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Math. 4103, homework 3, solutions
- Show that for any two complex numbers u, z and an integer n ≥ 1,
(u + z)n^ =
∑^ n
k=
n k
ukzn−k.
Hint: look at page 8 in the text.
Let
fn(u) = (u + z)n, gn(u) =
∑^ n
k=
n k
ukzn−k.
We want to show that
(1.1) fn(u) = gn(u)
for all n = 1, 2 , 3 , ....
Clearly, f 1 (u) = g 1 (u). Suppose we have this equality for n − 1. Then
gn(u)′^ =
∑^ n
k=
n k
kuk−^1 zn−k
∑^ n
k=
n! (n − k)!k! kuk−^1 zn−k
=n
∑^ n
k=
(n − 1)! ((n − 1) − (k − 1))!(k − 1)! uk−^1 z(n−1)−(k−1)
=n
∑^ n
l=
(n − 1)! ((n − 1) − l)!l! ulz(n−1)−l
=n
∑^ n
l=
n − 1 l
ulzn−^1 −l
=ngn− 1 (u) = nfn− 1 (u) = n(u + z)n−^1
=fn(u)′.
Thus (fn(u) − gn(u))′^ = 0. Clearly fn(0) − gn(0) = 0. Therefore fn(u) − gn(u) = 0.
- Show that for any complex number z,
√ 2 |z| ≥ |Re(z)| + |Im(z)|. 1
Set x = Re(z) and y = Im(z). We are supposed to show that
x^2 + y^2 ≥ |x| + |y|.
By squaring both sides of (2.1) we get an equivalent inequality,
(2.2) 2(x^2 + y^2 ) ≥ x^2 + 2|x||y| + y^2.
By moving all the variables on one side we see that (2.2) is equivalent to
(2.3) x^2 + y^2 − 2 |x||y| ≥ 0.
However the left hand side of (2.3) is equal to
(|x| − |y|)^2 ≥ 0.
- Show that for any polynomial f (z) = anzn^ + an− 1 zn−^1 + ... + a 1 z + a 0 , with real
coefficients (an ∈ R, an− 1 ∈ R, ... a 0 ∈ R),
f (z) = 0 if and only if f (z) = 0.
Since the coefficients ak are real, we have
f (z) =anzn^ + an− 1 zn−^1 + ... + a 1 z + a 0
=anzn^ + an− 1 zn−^1 + ... + a 1 z + a 0
=f (z),
which implies the claim.
- Prove the identity
1 + z + z^2 + ... + zn^ = 1 − zn+ 1 − z (z ∈ C; n = 0, 1 , 2 , ..).
- Show that if z 6 = 1 is an (n + 1)th root of unity then
1 + z + z^2 + ... + zn^ = 0.
This is obvious from Problem 4.
- Show that the following limit does NOT exist:
lim z→ 0 z z
Since
lim x→ 0 x x = (^) xlim→ 0 x x
and
y^ lim→ 0
iy iy = (^) xlim→ 0 − y y
are different, the limit for z → 0 does not exist.
- On which region is the function
f (z) = 3x + y + i(3y − x)
complex differentiable (in other words analytic, or holomorphic)?
Cauchy Riemann equations are satisfied on the whole complex plane, hence f is
holomorphic on the whole complex plane.
- Let u(x, y) = 2x(1 − y). Check that u is harmonic and find a harmonic function
v(x, y) such that u + iv is holomorphic.
Since
∂ x^22 x(1 − y) + ∂ y^2 2 x(1 − y) = 0 + 0 = 0,
u is harmonic. We need to find v such that Cauchy Riemann equations hold
∂xv = −∂y u = 2x,
∂y v = ∂xu = 2(1 − y).
From the first equation, v(x, y) = x^2 + f (y) and from the second one
f ′(y) = 2(1 − y).
Hence, f (y) = −(1 − y)^2. Therefore,
v(x, y) = x^2 − (1 − y)^2 + C
where C is an arbitrary constant. Hence,
f (z) = i(z − i)^2 + iC.
- Show that
|e(z
(^2) ) | ≤ e|z|
2 (z ∈ C).
This is straightforward:
|e(z (^2) ) | =
∑^ ∞
n=
(z^2 )n n!
∑^ ∞
n=
|z|^2 n n! = e|z| 2 .
- Show that for any complex number z
cos^2 (z) + sin^2 (z) = 1.
This is straightforward:
cos^2 (z) + sin^2 (z) =
eiz^ + e−iz 2
eiz^ − e−iz 2 i
e^2 iz^ + 2 + e−^2 iz 4
e^2 iz^ − 2 + e−^2 iz 4
If m = n then the integral is obviously equal to 2πi.
- Evaluate the integral (^) ∫
C
f (z) dz
where f (z) = (z + 2)/z and C is the semicircle z = 2 eiθ^ , 0 ≤ θ ≤ π.
We compute from the definition,
∫
C
f (z) dz =
C
(z + 2)/z dz =
C
1 dz +
C
2 z−^1 dz
∫ (^) π
0
2 ieiθ^ dθ +
∫ (^) π
0
2 eiθ^ 2 ieiθ^ dθ
=2(eiπ^ − e^0 ) + 2πi = 2πi − 4.
- Evaluate the integral (^) ∫
C
f (z) dz
where f (z) = z − 1 and C is the segment z = x, 0 ≤ x ≤ 2.
Since an antiderivative of z − 1 is z
2 2 −^ z, ∫
C
(z − 1) dz =
z^2 2 − z
0
