MATH 3160: APPLIED COMPLEX VARIABLES
TEST #1 (VERSION D)
1. Find all solutions z of the equation
tan z = sin z.
Solution:
Write it as
sin z
= sin z
cos z
and it is clear that we need sin z = 0 or cos z = 1. Write out
that
1 iz
1 = cos z =
e + e−iz
2
and you easily find
z = 2πN
with N any integer. Work out sin z = 0 and you get
z = πN
for N any integer. So the answer is
z = πN
for N any integer.
2. Calculate
3−i
3+i
in the form x + iy.
Solution:
4 3
− i
5 5
3. Show that
y
+ y2
is harmonic and find a harmonic conjugate v for it.
u=
x2
Date: February 19, 2002.
1
2
MATH 3160: APPLIED COMPLEX VARIABLES TEST #1 (VERSION D)
Solution:
Taking derivatives:
−2xy
+ y 2 )2
1
2y
uy = 2
−
2
2
x +y
(x + y 2 )2
8x2 y
2y
uxx =
−
3
(x2 + y 2 )
(x2 + y 2 )2
8y 3
−6y
+
uyy =
(x2 + y 2 )2 (x2 + y 2 )3
ux =
(x2
which gives
uxx + uyy = 0
so that u is harmonic. The conjugate is
x
v = c0 + 2
x + y2
where c0 is any real constant. This is easy to find in polar
coordinates.
4. At which points z does the function
f (z) = cos x cosh y − i tan x sinh y
satisfy the Cauchy–Riemann equations? Explain your answer.
Solution:
u = cos x cosh y
v = − tan x sinh y
so that
ux = − sin x cosh y
vy = − tan x cosh y
uy = cos x sinh y
vx = − 1 + tan2 x sinh y
The first Cauchy–Riemann equation is
0 = ux = vy
which says
sin x cosh y = tan x cosh y
MATH 3160: APPLIED COMPLEX VARIABLES
TEST #1 (VERSION D)
3
which, since cosh y > 0, says precisely that sin x = tan x. The
solutions are x = πN for N any integer. Plugging this in, the
second Cauchy–Riemann equation says
uy + v x = 0
or
cos x − 1 + tan2 x
Either sinh y = 0 or
sinh y = 0.
cos x = 1 + tan2 x .
In the first case, x = πN, y = 0, for N any integer. In the
second,
cos x = cos πN = (−1)N
while
1 + tan2 x = 1 + tan2 (πN ) = 1
(since sin πN = 0). So we need (−1)N = 1, i.e. N even.
Therefore there are two sets of solutions:
z = πN
for any integer N , and
z = 2πN + iy
for any value of y.
5. What is the region Re(1/z) < 1 in rectangular coordinates?
Describe it algebraically, and sketch it.
Solution:
2
1
1
+ y2 >
x−
2
4
The exterior of a circle of radius 1/2 about the point z = 1/2.
6. (a) Find all of the values of
sin−1 (z)
(the inverse
of the sine function, not the reciprocal) at
√
z = 1/ 2.
(b) Find all of the values of
√
at z = 1/ 2.
Solution:
d
sin−1 (z)
dz
4
MATH 3160: APPLIED COMPLEX VARIABLES TEST #1 (VERSION D)
√
(a) At z = 1/ 2, we find
sin w = z
gives
w=
π
+ 2πN
4
or
3π
+ 2πN
4
where N is any integer.
(b) Write w = sin−1 (z). Differentiating
w=
sin w = z
we find
cos w
dw
=1
dz
so that
dw
1
=
.
dz
cos w
Plugging in
w=
π
+ 2πN
4
w=
3π
+ 2πN
4
and
we find
( √
1/ 2
√
cos w =
−1/ 2
or
.
Therefore the two values of dw/dz are
dw
1
= ±√ .
dz
2
7. BONUS: State Stokes’ theorem in the plane (also known as
Green’s theorem, Gauß’s theorem, or the divergence theorem).
Solution: If U is a region in the plane with boundary given
by a curve C, then any functions f (x, y) and g(x, y) satisfy
Z
Z ∂g ∂f
f (x, y) dx + g(x, y) dy =
−
dx dy.
∂x ∂y
C
U
MATH 3160: APPLIED COMPLEX VARIABLES
TEST #1 (VERSION D)
5
In carrying out the integral over C, you travel along C so that
the region U is always on your left. Instead of this, you might
write that any vector field ~u on the plane satisfies
Z
Z
~u · ~t ds =
∇· ~u dx dy
C
U
where the notation is as follows:
(a) ~t is the unit length tangent vector to the curve C
(b) If we write our vector field as
~u = f~ı + g~
then the divergence ∇· is defined by
∂g ∂f
∇· ~u =
−
∂x ∂y
and is also often written as
div ~u = ∇· ~u.
(c) The symbol
ds means the element of arclength along C, so
R
that ds along a piece of C is the length of that piece.