7/29/2019 Mid 1 Sample Solution

http://slidepdf.com/reader/full/mid-1-sample-solution 1/2

Math282: Mid1 Sample Solution

1.Solution:

a). 3i + i

3= −3i + i

3= −8

3i.

b).sin(1+i) = ei(1+i)−e−i(1+i)

2i= e−1+i−e1−i

2i= e−1 sin(1)+e1 sin(1)

2− e−1 cos(1)−e cos(1)

2i.

c).log(3 − 3i) = log(3√ 

2) + i(−π4

+ 2kπ), k = 0,±1,±2, · · · .

d).ii = ei log(i) = ei(Log1+i(π

2+2kπ)) = e−(π2+2kπ), k = 0,±1,±2, · · · .

e).i1/4 = (eiπ

2 )1/4 = ei(π

2+2kπ

4 ), k = 0, 1, 2, 3.

2.Solution:

For this f , we find u(x, y) = ex2−y2 cos(2xy) and v(x, y) = ex2−y2 sin(2xy). Weverify the Cauchy-Riemann equations:

∂u

∂x= 2xex

2−y2 cos(2xy) − 2yex

2−y2 sin(2xy) =

∂v

∂y;

∂u

∂y= −2yex

2−y2 cos(2xy) − 2xex

2−y2 sin(2xy) = −∂v

∂x.

Hence f (z) is analytic, clearly it is defined for every z ∈ C, so it is entire.Moveover,

f (z) =∂u

∂x+ i

∂v

∂x

= 2xex2−y2 cos(2xy) − 2yex2−y2 sin(2xy) − (2yex2−y2 cos(2xy) + 2xex2−y2 sin(2xy))i.

3.Solution:Let f (z) = u(x, y) + iv(x, y), then f  = u(x, y)− iv(x, y). Both f  and f  are analytic,hence the pairs (u, v) and (u,−v) both satisfy Cauchy-Riemann equations, i.e.,

∂u

∂x=

∂v

∂y;

∂u

∂y= −∂v

∂x.

and∂u

∂x=

∂ (−v)

∂y;

∂v

∂y= −∂ (−v)

∂x.

It follows that each term is identically 0, consequently f (z) ≡ constant.

4.Solution:Using definition, the equation is equivalent to

eiz + e−iz

2= i

eiz − e−iz

2i.

Simply to get

e−iz = −e−iz.

Hence e−iz = 0. But there is no z ∈ C solving this. One easy way to see thisis that |e−iz| > 0 for any z. BTW, this is a useful fact to remember: the range of exponential function ez is all complex numbers except 0.

1

7/29/2019 Mid 1 Sample Solution

http://slidepdf.com/reader/full/mid-1-sample-solution 2/2

2

5.Solution:Notice first of all

log(2i − i) = Log1 + i( π2

+ 2kπ) = i( π2

+ 2kπ), k = 0,±1,±2, · · · .

So to have f (i) = π2

, we can just take the principal argument, i.e., we take

f (z) := Log(2z − i).

The domain of this branch is {z ∈ C|2z − i /∈ (−∞, 0]}, which is complex planewith a ray {x ≤ 0, y = 1

2} removed.

6.Solution:

Since w = 3zz−1

, by solving z, we get z = ww−3

. Therefore the image point wsatisfies

|w

w − 3 −2

|= 1.

Thus | ww−3

− 2(w−3)w−3

| = | 6−ww−3| = 1, i.e, |w − 6| = |w − 3|. Geometrically, this set

is all points equidistant to (6, 0) and (3, 0). Therefore it is a line x = 4.5.To show onto the line, it means each point on the line is being mapped. That is

clear: pick any point w on the line, z = ww−3 is its pre-image.

Remark: notice the map is not defined at z = 1, which is a point on the circle.So more precisely, the map maps the circle minus the point {z = 1} onto the line.