Math 210A Homework 2
Edward Burkard
Exercise 1. Let f : G → C be an analytic function, G = a region. Suppose f is real-valued. Prove that f must be a
constant function.
Proof. Decomposing f into real and imaginary parts we have f = u + iv = u since v ≡ 0 because f is real valued.
As f is analytic it satisfies the Cauchy-Riemann equations. The partial derivatives we can compute are:
vx = 0
and
vy = 0.
Thus by the Cauchy-Riemann equations we have: ux = 0 and uy = 0. Integrating both equations with respect to
the differentiated variable we get that
u(x, y) = g(x) = h(y).
Since g is independent of y and h is independent of x, we have that both g and h, hence u as well, must be constant
functions.
Exercise 2. Find the harmonic conjugates of the functions:
u1 (z) = x2 − y 2 + 1, u2 (z) = x3 − 3xy 2 + 10, and u3 (z) = x4 + xy + y 4 ,
respectively, if they exist.
Solution.
(u1 ) First let’s check if u is harmonic:
uxx = 2
and
uyy = −2
Thus ∆u = 2 + (−2) = 0, so that u is harmonic. Since u is harmonic on all of C, we can find its harmonic
conjugate:
=
Z x
ux (x, t) dt −
uy (s, 0) ds
0
0
Z y
Z x
2x dt −
0 ds
=
2xt|0 = 2xy
Z
v(x, y)
=
y
0
0
y
Thus v(x, y) = 2xy is the harmonic conjugate of u.
(u2 ) The second partials are:
uxx = 6x
and
uyy = −6x
Thus ∆u = 6x + (−6x) = 0 so that u is harmonic. Again, since u is harmonic on all of C, there is a
harmonic conjugate given by:
Z
v(x, y)
y
ux (x, t) dt −
=
Z0 y
=
=
x
Z
uy (s, 0) ds
0
Z
x
3x2 − 3t2 dt −
0 ds
0
0
y
3x2 t − t3 0 = 3x2 y − y 3
1
2
(u3 ) The second partials are:
uxx = 12x2
and
uyy = 12y 2
The Laplacian of u is:
∆u = 12x2 + 12y 2 = 12(x2 + y 2 )
which is zero only at the origin in C. Thus since u is not harmonic on C or some open disk, u does not have
a harmonic conjugate.
Exercise 3. Is u(x, y) = ln x2 + y 2 a harmonic function in C \ {0}? Can one find its harmonic conjugate in
C \ {0}? Explain your answer.
Solution. As
uxx =
−2(x2 − y 2 )
(x2 + y 2 )2
and
2(x2 − y 2 )
(x2 + y 2 )2
we clearly have that ∆u = 0 on C \ {0}. Thus u is indeed harmonic on C \ {0}. However, since the region C \ {0} is
not all of C or an open disk (in fact, the region is not even simply connected), we cannot find the harmonic conjugate
of u.
uyy =
Remark. Recall that a branch of log z is a function f : G → C such that
z = ef (z)
holds. In the specific case where G = C \ {z | Re z ≤ 0 and Im z = 0} we will refer to the branch cut as Log z. This
is the principal brach of log z.
√
1
Exercise 4. Describe a principal branch of the functions f (z) = z n (n ∈ N) and g(z) = 1 − z.
Solution.
(f ) First notice that if n = 1, then f is its own principal branch since it is not multivalued, so assume that n > 1.
Let us first rewrite f :
f (z)
=
log f (z)
=
f (z)
=
1
zn
1
log z
n
1
e n log z
With f in the rewritten form, we see that we need only replace log z with its principal branch to get the
principal branch of f . Thus the principal branch of f is:
1
fp (z) = e n Log z .
√
(g) Now let’s play this same game for the function g(z) = 1 − z:
√
g(z) =
1−z
1
log g(z) =
log 1 − z
2
1
g(z) = e 2 log 1−z
Let w = 1 − z. We need w to lie in C \ {w | Re w ≤ 0 and Im w = 0}. Hence the domain we will restrict to
is
C \ {w | Re w ≤ 0 and Im w = 0} = C \ {z = x + iy | 1 − x ≤ 0 and − y = 0} = C \ {z = x + iy | x ≥ 1 and y = 0}
to make f analytic.
Exercise 5. Describe a principal branch of the function f (z) = z z .
3
Solution. Using the same trick as above rewrite f as:
f (z) = ez log z .
Since z is single valued everywhere, we need only worry about log z. Replacing log z with Log z will give us a principal
branch of f , thus the principal branch of f is:
fp (z) = ez Log z