Solutions to MATH 300 Homework 4

EXERCISES 2.4
4 5 5 4
x3 y 3 x3 y 3
4. Let u(x, y) = 2 2
, v(x, y) = 2 . Then
x +y x + y2

∂u u(∆x, 0) − u(0, 0) 0
= lim = lim =0
∂x x=0,y=0
∆x→0 ∆x ∆x→ ∆x

and
∂u u(0, ∆y) − v(0, 0) 0
= lim = lim = 0.
∂y x=0,y=0
∆y→0 ∆y ∆y→ ∆y

∂v ∂v
Similarly = 0 and = 0. Hence the Cauchy-Riemann
∂x x=0,y=0 ∂y x=0,y=0
equations holds at z = 0. However, when ∆z → 0 through the real values
(∆z = ∆x),
f (0 + ∆x) − f (0)
lim = 0,
∆x→0 ∆x
while along the real line y = x (∆z = ∆x + i∆x)
(∆x)4/3 (∆x)5/3 +(∆x)5/3 (∆x)4/3
f (0 + ∆z) − f (0) 2(∆x)2 1
lim = lim = .
∆x→0 ∆z ∆x→0 ∆x(1 + i) 2

Therefore f is not differentiable at z = 0.

2 −y 2 2 −y 2
5. Let u(x, y) = ex cos(2xy), v(x, y) = ex sin(2xy). Then

∂u 2 2 ∂v
= 2ex −y [x cos(2xy) − y sin(2xy)] =
∂x ∂y
and
∂u 2 2 ∂v
= −2ex −y [y cos(2xy) − x sin(2xy)] = − .
∂y ∂x

1
f is entire since these first partials exist and are continuous for all x and y.
Since
2 2 2 2
f (z) = ex −y [cos(2xy) + i sin(2xy)] = e(x+iy) = ez ,
2
it follows that f 0 (z) = 2zez .

8. Let f (z) = f (x + iy) = u(x, y) + iv(x, y). Suppose that u(x, y) = c in D,

∂u ∂v ∂u
where c is a constant. Then = 0 and by Cauchy-Riemann − = = 0.
∂x ∂x ∂y
It follows that
∂u ∂v
f 0 (z) = +i = 0.
∂x ∂x
Therefore, f is constant in D.
If we suppose that Im(f ) = c, then by a similar argument we can prove the
result.
1
11. If both f and f¯ are analytic in D, then g := Re(f ) = (f + f¯) is
2
analytic and real-valued. That is, Im(g) = 0. Hence it follows from Exercise
8 that g = Re(f ) is constant in D. So f is constant by Exercise 8 again.

EXERCISES 2.5
∂ 2u x ∂ 2u ∂(ex cos y)
3. (b) Since = e sin y, and = = −ex sin y, we have
∂x2 ∂y 2 ∂y

∂ 2u ∂ 2u
+ = 0.
∂x2 ∂y 2
u is harmonic. Next, we find the harmonic conjugate of u, denoted by v,
which satisfies the Cauchy-Riemann equations. It follows from
∂u ∂v
= ex sin y =
∂x ∂y
that v(x, y) = −ex cos y + φ(x), where φ(x) is a differentiable real-valued
function of x. Also, from
∂u ∂v
= ex cos y = −
∂y ∂x
we have ex cos y = ex cos y − φ0 (x), i.e., φ(x) is constant. Therefore, v(x, y) =
−ex cos y + c, c ∈ C, is the harmonic conjugate of u.
Or: Since u(x, y) = ex sin y can be written as the real part of the function

2
−iex+iy = −iez , which is entire, it follows that the u is harmonic and a har-
monic conjugate of u is Im(−iez + c) = −ex cos y + c, c ∈ C.
i
(c) Since u(x, y) = xy − x + y can be written as Re(− z 2 − iz − z), which
2
is entire, it follows that the u is harmonic and a harmonic conjugate of u is
i 1
Im(− z 2 − iz − z + c) = − (x2 − y 2 ) − x − y + c, c ∈ C.
2 2
5. If f (x + iy) = u(x, y) + iv(x, y) is analytic, then −if (x + iy) = v(x, y) −
iu(x, y) is analytic. Thus −u is a harmonic conjugate of v.

7. Take z = x + iy. Since the region is bounded by x = −1 and x = 3,

naı̈vely, we consider the analytic function f (z) = z, whose real part is x. But
it does not satisfy the boundary conditions. Hence we take af (z) + b with
a, b ∈ R, which is also analytic. Then Re(af + b) = ax + b is harmonic. Using
the boundary values we have

a · (−1) + b = 0
,
a · (3) + b =4

which gives a = b = 1. Then φ(x) = x + 1.

