z −2i

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MATH 3007A
Test 2 Solutions
October 21, 2011
1. (a) Find all possible values of (−i) i .
Solution:
All possible values of log(−i) are − πi
+2πik, kɛZ. Hence, all possible values of
2
(−i) i = e i log(−i) πi i(−
are e 2 +2πik) = e π
2 −2πk .
(b) Find (−i) i if arg(z) ɛ [0, 2π).
Solution:
arg(z) ɛ [0, 2π) ⇒ (−i) i 3π −
= e 2 .
2. Evaluate the following limits.
z
(a) lim
z→2i
2 +4
z − 2i
Solution:
z
lim
z→2i
2 +4 (z − 2i)(z +2i)
= lim
=4i
z − 2i z→2i z − 2i
1
(b) lim
n→∞ (1 − i) n
Solution:
|1 − i| = √ 1
2 > 1 ⇒ lim =0
n→∞
n (1 − i)
z
z→∞
2 − 2z +3
2z2 − iz +2i
(c) lim
Solution:
z
lim
z→∞
2 − 2z +3
2z2 − iz +2i
z
(d) lim
z→2
5 − 32
z − 2
Solution:
= lim
z→∞
1 − 2/z +3/z 2
f(z) =z 5 ⇒ f ′ (z) =5z 4 ⇒ lim
z→2
3. Let f(z) =|z| 4 .
1
=
2 2 − i/z +2i/z 2
z 5 − 32
z − 2 = f ′ (2) = 80.
(a) Show that f is differentiable at z = 0 and compute f ′ (0).
Solution:
f ′ |z|
(0) = lim
z→0
4
z
= lim
z z→0
2 z 2
z =0
(b) Show that f is not differentiable at any z = 0.
Solution:
h(z, z) =z 2 z 2 ⇒ h z =2z 2 z = 0 for any z = 0.
(c) Is f analytic at 0? Justify your answer.
Solution:
No, since f is not differentiable in any neighbourhood of 0 except at 0.

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4. Let u(x, y) =x 4 − 6x 2 y 2 + y 4 .
(a) Show that u is harmonic on R 2 .
Solution:
u x =4x 3 − 12xy 2 , u xx =12x 2 − 12y 2 , u y = −12x 2 y +4y 3 ,
u yy = −12x 2 +12y 2 , u xx + u yy =0.
(b) Find a harmonic conjugate v. (You need not compute f(z).)
Solution:
v y = u x =4x 3 −12xy 2 ⇒ v(x, y) =4x 3 y−4xy 3 +g(x), v x =12x 2 y−4y 3 +g ′ (x) =
−u y =12x 2 y − 4y 3 ⇒ g ′ (x) =0 ⇒ g(x) =c ⇒ v(x, y) =4x 3 y − 4xy 3 + c, for
any constant c.
5. Let f(z) = 1
z
on an open set A not containing 0.
(a) Express f(z) asu(x, y)+iv(x, y).
Solution:
f(z) = 1
z =
1
x + iy
x − iy
=
x2 =
2 + y
x
x2 − i
2 + y
y
x 2 + y 2
(b) Employ the Cauchy-Riemann theorem to show that f is analytic on A.
Solution:
u =
2xy
x
x2 + y2 , ux = y2 − x 2
(x2 + y2 ) 2 , uy = −2xy
(x2 + y2 −y
, v =
2 ) x2 + y2 , vx =
(x2 + y2 ,
2 )
v y = y2 − x 2
(x 2 + y 2 ) 2 , all are continuous on A and u x = v y and u y = −v x. Hence, f is
analytic on A.
(c) Employ the Cauchy-Riemann theorem to find f ′ (z) for any zɛA, and express it
in terms of z.
Solution:
f ′ (z) =ux + ivx = y2 − x 2 + i(2xy)
.
(x2 + y2 ) 2
(x 2 + y 2 ) 2 = |z| 4 = z 2 z 2 , and
y 2 − x 2 2 z − z
+ i(2xy) =
f ′ (z) = −z2
z 2 z
1
= − .
2 2 z
2i
z + z
−
2
2
+2i
z + z z − z
2
2i
2
= −z 2 . Hence,