Problems 2 (due 22 January)

Problems 2 (due 22 January)
1. (a) Find the values of the constants a, b, c, d for which the polynomial
u(x, y) = ax3 + bx2 y + cxy 2 + dy 3
is harmonic, i.e.,
∂2u ∂2u
+ 2 = 0.
∂x2
∂y
(1)
Solution: First,
∂2u ∂2u
+ 2 = (6a + 2c)x + (2b + 6d)y.
∂x2
∂y
So, in order to satisfy Eq. (1), we must have c = −3a, b = −3d, and
u(x, y) = ax3 − 3dx2 y − 3axy 2 + dy 3
(b) Find v(x, y) such that f (z) = u + iv is analytic in the entire plane,
where z = x + iy.
Solution: The Cauchy-Riemann equations result in
and
∂u
∂v
=
= 3ax2 − 6dxy − 3ay 2
∂y
∂x
(2)
∂v
∂u
=−
= 3dx2 + 6axy − 3dy 2 .
∂x
∂y
(3)
Integrating the first of these leads to
v = 3ax2 y − 3dxy 2 − ay 3 + f (x),
for some function f (x) of x alone. For this to be compatible with Eq.
(3) requires that
3dx2 + 6axy − 3dy 2 = 6axy − 3dy 2 + f 0 (x),
meaning that f 0 (x) = 3dx2 and f (x) = dx3 +k, where k is a constant.
I.e.,
v = 3ax2 y − 3dxy 2 − ay 3 + dx3 + k,
for some constant k.
1
2. Find all singularities of these functions:
(a)
tanh(z) =
ez − e−z
ez + e−z
Solution: Since ez is entire, the singularities of this function are only
at zeros of the denominator ez + e−z . Writing z = x + iy, singularities will occur when ex eiy = e−x eiπ−iy . Equality of these requires
equality of the moduli: ex = e−x , or x = 0. Thus, any zeroes of
the denominator above must lie on the y-axis with y = π − y, up to
possible additional multiples of 2π. Thus,
1
+m
m = 0, ±1, ±2, . . . .
y=π
2
(b)
1
z(ez − 1)
Solution: There is obviously a singularity at z = 0. The term (ez −1)
also has additional zeros. Remember that ez “covers” C infinitely
many times. Writing z = x + iy, solutions to (ez − 1) correspond to
ex eiy = 1, which means both x = 0 and
m = 0, ±1, ±2, . . . .
y = 2mπ
(c)
z4
(z 2 + a2 )2
Solution: Since z 2 + a2 = (z + ia)(z − ia), the singularities are at
±ia.
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