Complex Analysis: Homework 1
Deadline: Tuesday 20 November 2018 at 13:30
- Do not forget to very clearly write your full name, student number, and tutorial group at the top of
the first page of your submission. If you use more than one paper sheets they should be stapled together.
Submissions not satisfying these requirements will not be accepted.
- Solutions should be complete and clearly present your reasoning.
- 10 points are “free” if you hand-in the assignment. There are 4 questions and the maximum number of
points is 100. The homework grade is the total number of points divided by 10.
- Hand-in your submissions at the lecture on Tuesday. Late submissions will not be accepted.
- If there are practical reasons that do not permit you to hand-in your submission at the day of the deadline
here are some alternatives in order of preference:
* Hand-in your submission at the Monday evening lecture before the deadline.
* Let a fellow student hand-in your submission.
* Send (before the deadline) a scan of your submission by e-mail as a PDF file to the teaching assistants
of your group with cc to K.Efstathiou@rug.nl and hand-in the hardcopy submission at the next
lecture. The e-mail addresses of the teaching assistants for each group and some tips for how to scan
your submission can be found on Nestor. Submissions for which the e-mail is sent after the deadline
or for which hardcopy has not been submitted will not receive a grade.
Question 1 (30 points)
Check if the following functions satisfy the Cauchy-Riemann equations.
2
(a) (15 points) w = f (z) = ez .
Solution
Let z = x + iy. Then
2
w = f (z) = e(x+iy)
2 −y 2
= ex
2 −y 2
= ex
e2xyi
cos(2xy) + iex
2 −y 2
sin(2xy).
So w = u(x, y) + iv(x, y) with
u(x, y) = ex
2 −y 2
cos(2xy) and v(x, y) = ex
2 −y 2
sin(2xy).
Note that
∂u
∂x
∂v
∂y
∂u
∂y
∂v
∂x
= 2xex
2 −y 2
cos(2xy) − 2yex
2 −y 2
2x cos(2xy) − 2y sin(2xy) ,
2 −y 2
− 2y sin(2xy) + 2x cos(2xy) ,
2 −y 2
sin(2xy) = ex
2 −y 2
− 2y cos(2xy) − 2x sin(2xy) ,
sin(2xy) + 2xex
= −2yex
2 −y 2
cos(2xy) − 2xex
sin(2xy) + 2yex
cos(2xy) = ex
2 −y 2
2 −y 2
2 −y 2
2 −y 2
= −2yex
= 2xex
sin(2xy) = ex
2 −y 2
cos(2xy) = ex
2 −y 2
2x sin(2xy) + 2y cos(2xy) ,
so indeed the Cauchy-Riemann equations,
∂u
∂v
∂u
∂v
=
and
=− ,
∂x
∂y
∂y
∂x
hold.
2
Thus the function w = ez satisfies the Cauchy-Riemann equations.
Page 1 of 4
(b) (15 points) w = f (z) =
z
.
z + z̄
Solution
Let z = x + iy. Then
w = f (z) =
1
y
x + iy
= +i ,
x + iy + x − iy
2
2x
so w = u(x, y) + iv(x, y) with
u(x, y) =
1
y
and v(x, y) =
.
2
2x
Here we already see that
∂u
1
∂v
= 0 6=
=
,
∂x
2x
∂y
and hence w =
z
z+z̄
does not satisfy the Cauchy-Riemann equations.
Question 2 (20 points)
Show that sin x sinh y + i cos x cosh y satisfies the Cauchy-Riemann equations and find an expression for this function only in terms of z (where “only” here means “no z̄”).
Solution
Let w = sin(x) sinh(y) + i cos(x) cosh(y), i.e., w = u(x, y) + iv(x, y) with u(x, y) = sin(x) sinh(y)
and v(x, y) = cos(x) cosh(y). We see that
∂u
∂v
= cos(x) sinh(y) =
,
∂x
∂y
and
∂u
∂v
= sin(x) cosh(y) = −(− sin(x) cosh(y)) = − .
