MATH 511
Spring 2011
Midterm Exam #2
Explain your answers carefully!
1. (15 points) Evaluate
I
cos (z + 1)
dz
3
|z|=1 z − 4z
so that f is analytic inside |z| = 1. By
Solution: Let f (z) = cosz 2(z+1)
−4
the Cauchy integral formula it follows that
I
I
f (z)
πi cos 1
cos (z + 1)
dz
=
=
2πif
(0)
=
−
3
2
|z|=1 z
|z|=1 z − 4z
Alternatively, a partial fractions decomposition of the integrand could
be used.
2. (15 points) If f is an entire function and |f (z)| ≤ ex if z = x + iy, show
that f (z) = Cez for some constant C, |C| ≤ 1.
z
Solution: Let g(z) = fe(z)
z . Since e 6= 0 for any z, it follows that g is
entire and
|f (z)| |f (z)|
|g(z)| =
=
≤1
|ez |
ex
for all z. By Liouville’s theorem g must be a constant C so f (z) = Cez
and clearly |C| ≤ 1 must hold.
3. (20 points) Let G = {z : |z| < 1, Im z > 0} be the upper half of B(0, 1).
If S is the Möbius transformation
iz
S(z) =
z−1
what is the image under S of G?
Solution: The boundary of G may be regarded as the union of E1
and E2 where E1 is the real axis from −1 to 1 and E2 is the upper half
MATH 511
Spring 2011
Midterm Exam #2
of the unit circle. The image of either this line or this circle under S is
a line since both contain z = 1 which gets mapped to ∞. Computing
the images of 0, 1, ∞ we see that E1 gets mapped to a part of the line
through 0, i, that is, the imaginary axis, and computing the image
of −1, i, 1 we see that E2 is mapped to a part of the horizontal line
through i/2 and (1 + i)/2. The image of G then must be one of the
four quadrants these two lines divide the plane into. By considering
the orientations, or just by checking a point, we see that it is the lower
right quadrant, i.e.
1
S(G) = {z = x + iy ∈ C : x > 0, y < }
2
4. (15 points) Find and classify all isolated singularities of the function
f (z) =
z
e2z − 1
Solution: The isolated singularities of f are those points z for which
e2z = 1, namely z = kπi for any integer k. For k = 0 we notice that
lim f (z) =
z→0
1
2
so this singularity is removable. For all other choice of k we have
lim (z − kπi)f (z) =
z→kπi
kπi
2
so that these are all simple poles.
5. (20 points) Suppose that f is holomorphic in B(0, 1) and satisfies
|f (z)| ≤
1
1 − |z|
|z| < 1
00
Show that |f 00 (0)| ≤ 27
2 . (Hint: first show that |f (0)| ≤
r ∈ (0, 1), then optimize with respect to r.)
Page 2
2
r2 (1−r)
for any
MATH 511
Spring 2011
Midterm Exam #2
Solution: Use the generalized Cauchy integral formula with n = 2
and a = 0,
I
2
f (w)
00
f (0) =
dw
0<r<1
2πi |z|=r w3
to get
1
f
(w)
|f 00 (0)| ≤ max 3 2πr
π |w|=r w
Using the assumption on f it follows that
|f 00 (0)| ≤
2
r2 (1 − r)
for any r < 1. Now make a choice of r to minimize the right hand
side (equivalently maximize r2 (1 − r)). The optimal value is r = 2/3
leading to |f 00 (0)| ≤ 27
2.
6. (15 points) If f is continuous on an interval [a, b] ⊂ R, use Morera’s
Theorem to show that
Z b
F (z) =
e−tz f (t) dt
a
is an entire function of z.
Solution: If T is any triangle in C we have
I
I
I Z b
Z b
−tz
−tz
F (z) dz =
e f (t) dt =
f (t)
e dz dt
T
T
a
a
T
by Fubini’s Theorem.
Since e−tz is an entire function
of z for any fixed
H −tz
H
t, it follows that T e dz = 0 for all t, hence T F (z) dz = 0 for any
such triangle. F is clearly continuous, so by Morera’s Theorem it is
analytic on any open set. (Note that F is the Laplace transform of f
if we regard f as defined to be zero outside of [a, b].)
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