z) = z 2 + 1 is holomorphic but f ′ (z) = 2z = 0 only when z = 0. So f is conformal in C − {0}
Consider the line segments γ1(t) = te iθ1 , γ2(t) = te iθ2 , t ∈ [0, 1], θ1 = θ2 and their images.
Q.22 Let f(z) = z 2 + 2z. Show that f is conformal at z = i and describe the effect on the tangent vectors of
curves passing through the point.

Complex Analysis II, Michaelmas 2012. Sheet 3: Harmonic Fns and Conformal Maps page 7
S.22 First note
f ′ (z) = 2z + 2
f ′ (i) = 2 + 2i = 0
so that f(z) is conformal at z = i. Now arg f ′ (i) = π
π
, so the tangent vectors at z = i are rotated by
4 4 .
The dilation factor is
f ′ (i) = √ 4 + 4 = 2 √ 2.
Q.23 Let f(z) = 2z 3 + 3z 2 . Show that f is conformal at z = i and describe the effect on the tangent vectors of
curves passing through the point.
S.23 First note
f ′ (z) = 6z 2 + 6z
f ′ (i) = −6 + 6i = 0
so that f(z) is conformal at z = i. Now arg f ′ (i) = 3π
3π
, so the tangent vectors at z = i are rotated by
4 4 .
The dilation factor is
f ′ (i) = 62 + 62 = 6 √ 2.
Q.24 Let f(z) = z 4 + 4z. Show that f is conformal at z = i and describe the effect on the tangent vectors of
curves passing through the point.
S.24 First note
f ′ (z) = 4z 3 + 4
f ′ (i) = 4 − 4i = 4(1 − i)
so f(z) is conformal at z = i.
Now arg f ′ (i) = − π
or
4
7π
, so that the tangent vectors at z = i are rotated by −
4
π
4
dilation factor is √
f ′
(i) = 4 2.
= 0
or 7π
. The
4
Q.25 Is the following true or false? If f, g are conformal at a point z0 then f + g is conformal at z0. Give a proof
or a counter-example.
S.25 False. For let f(z) = z + z 2 , g(z) = −z and z0 = 0. Then f ′ (0) = 1, g ′ (0) = −1 so both f and g are
conformal at z = 0. But f(z) + g(z) = z 2 , which is not conformal at z = 0 because its derivative is 0
there.