MATH 511
Spring 2011
Midterm Exam #1
Explain your answers carefully!
1. (20 points) (a) Compute (−4 + 4i)1/3
(b) Find the real and imaginary parts of the function
f (z) =
i
z2
19π
Solution: a) 25/6 eiφ for φ = π4 , 11π
12 , 12 .
b) Re f =
2xy
(x2 +y 2 )2
, Im f =
x2 −y 2
(x2 +y 2 )2 ,
2. (20 points) We say that w = arcsin(z) if sin w = z. Show that
2 1/2
arcsin(z) = −i log iz + (1 − z )
(Suggestion: If sin w = z then eiw is a root of a certain quadratic.) Find
all possible values of arcsin(2).
Solution: The equation sin w = z may be arranged as
e2iw − 2izeiw − 1 = 0
Viewing this as a quadratic to solve for eiw we get
eiw = iz + (1 − z 2 )1/2
(Here the multivalued version of the (1 − z 2 )1/2 is meant which takes
care of the usual ± in the quadratic formula.) The stated formula
then follows from the definition of logarithm. Substituting z = 2 gives
√
√
√
arcsin(2) = −i log (2 ± 3)i = −i(log |2 ± 3| + i arg (2 ± 3)i)
√
Since 2 ± 3 > 0 we get
√
π
arcsin(2) = + 2kπ − i log |2 ± 3|
k∈Z
2
MATH 511
Spring 2011
Midterm Exam #1
3. (15 points) Show that f (z) = z Re z is differentiable at z = 0. Using the
Cauchy-Riemann equations show that f is not analytic anywhere.
Solution: Since f (0) = 0 we have
f (h)
= lim Re h = 0
h→0 h
h→0
f 0 (0) = lim
so f is differentiable at z = 0. Since f = u+iv with u = x2 and v = xy,
if f were analytic anywhere there would have to exist an open set on
which the Cauchy-Riemann equations were satisfied, ux = 2x = vy = x
and uy = 0 = −vx = y. But these equations are satisfied only for
x = y = 0 so there is no such open set.
4. (15 points) Discuss the validity of the statement that
z, α ∈ C.
d α
dz z
= αz α−1 for
Solution: For differentiability to make sense we must mean a branch
g(z) of z α , namely eαf (z) where f is any specific branch of the logarithm. We already know that f 0 (z) = z1 for any such branch so by the
chain rule
eαf (z)
1
g 0 (z) = αeαf (z) = α f (z) = αe(α−1)f (z)
z
e
α−1
which is a branch of αz .
any bounded sequence of complex numbers.
5. (15 points) Let {an }∞
n=0 be P
n
Show that the power series ∞
n=0 an z is convergent in B(0, 1).
Solution: If |an | ≤ M for all n and some M > 0 then
1
1
lim sup |an | n ≤ lim sup M n = 1
n→∞
n→∞
Thus the radius of convergence R of the series must satisfy R ≥ 1, so
in particular the series converges in B(0, 1).
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MATH 511
Spring 2011
Midterm Exam #1
6. (15 points) Let E be the line through −i and 1. Find the image of E by
the mapping w = z − z 2 .
Solution: The line E is given by y = x − 1 and the mapping may be
expressed as
u = x + y 2 − x2
v = y − 2xy
Substituting y = x − 1 gives the image curve in parametric form
u=1−x
v = 3x − 2x2 − 1
Eliminating the parameter x then gives v = −2u2 + u, which is the
equation of a parabola.
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