8/11/2019 complex analysis Solution to Homework Assigment

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20 #6

z= a + R z =a R 2R

d(z, z) = 2 |z z|(1 + |z|2) (1 + |z|2)

= 2 |2R|

1 + |a|2 + |R|2 + 2

aR

1 + |a|2 + |R|2 2

aR

= 4 |R|

(1 + |a|2 + |R|2)2 4

aR

2

1

2d(z, z) =

2 |R|(1 + |a|2 + |R|2)

2 4

aR

2

28 #1

g(w) = r(w) +is(w) f(z) = u(z) +iv(z)

g (f(z)) = g(u, v)

= r (u(x, y), v(x, y)) +is (u(x, y), v(x, y))

g (f(z))

r

x

s

y =

r

u

u

x

s

u

u

y (1)

+ r

v

v

x

s

v

v

y (2)

g

f

(1) = r

u

u

x

s

u

u

y =

r

u

u

x

r

u

u

x= 0

(2) = r

v

v

x

s

v

v

y =

r

v

v

x

r

v

v

x

= 0

r

x

s

y = 0 =

r

x=

s

y

8/11/2019 complex analysis Solution to Homework Assigment

2/6

r

y+

s

x=

r

u

u

y+

s

u

u

x

(3)+

r

v

v

y+

s

v

v

x

(4) (3) (4)

r

y+

s

x= 0 =

r

y =

s

x

g(f(z))

28 #4

|f(z)|2 =u2 + v2 =c f(z) = u(x, y) + iv(x, y) f(z)

f(z) = 0 x

(f(z))2

2uu

x+ 2v

v

x= 0

f

uu

x v

u

y = 0

y

2uu

y 2v

v

y = 0

= u uy

+v ux

= 0

u vv u

uxuy

=

00

u2 + v2 = 0 u2 0 v2 0 u= v = 0 = f(z) = 0

u2 +v2 =c c

uxuy

= 00= f(z) = 0 = f(z)

8/11/2019 complex analysis Solution to Homework Assigment

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28 #5

f(z) = u(x, y) +iv(x, y) ux

= vy

uy

= vx u

x = (v)

(y) u

(y) = (v)x

u(x,y) iv(x,y) f(x,y) = f(z)

28 #7

u(z) = u(x+iy)

u

zz =

z

1

2

u

x+ i

u

y

= 1

4

2u

x2+i

2u

xy i

2u

xy i2

2u

y2

= 1

4 2u

x2+

2u

y2 0

u

= 0

32 33 #1

z4

z3 1=z +

z

z3 1

0 z3 1

H

0+

1

=

03 +2

3202 + 30+ 1

= 1

30+Q,

Q k k z3 1

Gk() = /3j

z4

z3 1=z +

3k=1

1

3k(z k).

1

z(z+ 1)2(z+ 2)3 .

8/11/2019 complex analysis Solution to Homework Assigment

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1 = 0 2 = 1 3 = 2 R(z)

1

1 1+ 12 1+ 13

= 6

(+ 1)2

(2+ 1)3

=

8

+ H1().

11 2

1 1

2 1

3 = 6

(2+ 1)(+ 1)2 =2 + 2+H2().

11 2

1 1

1

3 = 6

(1 2)(1 )2 =

3

2

52

4

17

8 +H3().

1

z(z+ 1)2(z+ 2)3 =

1

8z

1

(z+ 1)2+

2

(z+ 1)

1

2(z+ 2)3

5

4(z+ 2)3

17

8(z+ 2),

32 33 #2

Q 1, . . . , n P < n

limzk

(z k)P(z)

Q(z) = lim

zk

P(z) + (z k)P(z)

Q(z) =

P(k)

Q(k).

Q(k)= 0

P(z)

Q(z)

n

k=1P(k)

Q

(k)(z k) P < n P < n

P(z)

Q(z)=

nk=1

P(k)

Q(k)(z k).

32 33 #3

P(z) = Q(z)

nk=1

P(k)Q(k)(z k)

P(k) = ck

8/11/2019 complex analysis Solution to Homework Assigment

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P(z) = Q(z)

nk=1

ckQ(k)(z k)

z k z k Q(z) < n

< n S S(k) =ck P(z) S(z) < n

< n

P(i) = S(i) = n P(z) S(z) P(z) S(z) = 0 < n P(z) = S(z)

32 33 #4

R(z) z |z| = 1 1 = |R(z)| =R(z)R(z) = R(z)R(1/z) |z| = 1 z = 1/z R(z)R(1/z)1 R(1/z) = 1/R(z) =R(1/z) = 1/R(z) z n n 1/

32 33 #5

f(z) = g

1

z

.

f(z) = g(z)

f(z) z0 f(1/z0)

z0 f z0 1/z0

32 33 #6

R(z) =P(z)/Q(z) P p Q q R

n= max(p, q)

R(z) =P(z)Q(z) Q(z)P(z)

Q(z)2 .

8/11/2019 complex analysis Solution to Homework Assigment

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P, Q

R = (p 1) +q p+ (q 1)

2q = max(p+q 1, 2q) .

P Q

R =0 0 + (q 1)

2q = 2q .

P

Q

R = (p 1) + 0 0

2 0 = p 1 .

P, Q R = 0 R 2n q= n n 1 p= n, q= 0