8. (a) Yes, because ∇2 (u + v) = ∇2 u + ∇2 v = 0 if both u and v are

harmonic.
∂f
(b) No. Set = fx . Then ∇2 (uv) = v(uxx + uyy ) + u(vxx + uyy ) + 2(ux vx +
∂x
uy vy ) = 2(ux vx + uy vy ). So unless ux vx + uy vy = 0, uv is not harmonic.
Take u = x, v = xy as an example. Both of them are harmonic in R2 . But
∇2 (uv) = 2y, which means that it is harmonic only on the line y = 0.
∂ ∂
(c) Yes, because ∆(ux ) = uxxx + uxyy = uxxx + uyyx = (∆u) = (0) = 0.
∂x ∂x
15. Take z = reiθ . Denote the annulus {z ∈ C 1 6 |z| 6 2} by A. We con-
sider on A the analytic function f (z) = az n +bz −n +c with a, b, c ∈ R, n ∈ N,
whose real part is arn cos nθ + br−n cos(−nθ) + c. Set n = 3 to agree with
the cosine argument on |z| = 2.

φ(reiθ ) = ar3 cos 3θ + br−3 cos(−3θ) + c.

= (ar3 + br−3 ) cos 3θ + c

3
If r = 1, then φ(eiθ ) = 0 ⇒ a + b = 0, c = 0. If r = 2, then φ(2eiθ ) =
5 cos 3θ ⇒ (8a + b/8) cos 3θ = 5 cos 3θ. So a = 40/64, b = −40/63 and
40 3
φ(reiθ ) = (r − r−3 ) cos 3θ,
63
40
φ(z) = Re(z 3 − z −3 ).
63

EXERCISE 3.1
2. (a) Since deg(f g) = deg(f ) + deg(g), it follows that
r
X r
X
di
n = deg(p(z)) = deg((z − zi ) ) = di .
i=1 i=1

(b) an−1 is the coefficient corresponding to z n−1 . Observe from the expression

p(z) = an (z − z1 ) · · · (z − z1 ) (z − z2 ) · · · (z − z2 ) · · · (z − zr ) · · · (z − zr )
| {z }| {z } | {z }
d1 copies d2 copies dr copies

that we need to choose −zi from one of the factors once and z from the rest
of the factors n − 1 times, in order to give a term of degree n − 1. Repeat
the process for i = 1, 2, . . . , r. Then since there are di copies of (z − zi ), we
have di (−zi )’s. Therefore,

an−1 = an ((−z1 )d1 + (−z2 )d2 + · · · + (−zr )dr ) = −an (d1 z1 + d2 z2 + · · · + dr zr ).

(c) By a similar consideration, a0 is the constant term. Hence we need to

choose −zi from each of the factors di times and z without once. Therefore,

a0 = an (−z1 )d1 (−z2 )d2 · · · (−zr )dr = an (−1)d1 +d2 +···+dr z1d1 z2d2 · · · zrdr
= an (−1)n z1d1 z2d2 · · · zrdr .

3. (b) z 4 − 16 = (z 2 + 4)(z 2 − 4) = (z + 2i)(z − 2i)(z + 2)(z − 2).

(c)

z7 − 1
1 + z + z2 + z3 + z4 + z5 + z6 =
z−1
= (z − ζ7 )(z − ζ72 ) · · · (z − ζ76 ),

4
where ζ7 = e2πi/7 .

5. (c) (z − 1)(z − 2)3 = ((z − 2) + 1)(z − 2)3 = (z − 2)4 + (z − 2)3 .

13. (b)
2z + i 2z + i
3
=
z +z z(z + i)(z − i)
(1) (2) (3)
A A A
= 0 + 0 + 0
z z+i z−i
i i/2 3i/2
= + − ,
z z+i z−i
(1) 2z + i (2)
since by using (21), we have A0 = lim z · 3
= i, A0 = lim (z + i) ·
z→0 z +z z→−i
2z + i i (3) 2z + i 3i
3
= , A0 = lim(z − i) · 3 =− .
z +z 2 z→i z +z 2
(d)

5z 4 + 3z62 + 1 5 2 15 47 − 111
8
z − 39
8
= z − z + +
2z 2 + 3z + 1 2 4 8 (z + 1)(2z + 1)
(1) (2)
5 15 47 1 A0 A0
= z2 − z + + · 1 +
2 4 8 2 z+ 2
z+1
33
5 15 47 9
= z2 − z + + 16 1 − ,
2 4 8 z+ 2
z+1

(1) 1 − 111
8
z − 39
8 33
since by using (21), we have A0 = lim1 (z + ) · = ,
z→− 2 2 (z + 1)(2z + 1) 16
(2) − 111
8
z − 39
8
A0 = lim (z + 1) · = −9.
z→−1 (z + 1)(2z + 1)