Math 208 Solutions to Homework Assignment 2

p37. 1. (a)

f (z) =

1

z

2

+1

is de…ned everywhere except for the points which satisfy z

2

+ 1 = 0; that is,

all the points except for z =

i:

(b)

f (z) = Arg

1
z

is de…ned everywhere except for z = 0.

(c)

f (z) =

z

z+z

is de…ned everywhere except for z + z = 0, or Re(z) = 0 or the axis of real numbers.

(d)

f (z) =

1

z

2

+1

is de…ned everywhere 1

jzj

2

= 0; or jzj

2

= 1; since e

i

= jcos + i sin j = 1;

jzj

2

= re

i

2

= jrj

2

= 1 and jzj > 0; we get r = 1 which implies the unit circle on the complex plane.

3.

Suppose f (z) = x

2

y

2

2y + i(2x

2xy); where z = x + iy: Using the expressions

x = Re (z) =

z + z

2

and y = Im (z) =

z

z

2i

we will obtain f (z) as

f (z)

=

x

2

y

2

2y + i(2x

2xy)

=

z + z

2

2

z

z

2i

2

2

z

z

2i

+ i 2

z + z

2

2

z + z

2

z

z

2i

=

(z + z)

2

+ (z

z)

2

2

+ i (z

z) + i (z + z)

z

2

2

+

z

2

2

=

z

2

2

+

z

2

2

+ 2iz

z

2

2

+

z

2

2

=

z

2

+ 2iz

4.

Let f (z) = z +

1
z

(z 6= 0) be given. We want to write it in the form f(z) = u (r; ) + iv (r; ) : Using

…rst z = re

i

then e

i

= cos + i sin

f (z)

=

z +

1
z

= z + z

1

= re

i

+ r

1

e

i

=

r (cos + i sin ) +

1
r

(cos (

) + i sin (

))

=

r (cos + i sin ) +

1
r

(cos

i sin )

=

r cos + ir sin +

1
r

cos

i

1
r

sin

=

r cos +

1
r

cos + ir sin

i

1
r

sin

=

r +

1
r

cos + i r

1
r

sin

p44. 2.

Consider the hyperbolas

x

2

y

2

=

c

1

;

(c

1

< 0) ;

2xy = c

2

;

(c

2

< 0)

x

2

=

c

1

+ y

2

;

x =

c

2

2y

x

=

p

c

1

+ y

2

;

x =

c

2

2y

If we write them with complex coordinates z = x + iy

z =

p

c

1

+ y

2

+ iy;

z =

c

2

2y

+ iy

1

then apply the transformation w = z

2

; they would be mapped to

w

=

z

2

= c

1

+ y

2

y

2

+ i2y

p

c

1

+ y

2

;

w = z

2

=

c

2

2y

2

y

2

+ i2

c

2

2y

y

w

=

c

1

+ i2y

p

c

1

+ y

2

;

( 1 < y < 1) ;

w =

(c

2

)

2

4y

2

y

2

!

+ ic

2

;

( 1 < y < 0)

we can sketch these hyperbolas and their images under the transformation as below,

w = z

2

!

x

2

y

2

= c

1

;

(c

1

< 0)

w = c

1

+ i2y

p

c

1

+ y

2

;

( 1 < y < 1)

w = z

2

!

2xy = c

2

;

(c

2

< 0)

w =

(c

2

)

2

4y

2

y

2

+ ic

2

;

( 1 < y < 0)

2

3.

Consider the region given by the sector r

1;

0

4

:

(a)

It will be mapped by w = z

2

onto the region r

1 ) r

2

r

1;

0

2

4

=

2

:

3

(b)

It will be mapped by w = z

3

onto the region r

1 ) r

3

r

1;

0

3

4

=

3

4

:

(c)

It will be mapped by w = z

4

onto the region r

1 ) r

4

r

1;

0

4

4

= :

4

7.

Consider the semi-in…nite strip x

0; 0

y

and to …nd its image under the transformation w = e

z

let’s check where its boundary is mapped to:

x = 0;

0

y

) w = e

0+iy

= cos y + i sin y;

0

y

the upper half unit circle;

x

0;

y = 0 ) w = e

x+i0

= e

x

+ i0;

x

0

the half ray [1; 1) which lies on the x-axis;

x

0;

y =

) w = e

x+i

= e

x

(cos

+ i sin ) =

e

x

+ i0;

x

0

the half ray (1; 1] which lies on the x-axis.

Finally a point in this region is z = ln 2 + i

2

and it’s mapped to w = e

ln 2+i

2

= 2 cos

2

+ i sin

2

= 2i

which means the strip will be mapped to the region which lies on the upper side of this image of the boundary.
Thus the image of the semi-in…nite strip will be the following region.

p55. 5.

Consider the function

f (z) =

z
z

2

:

We want to show that the limit of this function as z tends to 0 does not exist. By calculus, if the limit exists
then it should be independent of the choice of path, so let’s approach to 0 with z = x + i0 …rst.

lim

(x;0)

!(0;0)

x + iy

x + iy

2

=

lim

(x;0)

!(0;0)

x + i0
x

i0

2

=

x
x

2

= 1

2

= 1

5

then let’s approach to 0 with z = x + ix; then we have

lim

(x;x)

!(0;0)

x + ix

x + ix

2

=

x + ix
x

ix

2

=

x (1 + i)
x (1

i)

2

=

(1 + i)
(1

i)

2

=

(1 + i) (1 + i)
(1

i) (1 + i)

2

=

2i

2

2

=

1

choosing two di¤erent paths gives two di¤erent limits, which means the limit does not exist.

10. (a)

If we use theorem (2) of section 17

lim

z

!0

4

1
z

2

1
z

1

2

= lim

z

!0

4

(1

z)

2

=

4

(1

0)

2

= 4

Thus lim

z

!1

4z

2

(z

1)

2

= 4:

(b)

If we use theorem (1) of section 17

lim

z

!1

1

1

(z

1)

3

= lim

z

!1

(z

1)

3

= (1

1)

3

= 0

Thus lim

z

!1

1

(z

1)

3

= 1:

(c)

If we use theorem (3) of section 17

lim

z

!1

1

(

1
z

)

2

+1

(

1
z

)

1

= lim

z

!0

1

1

z2

+1

1

z

z

= lim

z

!0

1
z

1

1

z

2

+ 1

= lim

z

!0

1 z

z

1+z

2

z

2

= lim

z

!0

z

2

(1

z)

(1 + z

2

) z

= lim

z

!0

z (1

z)

(1 + z

2

)

=

0 (1

0)

(1 + 0)

=

0
1

= 0

Thus lim

z

!1

z

2

+1

z

1

= 1:

6