COMPLEX ANALYSIS: SOLUTIONS 3
1. Prove that the following functions f : C → C are not analytic
(i) f (z) = z,
(ii) f (z) = =(z),
(iii) f (z) = <(z).
Solution: (i) We have for all z
f (z + h) − f (z)
z+h−z
h
=
= =
h
h
h
(
1
−1
if h ∈ R
if h ∈ iR.
Therefore f cannot be analytic since the derivative must be the same no matter
how h approaches zero. Alternatively, one could check that the Cauchy–Riemann
equations are not satisfied.
(ii) As a function of x, y we have f (x + iy) = y. So writing f = u + iv gives
u(x, y) = y and v(x, y) = 0. We check if these functions satisfy the Cauchy–Riemann
equations. We have
ux (x, y) = 0 = vy (x, y)
but
uy (x, y) = 1 6= 0 = −vx (x, y)
so f cannot be analytic.
(iii) This follows similarly to (ii). This time
ux (x, y) = 1 6= 0 = vy (x, y)
and so f is not analytic.
2. Let f : C → C be analytic. Prove that f (z) is analytic.
Solution: Since f is analytic its real and imaginary parts, u and v, have continuous
partial derivatives which satisfy the equations
ux (x, y) = vy (x, y),
uy (x, y) = −vx (x, y).
Writing g(z) = f (z) = µ(x, y) + iν(x, y) we see that
µ(x, y) = u(x, −y),
ν(x, y) = −v(x, −y).
1
COMPLEX ANALYSIS: SOLUTIONS 3
2
The functions µ, ν have continuous partial derivatives and satisfy the equations
µx (x, y) = ux (x, −y) = vy (x, −y) =
∂
(−v(x, −y)) = νy (x, y)
∂y
and
µy (x, y) =
∂
(u(x, −y)) = −uy (x, −y) = vx (x, −y) = −νx (x, y).
∂y
Hence, g is analytic.
3. Show that the following functions u : R2 → R are harmonic. In each case find a
harmonic conjugate v and hence construct an analytic function f : C → C, f = u+iv,
writing it as a function of z.
(i) u(x, y) = x2 − y 2 ,
(ii) u(x, y) = sinh x sin y.
Solution: (i) uxx (x, y) = 0 and uyy (x, y) = 0 so ∇u = 0 and u is harmonic. A
harmonic conjugate v of u must satisfy the Cauchy-Riemann equations, so we require
vx (x, y) = −uy (x, y) = 2y
and
vy (x, y) = ux (x, y) = 2x.
The first of these implies that v(x, y) = 2xy + f (y) for some function f , and the
second implies v(x, y) = 2xy + g(x) for some function g. Hence, v(x, y) = 2xy + c
and we see that
f (z) = u(x, y) + iv(x, y) = x2 − y 2 + 2ixy + C = (x + iy)2 + C = z 2 + C.
(ii) Here, ux x(x, y) = sinh x sin y and uy y(x, y) = − sinh x sin y so ∇u = 0 and u
is harmonic. The Cauchy–Riemann equations read
vx (x, y) = − sinh x cos y
and
vy (x, y) = cosh x sin y.
COMPLEX ANALYSIS: SOLUTIONS 3
3
Together these imply that v(x, y) = − cosh x cos y + c. Hence
f (z) =u(x, y) + iv(x, y)
= sinh x sin y − i cosh x cos y + C
= − i sin(ix) sin y − i cos(ix) cos y + C
=i sin(ix) sin(−y) − i cos(ix) cos(−y) + C
=i cos(ix − y) + C
=i cos(iz) + C.
4. Find a mapping which maps the region {reiθ ∈ C : r > 0, −π/4 6 θ 6 π/4}
onto the closed unit disk {z ∈ C : |z| 6 1}. At what points is this mapping not
conformal?