∂y
∂x
Hence w = sin(x) sinh(y) + i cos(x) cosh(y) satisfies the Cauchy-Riemann equations.
Using the identities
eix − e−ix
,
2i
eix + e−ix
cos(x) =
,
2
ey − e−y
sinh(y) =
,
2
ey + e−y
cosh(y) =
,
2
sin(x) =
we can rewrite w as follows:
w = sin(x) sinh(y) + i cos(x) cosh(y)
=
eix − e−ix ey − e−y +i
eix + e−ix ey + e−y 2i
2
2
2
1 ix+y
ix−y
−ix+y
−ix−y
ix+y
=
e
−e
−e
+e
−e
− eix−y − eix+y − e−ix−y
4i
eix−y + e−ix+y =i
.
2
Page 2 of 4
Now let z = x + iy. Then, iz = ix − y, so
eix−y + e−ix+y w=i
eiz + e−iz =i
2
2
= i cos(z).
In conclusion, by letting z = x + iy, we can write w = f (z) = i cos(z).
Question 3 (20 points)
Show that the function f (z) = x2 +y +i(2y −x) is differentiable on the line x = 1 but is nowhere
analytic.
Solution
For the function to be differentiable it must satisfy the Cauchy-Riemann equations and the first
partial derivatives must be continuous. Letting u = x2 + y and v = 2y − x we find
and
∂u
= 2x
∂x
∂u
= 1,
∂y
∂v
= −1
∂x
∂v
= 2.
∂y
The Cauchy-Riemann equations become
2x = 2 and 1 = −(−1).
Therefore, they are satisfied exactly when x = 1. Moreover, the partial derivatives are continuous and thus the function is differentiable on the line x = 1.
If z is a point on the line x = 1 then any open disk Dε (z) contains points where the function is
not differentiable. Therefore, the function is nowhere analytic.
Question 4 (20 points)
Show that if f is analytic on a domain D, and if |f | is constant, then f is constant.
Solution
Let f = u + iv. If |f | is constant then |f |2 = u2 + v 2 is also constant. This implies
0=
∂u
∂|f |2
∂v =2 u
+v
,
∂x
∂x
∂x
(1)
0=
∂u
∂|f |2
∂v =2 u
+v
.
∂y
∂y
∂y
(2)
and
Since f is analytic we can use the Cauchy-Riemann equations to rewrite Eq. (1) as
∂u
0=2 u
∂x
−v
∂u .
∂y
(3)
+v
∂u .
∂x
(4)
and Eq. (2) as
∂u
0=2 u
∂y
Page 3 of 4
Multiplying Eq. (3) by u, Eq. (4) by v, and adding them together we find
2(u2 + v 2 )
∂u
= 0.
∂x
Similarly, multiplying Eq. (1) by v, Eq. (3) by u, and subtracting the first from the second we
find
∂u
2(u2 + v 2 )
= 0.
∂y
This implies that, at each point in D, either u2 + v 2 = 0 or ∂u/∂x = ∂u/∂y = 0.
Suppose that there is a point z0 = x0 + iy0 ∈ D where ∂u/∂x 6= 0. Then, from continuity, there
must be an open disk Dε (z0 ) such that ∂u/∂x 6= 0 for all points in the open disk. But then we
must have u(x, y)2 +v(x, y)2 = 0 for all points x+iy ∈ Dε (z0 ). This implies u(x, y) = v(x, y) = 0
for all points x + iy ∈ Dε (z0 ). Since u is constant throughout the open disk Dε (z0 ) this means
that ∂u/∂x = 0 throughout the same open disk. But this contradicts the assumption that
∂u/∂x 6= 0 at x0 + iy0 and we conclude that ∂u/∂x = 0 for all points in D.
Similarly, we have that ∂u/∂y = 0 for all points in D. Therefore, u is constant in D.
From the Cauchy-Riemann equations we also find that ∂v/∂x = ∂v/∂y = 0 in D and thus v is
constant in D.
Since both u and v are constant in D we conclude that f = u + iv is also constant in D.
Page 4 of 4