HOMEWORK 3:
SOLUTIONS
1. Find exactly all roots of the equations
a) sin z = i
b) z 4 = −1 +
√
3i
c) ez = 3
Solution:
a) Write z = x + iy and separate sin(z) into real and imaginary parts:
(
sin(z) = sin(x + iy) = sin(x) cos(iy) + cos(x) sin(iy)
sin(x) cosh(y) = 0
⇒
= sin(x) cosh(y) + i cos(x) sinh(y) = i
cos(x) sinh(y) = 1
The first equation gives x = nπ, n = 0, ±1, ±2, . . . (it tells us nothing about y, cosh(y) 6= 0 for
any y). Plugging x = nπ into the second equation and solving for y gives y = arcsinh[(−1)n ].
So
z = nπ + i arcsinh[(−1)n ],
n = 0, ±1, ±2, . . .
b) The trick here is to use polar coordinates
√
z 4 = −1 + 3i = 2e2iπ/3+2iπk
z = 21/4 eiπ/6+iπk/2 ,
k = −2, −1, 0, 1
any 4 consecutive values of k will do, only 4 because the 5th number and the 1st differ by a
factor of e2iπ so they are the same!
z = 21/4 eiπ/6 ,
z = 21/4 e2iπ/3 ,
z = 21/4 e−iπ/3 ,
z = 21/4 e−5iπ/6
c) This one is the easiest, just take the (complex) logarithm of both sides:
z = log 3 = log 3 + 2ikπ,
k = 0, ±1, ±2, . . .
were we remember that for a complex number log z = log |z| + i arg(z), and arg(z) is not
unique.
2. Show that the function f (z) = z 1/3 is multivalued by determining u(x, y) and v(x, y) explicitly.
Then show that if we take the principal branch, −π < arg z ≤ π, that f (z) is not continuous on
the negative real line, by explicitily computing the values of f (z) on either side of the branch.
Solution:
f (z) = z 1/3 = r1/3 eiθ/3 = r1/3 cos(θ/3) + ir1/3 sin(θ/3)
1
HOMEWORK 3:
SOLUTIONS
2
where θ is measured in the interval −π < θ ≤ π. Now if we take the limit as θ goes to the branch
cut from above or below we get:
√ !
1 i 3
iθ
1/3
+
lim f (re ) = r
θ→π
2
2
√ !
1 i 3
lim f (reiθ ) = r1/3
−
θ→−π
2
2
since the limits are not the same, the function is not continuous along the branch cut.
3. Show that u(x, y) = x3 y − xy 3 is a harmonic function (i.e. ∆u = 0). Use the Cauchy-Riemann
equations to construct a function v(x, y) such that f (z) = u(x, y) + iv(x, y) is analytic. The
function v is called the harmonic conjugate of u.
Solution:
∆u = uxx + uyy = 6xy + (−6xy) = 0
so u is harmonic.
Now we choose v according to the Cauchy Riemann equations.
vy = ux = 3x2 y − y 3
−vx = uy = x3 − 3xy 2
3 2 2 1 4
x y − y + c1 (x)
2
4
1 4 3 2 2
v = − x + x y + c2 (y)
4
2
v=
Comparing the two equations we find that
1
3
1
v(x, y) = − x4 + x2 y 2 − y 4 .
4
2
4
The functions u and v are the real and imaginary parts of the function f (z) = − 4i z 4 .
∂
∂
∂
∂
¯ for any analytic function
4. Let ∂ = 12 ∂x
− i ∂y
and ∂¯ = 12 ∂x
+ i ∂y
. Compute ∂f and ∂f
f (z) = u(x, y) + iv(x, y).
Solutions:
∂
u x + vy
vx − uy
1 ∂
(u + iv) =
−i
+i
= ux + ivx = vy − iuy
2 ∂x
∂y
2
2
¯ = 1 ∂ + i ∂ (u + iv) = ux − vy + i vx + uy = 0
∂f
2 ∂x
∂y
2
2
∂f =
5. Compute, without theorems,
z = −1 to z = 1:
a) the straight line
´
C
f (z)dz for f (z) =
1
z−i/2
and C each of the following paths from
b) the upper semi-circle
c) the lower semi-circle.