Solution: We first map the region A = {reiθ ∈ C : r > 0, −π/4 6 θ 6 π/4} on to
the half plane <(z) > 0. Writing the half plane as the set of reiθ ∈ C with r > 0
and −π/2 6 θ 6 π/2, we see that the map z 7→ z 2 does the job since θ 7→ 2θ. We
then compose this with the map given in the lectures which took the half plane to
the unit disk. This was given by z 7→ i(1 − z)/(1 + z). Hence, the map
f (z) := i
1 − z2
1 + z2
has the desired property.
Now, f is the composite of analytic functions hence is analytic on A. Thus, f is
conformal at all points z0 for which f 0 (z0 ) 6= 0. We have
f 0 (z) = −i
4z
(1 + z 2 )2
and this equals zero only when z = 0.
5. Let S(z) = az+b
with a, b, c, d ∈ R, ad−bc 6= 0. Give a condition on the coefficients
cz+d
which guarantees that S preserves the upper half plane H = {z ∈ C : =(z) > 0}.
Solution: We let z = x + iy with y > 0 and look at =S(z). If c = 0 then, since the
coefficients are real,
a
a a
b
=S(z) = = z +
= = z = y.
d
d
d
d
This is nonnegative iff a, d have the same sign – a condition which can be expressed
as ad > 0.
COMPLEX ANALYSIS: SOLUTIONS 3
4
If c 6= 0, then we may use the expression for S(z) involving only one z, as given in
the lectures. Then,
a
ad − bc −
=S(z) ==
c c(cz + d)
ad − bc =0 − =
c(cz + d)
ad − bc
=−=
(cz
+
d)
c|cz + d|2
ad − bc z +0
=−=
|cz + d|2
ad − bc
=
y.
|cz + d|2
This last quantity is nonnegative iff ad − bc > 0. Hence, all in all, this last condition
is the required condition whether c = 0 or not.
6. Let fn : G → C be a sequence of continuous functions which converges uniformly
to f : G → C. Prove that
Z
Z
lim
fn (z)dz = f (z)dz
n→∞
γ
γ
for all piecewise differentiable curves γ : [a, b] → G. What does this mean in the case
of uniformly convergent series?
Solution: By the estimation lemma we have
Z
Z
Z
[f
(z)
−
f
(z)]dz
f
(z)dz
=
f
(z)dz
−
6 L(γ) max |fn (z) − f (z)|
n
n
γ
z∈γ
γ
γ
where L(γ) is the length of the curve. Since fn → f uniformly, maxz∈γ |fn (z) −
f (z)| → 0 as P
n → ∞ and the
Presult follows.
If fn (z) = nj=1 uj (z) → ∞
j=1 uj (z) = f (z) uniformly, then the result says
Z X
Z X
∞
n
n Z
∞ Z
X
X
uj (z)dz = lim
uj (z)dz = lim
uj (z)dz =
uj (z)dz,
γ j=1
n→∞
γ j=1
n→∞
j=1
γ
n=1
γ
i.e. if a series is uniformly convergent then we can interchange the integral with the
sum.
7. Let γR be the semicircular arc traced counterclockwise from R to −R, R > 0.
Find the limit as R → ∞ of the following quantities
COMPLEX ANALYSIS: SOLUTIONS 3
(i)
(ii)
R
RγR
γR
5
dz
z2
eiz
dz
z
Solution: (i) We apply the estimation lemma. The curve is of length πR and hence
Z dz 1
π
→ 0.
6 πR max 2 =
2
z∈γ
R
R |z|
γR z
(ii) Here, the integrand gets small as R gets large so we can expect the limit to be
zero also. However, an immediate application of the estimation lemma gives
Z eiz eiz 1
dz 6 πR max = πR · max |eiz | = π
z∈γ
z
R z∈γR
R
γR z
so we need to do something more subtle. There’s a few ways to do this but each way
needs some justification/explanation/waffle.
