MATH 621. SPRING 2015. MIDTERM I - SOLUTIONS
1. Let f (z) be a holomorphic function on an open set Ω ⊂ C. Write
f (x + iy) = u(x, y) + iv(x, y).
(a) Show that u and v are infinitely-differentiable.
(b) Show that u and v are harmonic, that is, ∆u = ∆v = 0, where
∆=
∂2
∂2
+
.
(∂x)2 (∂y)2
Proof. (a) A holomorphic function is infinitely differentiable as a function of
complex variable, i.e. f (n) (z) exists and is continuous. We have f 0 = ux +ivx
∂nu
∂nv
and, so, inductively, f (n) = ū + iv̄, where ū = (∂x)
n and v̄ = (∂x)n are
continuous. We keep differentiating, but now use CR equations to write
f (n+1) = v̄y − iūy , f (n+2) = −ūyy − iv̄yy , and so on, inductively. This shows
∂ m v̄
∂ m ū
that (∂y)
m and (∂y)m exist and are continuous, i.e. u and v have continuous
partial derivatives
∂ n+m ū
(∂x)n (∂y)m
and
∂ n+m v̄
(∂x)n (∂y)m
of all orders, i.e. are infinitely
1
differentiable. (b) ∆u = uxx + uyy = (vy )x − (vx )y = 0 by Clairaut theorem,
because v has continuous second partials. Similar for v.
2. (a) Find the power series expansion of sin z with center at z = π/4.
What is its radius of convergence?
(b) Let log z be the logarithm function defined on the open set C\(−∞, 0].
Using the fact that (log z)0 = 1/z, find the power series expansion of log z
with center at z = 1. What is its radius of convergence?
Proof. (a)
1
sin(z) = sin((z − π/4) + π/4) = √ (sin(z − π/4) + cos(z − π/4)) =
2
(z − π/4)2 (z − π/4)3
1
=√
1 + (z − π/4) −
−
+ ... =
2!
3!
2
∞
1 X ak (z − π/4)k
=√
,
k!
2 k=0
where ak = 1 for k ≡ 0, 1 mod 4 and ak = −1 for k ≡ 2, 3 mod 4.2 For the
radius of convergence, we can use Hadamard’s formula
ak+1 1
1
(k+1)! = lim sup ak = lim sup
= 0,
R
k→∞ k + 1
k→∞ k! 1It is also easy to show that u, v are analytic (and hence infnitely-differentiable), i.e. can
be represented as converging power series in (x − a)n (y − b)m near every point a + ib in Ω.
2One can also get these coefficients by applying Taylor’s formula
1
2
MATH 621. SPRING 2015. MIDTERM I - SOLUTIONS
so R = ∞. 3
(b)
1
1
=
= 1 − (z − 1) + (z − 1)2 − (z − 1)3 + . . .
z
1 + (z − 1)
with radius of convergence 1. Integrating,
(z − 1)2 (z − 1)3 (z − 1)4
+
−
+
2
3
4
with the same radius of convergence as its derivative series, i.e. 1.
log z = (z − 1) −
3. (a) State and prove the residue formula.
(b) Let a, b ∈ C and R ∈ R such that R > 0, R 6= |a|, R 6= |b|, and |a| ≥ |b|.
Let γ be a circle of radius R centered at 0. Compute
Z
dz
γ (z − a)(z − b)
(the answer will of course depend on R, a, and b).
Proof. (a) See Theorem 2.1 and Corollary 2.3 in Chapter 2.
1
is holomorphic for z 6= a, b, and has a
(b) The function f (z) = (z−a)(z−b)
pole of order 1 at a and b if a 6= b, in which case
1
1
, resb f = lim (z − b)f (z) =
.
resa f = lim (z − a)f (z) =
z→a
z→b
a−b
b−a
If a = b then f (z) has a pole of order 2 at a and the residue is obviously
1
equal to zero (the principal part of f (z) at a is (z−a)
2 ). In both cases, by the
residue formula,


R < |b|
Z
0
2πi
f (z) dz = b−a
|b| < R < |a|

γ
 2πi
2πi
b−a + a−b = 0 R > |a|
4. Compute the integral
Z
∞
−∞
x sin(x)
dx
(1 + x2 )2
Justify any estimates you use.
R ∞ xeix
Proof. We are going to compute −∞ (1+x
2 )2 dx and then take the imaginary
part. Let γ be an upper semicircle centered at 0 of radius R. Let
f (z) =
zeiz
.