HOMEWORK 3:
SOLUTIONS
3
Explain why some of your answers are equal and others are different.
Solutions: Skip this one!
ˆ
6. Determine an upper bound for 1 for C the semicircle of radius R in the upper halfplane.
z2 + 1 C
Calculate it’s limit as R → ∞.
Solution: Parameterize the path by z = Reit .
ˆ
ˆ π
ˆ
|Rieit dt|
1
dz ≤
≤
z2 + 1 |R2 e2it + 1|
0
C
0
π
R
πR
dt = 2
.
−1
|R − 1|
R2
So if we take the limit as R → ∞:
ˆ
πR
1
≤ lim
dz
lim R→∞ |R2 − 1| = 0.
2
R→∞
+
1
z
C
˛
7. Compute
C
z 3 e−z
where C is the c.c.w. oriented circle
z3 + 8
(a) |z| = 1,
(b) |z + i| = 2
(c) |z + 2| = 1
Solution: The three roots of z + 8 = 0 are z1 = −2, z2 = 2eiπ/3 , z3 = 2e−iπ/3 .
3
Now we compute the three integrals:
(a) All three roots are outside the circle |z| = 1 so
¸
z 3 e−z
|z|=1 z 3 +8
= 0.
(b) The root z3 = 2e−iπ/3 is inside the circle |z + i| = 2 and the others are outside. So we can
use the C.I.F. to write
˛
z 3 e−z
4iπ −iπ/3 −1+i√3
z 3 e−z
=
=
2iπ
·
e
e
3
iπ/3
3
(z + 2)(z − 2e
) z=2e−iπ/3
|z+i|=2 z + 8
(c) This is the same as the previous example except now the root inside the circle is z1 = −2:
So we can use the C.I.F. to write
˛
z 3 e−z
z 3 e−z
4iπ 2
=
2iπ
·
=−
e
3
iπ/3
−iπ/3
3
(z − 2e
)(z − 2e
) z=−2
|z+2|=1 z + 8
8. Suppose C1 and C2 are oriented curves in C with a nontangential intersection at z0 and suppose
that f (z) is analytic in a neighborhood of z0 . Let z1 (t) and z2 (t) be parameterizations of C1
and C2 respectively, such that z10 (t) 6= 0 and z20 (t) 6= 0 and z1 (t0 ) = z2 (t0 ) = z0 , and let
w1 (t) = f (z1 (t)) and w2 (t) = f (z2 (t)) be the image of each curve under the mapping w = f (z).
Show that if f 0 (z0 ) 6= 0 then
a) wj0 (t0 ) 6= 0.
HOMEWORK 3:
SOLUTIONS
4
b) arg wj0 (t0 ) − arg zj0 (t0 ) = arg f 0 (z0 ), for j = 1, 2.
c) arg z20 (t0 ) − arg z10 (t0 ) = arg w20 (t0 ) − arg w10 (t0 ).
Draw a diagram that includes the curves and derivatives (tangent vectors) at z0 and w0 = f (z0 ).
Since zj0 (t) and wj0 (t) are oriented tangent vectors to the curves zj (t) and wj (t) the last result
shows that analytic functions f (z) are conformal, i.e., they preserve the angles between curves
in both magnitude and direction wherever f 0 (z) 6= 0.
Solution
a) By the chain rule wj0 (t0 ) = f 0 (zj (t0 )zj0 (t0 ) = f 0 (z0 )zj0 (t0 ). Since zj0 (t0 ) 6= 0 and f 0 (z0 ) 6= 0 it
follows that wj0 (t0 ) 6== 0.
b) Take the argument of each side:
arg wj0 (t0 ) = arg f 0 (z0 )zj0 (t0 ) = arg f 0 (z0 ) + arg zj0 (t0 )
rearranging gives
arg wj0 (t0 ) − arg zj0 (t0 ) = arg f 0 (z0 ).
c) To get this simply subtract the above equation with j = 2 from the same equation with j = 1:
(arg w20 (t0 ) − arg z20 (t0 )) − (arg w10 (t0 ) − arg z10 (t0 )) = arg f 0 (z0 ) − arg f 0 (z0 )
Simplifying gives
arg z20 (t0 ) − arg z10 (t0 ) = arg w20 (t0 ) − arg w10 (t0 ).