• The first method involves splitting the curve up. This should work since the
majority of mass is concentrated in a small set near the real axis. If we go further
away from the real axis we have exponential decay: |eiz | = |eix−y | = e−y . The
problem with the estimation lemma is that we lose this information. Thus, we split
the curve up into several parts. On one part we will keep the exponential decay.
The other parts will be of relatively short length which will allow for the more brutal
estimate |eiz | 6 1.
Write γR = γR,1 + γR,2 + γR,3 where the components are as follows. Let γR,1√be
the curve obtained by starting at R and tracing a continuous curve of length R
along γR . Let γR,3 be the same but at the other endpoint −R and let γR,2 be the
remaining curve.
On γR,1 we have
eiz √
Z
1
1
eiz √
dz 6 R max = R · = √ → 0
z∈γ
z
R
R,1
R
γR,1 z
and similarly for γR,3 .
√
√
The curve γR,2 is the set of z =√x + iy = Reiθ with 1/ R 6 θ 6 π − 1/ R.
We have y = R sin θ > R(θ/2) > R/2 where we have used sin θ > θ/2 valid for
θ 6 π/2. By symmetry,√ this inequality holds for all y under consideration. Thus, on
γR,2 we have |eiz | 6 e− R/2 . Since γR,2 is of length 6 πR we have
Z
eiz √
eiz dz 6 πR max = πe− R/2 → 0.
z∈γR,2 z
γR,2 z
COMPLEX ANALYSIS: SOLUTIONS 3
6
Thus,
Z
γR
Z
3 Z
3
eiz X eiz eiz X
dz = dz 6
dz → 0.
z
z
γR,j z
j=1 γR,j
j=1
This should be compared with the proof of Jordans Lemma. Where are the savings
made here?
• An alternative way, which also takes advantage of the exponential decay, is to
deform the curve to a rectangle ΓR with vertices R, R + iR, −R + iR, −R. On the
area between the semicircle and the rectangle the integrand is analytic, hence the
deformation is valid by Cauchy’s theorem. Thus,
Z R+iR Z −R+iR Z −R iz
Z
Z
e
eiz
eiz
+
+
dz =
dz =
dz
z
R
R+iR
−R+iR
γR z
ΓR z
where in these last integrals it is understood that the curve of integration is the
straight line joining the lower limit of integration to the upper limit.
To estimate these integrals we write them in parametrised form. For the horizontal
segment we have
Z
Z −R+iR eiz Z −R ei(t+iR) Z R eit−R e−R R
dz = dt 6
1 · dt = 2e−R .
dt 6
z
t
+
iR
t
+
iR
R
R+iR
R
−R
−R
For the vertical segments;
Z
Z R+iR eiz Z R ei(R+it) Z R eiR−t 1
1 R −t
e dt < .
dz = i
dt 6
dt 6
z
R + it
R + it
R 0
R
0
0
R
Both of these quantities tend to zero.
• Yet another approach involves integrating by parts to take advantage of the
oscillations of eiz . This is a standard trick for oscillating integrands (see here). Let’s
first check how integration by parts works for line integrals. If g has primitive G and
γ has endpoints α, β then
Z
Z
f (z)g(z)dz =
γ
b
f (γ(t))g(γ(t))γ 0 (t)dt
a
t=b Z b
= f (γ(t))G(γ(t)) −
f 0 (γ(t))γ 0 (t)G(γ(t))dt
t=a
a
z=β Z
= f (z)G(z)
− f 0 (z)G(z)dz,
z=α
as expected.
γ
COMPLEX ANALYSIS: SOLUTIONS 3
7
We integrate the thing that oscillates; eiz , and differentiate the thing that decays;
1/z. This gives
Z
Z
eiz z=−R 1
eiz
eiz
dz =
+
dz.
iz z=R
i γR z 2
γR z
We can now do some estimates. The absolute value of the first term on the right is
6 2/R and the absolute value of the integral is 6 π/R by the estimation lemma.