(1 + z 2 )2
f (z) is holomorphic along and inside γ except for a pole of order 2 at i. We
have
d
d
zeiz
2
resi f (z) = lim
f (z)(z − i) = lim
=
z→i dz
z→i dz
(z + i)2
3Alternatively, sin z is an entire function, and therefore the radious of convergence is ∞
at every point.
MATH 621. SPRING 2015. MIDTERM I - SOLUTIONS
3
(after some algebraic manipulations)
1
.
4e
R
πi
= 2e
. Let γ 0 be an upper arc of a
The residue formula gives γ f (z) dz = 2πi
4e
R
semicircle. If we can show that limR→∞ γ 0 f (z) dz = 0, then we will show
R∞
R ∞ x sin(x)
πi
π
that −∞ f (x) dx = 2e
, and therefore −∞ (1+x
2 )2 dx = 2e .
It remains to prove the claim. Using the standard parametrization of a
circle arc, we have
Z π
Z
iθ
Reiθ eiRe
f (z) dz =
iReiθ dθ ≤
iθ )2 )2
(1
+
(Re
0
−π
γ
iReiθ R e
2π sup
2R ≤
θ |1 + (Reiθ )2 |
(by the reverse triangle inequality, 1 + (Reiθ )2 ≥ |R2 − 1|)
iθ R2 eiRe ≤
2π sup
|R2 − 1|2
θ
iθ (we have eiRe = eiR cos θ−R sin θ = e−R sin θ ≤ 1)
2π
R2
→0
|R2 − 1|2
as R → +∞.
5. (a) State the theorem on zeros of a holomorphic function.
(b) Compute the number of zeros (counting mutiplicities) of the function
f (z) = 3z 3 − cos z in the square
Ω = {z ∈ C | − 1 < Re(z) < 1 and − 1 < Im(z) < 1}.
Proof. (a) See Theorem 1.1 in Chapter 3.
p (b) Let Γ be the unit square (the boundary of Ω). Let z = x+iy. Then |z| =
x2 + y 2 ≥ 1 along Γ. Therefore, |3z 3 | ≥ 3 along Γ. We also have |−cos z| =
4
1 −y+ix
+ ey−ix | ≤ 21 |ey + e−y | ≤ 2e
2 |e
2 = e < 3. . Therefore, by Rouche’s
theorem, the number of zeros of f (z) in Ω (counted with multiplicities) is
equal to the number of zeros of 3z 3 , which is three.
6. Let f and g be entire functions such that |f (z)| ≤ |g(z)| for all z ∈ C.
Show that f (z) = cg(z) for some c ∈ C.
Proof. If g(z) is identically zero then f (z) is identically zero as well, and
there is nothing to prove (take any c). If f (z) is identically zero then we can
take c = 0. So we can assume that f (z) and g(z) are not identically zero.
Let h(z) = f (z)/g(z). We claim that h(z) can be extended to an entire
function. Given the claim, by Liouville’s theorem, h(z) is constant, hence
f (z) = cg(z) for some c ∈ C.. To prove the claim, we first notice that h(z)
is holomorphic at any point z0 where g(z0 ) 6= 0. Consider a point where
4One can prove a better bound
1
(ey
2
+ e−y ) ≤ 21 (e + e−1 ) but we don’t need it
4
MATH 621. SPRING 2015. MIDTERM I - SOLUTIONS
g(z0 ) = 0.5 By a theorem on zeros, we can write g(z) = (z−z0 )n g0 (z), where
g0 (z) is holomorphic and g0 (z0 ) 6= 0. So we have |f (z)| ≤ |(z − z0 )n g0 (z)|
for any z. Plug-in z = z0 to get f (z0 ) = 0. By a theorem on zeros, we
have f (z) = (z − z0 )m f0 (z), where f0 (z) is holomorphic and f0 (z0 ) 6= 0.
To show that h(z) is holomorphic at z0 , it suffices to show that m ≥ n.
Arguing by contradiction, suppose m < n. Then we have |(z −z0 )m f0 (z)| ≤
|(z − z0 )n g0 (z)| for any z, and therefore |f0 (z)| ≤ |(z − z0 )n−m g0 (z)| for any
z 6= z0 , and therefore at z = z0 as well, by continuity. Therefore f0 (z0 ) = 0,
a contradiction.
5A sneakier proof: notice that |h(z)| ≤ 1 in a deleted neighborhood of z . Therefore,
0
h(z) has a removable singularity at z0 by Riemann’s singularity theorem, and so h(z) can
be extended to an entire function.