MATH 3160: APPLIED COMPLEX VARIABLES TEST #1 (VERSION A) 3 + i

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MATH 3160: APPLIED COMPLEX VARIABLES
TEST #1 (VERSION A)
1. (10 points) Calculate
3+i
√
√
2 − 2i + |1/ 2 + i/ 2|
Solution:
7
9
+ i
13 13
2. (10 points) Prove that
Re z ≤ |z|
for any complex number z.
Solution:
Re z = x ≤ |x| ≤
p
x2 + y 2 = |z|
3. (10 points) Find the roots z of
z4 + 9 = 0
Solution:
±
1√
1√
6±
6i
2
2
4. (10 points) What is the value of the limit
z2
?
z→0 |z|2
lim
Explain your answer.
Solution: If z = x is real, then you get
x2 /x2 = 1
while if z = iy then you get
−y 2 /y 2 = −1
so the limit does not exist as z → 0.
Date: February 6, 2001.
1
2
MATH 3160: APPLIED COMPLEX VARIABLES TEST #1 (VERSION A)
5. (10 points) Find all complex numbers z which solve the equation
√
2 = cos z .
Solution:
Using the definition
cos z =
eiz + e−iz
2
we find the equation
√
eiz + e−iz = 2 2 .
Try writing
q = eiz .
Then our equation is
√
1
=2 2
q
so if we multiply by q then
√
q 2 + 1 = 2 2q
q+
a quadratic equation. Solve it to find
√
q = 2±1.
But now return to
q = eiz
to find z:
z = −i Log q
or
√
z = −i ln( 2 ± 1) + 2πN
for any integer N .
6. (10 points) What are all of the values of ii ?
Solution: By definition
ii = exp(i Log i)
and so we work out
Log i = i
π
+ 2πN
2
where N can be any integer. Then we find
π
ii = exp − − 2πN .
2
MATH 3160: APPLIED COMPLEX VARIABLES
TEST #1 (VERSION A)
3
7. (10 points) Bonus: Derive the polar coordinate expressions
of the Cauchy–Riemann equations from the rectangular ones
given on page 1 above.
Solution: We will use the equations
x = r cos θ
y = r sin θ
as follows:
u r uθ
ux
=
vr vθ
vx
ux
=
vx
xr xθ
yr yθ
uy
cos θ −r sin θ
vy
sin θ r cos θ
uy
vy
Therefore to solve for the rectangular coordinate derivatives:
−1
u x uy
ur uθ
cos θ −r sin θ
=
vx vy
vr vθ
sin θ r cos θ
cos θ sin θ
ur uθ
=
vr vθ
− sinr θ cosr θ
ur cos θ − 1r uθ sin θ ur sin θ + 1r uθ cos θ
=
vr cos θ − 1r vθ sin θ vr sin θ + 1r vθ cos θ
Then the usual Cauchy–Riemann equations say
1
1
ur cos θ − uθ sin θ = vr sin θ + vθ cos θ
r
r
1
1
ur sin θ + uθ cos θ = − vr cos θ − vθ sin θ
r
r
Another way to write this:
1
ur = v θ
r
1
v r = − uθ
r
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