(1)SOLUTIONS FOR FINAL OF COMPLEX ANALYSIS 1

SOLUTIONS FOR FINAL OF COMPLEX ANALYSIS

1.

(1) Let

T(z) = (z−i)(1+i−(2−i))

(z−(2−i))(1+i−i) = (z−i)(2i−1) (z−2+i)

= S(w) = ^{(w−0)(1−∞)}_{(w−∞)(1−0)} =w Sow=f(z) = (z−i)(2i−1)

(z−2+i) . Thenf(0) = (0−i)(2i−1)

0−2+i = ⁻³⁻⁴ⁱ₅ .

(2) We know thati =e^iπ2. Hence,i^1/3 = ei(π/2+2kπ)/3 fork = 0,1,2. Thus the cubic roots are e^πi/6= ^√₂³ +₂ⁱ,e^5πi/6=−^√₂³+ ₂ⁱ,e^3πi/2 =−i.

2.

(1) sin(2i) = ^ei(2i)−e_2i⁻ⁱ⁽²ⁱ⁾ = ^e−2_2i^−e2 = ^(e2−e₂⁻²⁾i.

(2) Nowz(t) = 3e^itfor−^π₂ ≤t≤ ^π₂. Hence, Z

C

z dz = Z _π/2

−π/2

3e^−it(3ie^it)dt= Z _π/2

−π/2

9i dt= 9i hπ

2 −(−π 2)

i

= 9πi.

3.

(1) Nowtanz = _cos^sinz_z. We know ifcosz = 0, thenz = ^(2n+1)π₂ forn ∈ Zand all the zeros are simple. Hence, the functiontanzhas simple poles at ^(2n+1)π₂ forn∈Z.

(2) The poles oftanzin the region|z|<2πare−3π/2,−π/2,π/2,3π/2. Hence, I

|z|=2π

tanz dz = 2πi[Res(−3π/2) + Res(−2π/2) + Res(π/2) + Res(3π/2)]

= 2πi[(−1) + (−1) + (−1) + (−1)]

= −8πi.

WhereRes(−3π/2) = _(cos^sin_z)^z0

¯¯

¯z=−3π/2=−1.SimilarlyRes(−2π/2) = Res(π/2) = Res(3π/2) =

−1.

4.

(1) Sincefis analytic inside and onΓ, andz₀lies outsideΓ, we know that _(z−z^f(z)

0)² is analytic inside Γ. Hence,

1 2πi

Z

Γ

f(z)

(z−z₀)² dz= 0 (2)

G(z) = Z

C

ζ²−ζ+ 2

ζ−z dz = 2πi(z²−z+ 2) by Cauchy integral formula. Hence,G⁰(z) = 2πi(2z−1)and

G⁰(i) = 2πi(2i−1) =−2π(2 +i).

1

2 SOLUTIONS FOR FINAL OF COMPLEX ANALYSIS

5.

(1) Letg(z) = z⁶+ 5z² −1and letCbe the curve|z|= 1. Letf(z) = 5z²−1andh(z) =z⁶. The functionf(z)has exactly two zeros insideC. Now

|h(z)| = |z⁶|=|z|⁶= 1 on C

|f(z)| = |5z²−1| ≥5|z|²−1 = 4 on C

Hence,|h(z)| <|f(z)|forz ∈ C. Thusg(z) =f(z) +h(z) =z⁶ + 5z−1has exactly two zeros in|z|<1by Rouch´e theorem.

(2) We know that

e^1/z = X∞

j=0

1 j!z^−j.

Soz= 0is an essential singularity fore^1/z. Thuse^1/z is not a meromorphic function onC.

6.

(1) Letf(z)be the M¨obius transformation such thatf(2) = 0, f(1 +i) = 1, f(0) =∞. Hence, f(z) = (z−2)(1 +i)

z(1 +i−2) = (z−2)(1 +i)

z(i−1) = −i(z−2)

z .

(2) We know that there is a one-to-one functionf:D^∗ → {z ∈C| |z| ≤ 1}Andf is analytic on D^∗ by Riemann Mapping Theorem. Assumeg:D^∗ → Cis analytic and onto. Thengf⁻¹ is analytic and onto. Then we get a contradiction. Thus there are no analytic function that maps D^∗ontoC.

7.

(1) Letφ(z) =ALog|z−1|+B. ThenALog 1+B = 0andALog 2+B = 20. Hence,A= _{Log 2}²⁰ andB = 0. Therefore,φ(z) = _{Log 2}²⁰ Log|z−1|and

φ(x, y) = 20

Log 2Logp

(x−1)²+ (y−1)²= 10

Log 2Log £

(x−1)²+ (y−1)²¤ . (2) f(z) =P_∞

k=0(k³/3^k)z^k. Thenf⁰(z) =P_∞

k=1(k⁴/3^k)z^k−1. H

|z|=1 f(z)

z² dz = 2πif⁰(0) by Cauchy integral formula

= 2πi(1/3)

= ²₃πi 8.

(1) Letf(z) = ^cosz_z2 Thenf(_w¹) = w²cos_w¹. Hence,f(_w¹)has an essential singularity at w = 0.

Therefore,f(z)has an essential singularity at∞.

(2) Suppose thatf is an analytic function on C. Thenb f is analytic at∞. Thus it is bounded for

|z|> M. By continuity,f is also bounded for|z| ≤M. Hence,f is a bounded entire function.

Therefore,f is constant by Liouville’s theorem.

SOLUTIONS FOR FINAL OF COMPLEX ANALYSIS 3

9. Z ^π

2

0

1

1 + sin²θdθ= Z ^π

2

0

1

1 +^{1−cos 2θ}₂ dθ= Z ^π

2

0

2

3−cos 2θdθ= Z _π

0

1

3−cosθdθ.

LetI =R_π

0 1

3−cosθdθ=R_2π

π 1

3−cosθdθ. Then 2I =

Z _2π

0

1 3−cosθdθ

= Z

C

1

3− ¹₂(z+¹_z)iz dz where C :|z|= 1

= 2i Z

C

1

z²−6z+ 1dz

= 2i Z

C

1 [z−(3 + 2√

2)][z−(3−2√ 2)]dz

= 2i(2πi)Res [f(z),3−2√

2] where f(z) = 1

z²−6z+ 1

= −4π( 1

−4√ 2)

=

√2 2 π ThusR^π

02 1

1+sin²θdθ = ^√₄²π.

10. It is easy to see that ^sin_x^mx is an even function ofx. Now Z _∞

0

sinmx

x dx= 1 2

Z _∞

−∞

sinmx

x dx= 1 4i

µZ _∞

−∞

e^imx x dx−

Z _∞

−∞

e^−imx x dx

¶

and Z _∞

−∞

e^imx

x dx= lim

ρ→∞,r→0⁺

µZ _−r

−ρ

e^imx x dx+

Z _ρ

r

e^imx x dx

¶

It is known that ÃZ _−r

−ρ

+ Z

Sr

+ Z _ρ

r

+ Z

Cρ⁺

! e^imx

x dx= 0, andlim_ρ→∞R

C⁺ρ

e^imx

x dx= 0by Jordan’s Lemma. Hence, Z _∞

−∞

e^imx

x dx= lim

ρ→∞,r→0⁺

Ã

− Z

Sr

e^imx x dx−

Z

Cρ⁺

e^imx x dx

!

=−(−πi)Res (0)−0 =πi.

Similarly Z _∞

−∞

e^−imx

x dx=−πiRes (0) =−πi.

Thus Z _∞

0

sinmx

x dx= 1

4i(πi−(−πi)) = π 2.