MAT3705. Tutorial Letter 103/3/2015. Complex Analysis. Semesters 1 & 2. Department of Mathematical Sciences MAT3705/103/3/2015

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Tutorial Letter 103/3/2015

Complex Analysis

MAT3705

Semesters 1 & 2

Department of Mathematical Sciences

This tutorial letter contains the memorandum for the October 2012

examination paper.

MAT3705/103/3/2015

QUESTION 1 1.1

(z − 4

z + 4) = (z − 4)(z + 4) (z + 4)(z + 4)

= (z − 4)(z + 4)

|z + 4|²

= zz + 4z − 4z − 16

|z + 4|²

= |z|²− 16

|z + 4|² + 4(z − z)

|z + 4|²

Since z − z is purely imaginary (take z = x + iy then z − z = 2iy) we have that

Re(z − 4

z + 4) = |z|²− 16

|z + 4|² and if

|z|²− 16

|z + 4|² = 0 then |z| = 4.

1.2

Suppose that z = x + iy where x, y ∈ R Then

z + 4 z − 4    

< 1 ⇔ |z + 4| < |z − 4|

⇔ p

(x + 4)²+ y² <p

(x − 4)²+ y²

⇔ (x + 4)²+ y² < (x − 4)²+ y²

⇔ 8x < −8x

⇔ 16x < 0

⇔ x < 0

So the region is {x + iy | y ∈ R,x < 0 } which is the region left of the y=axis excluding the y=axis.

The set is open since it contains none of its boundary points.

MAT3705/103

QUESTION 2 2.1

sin z = i ⇐⇒ 1

2i(e^iz − e^−iz) = i ⇐⇒ e^iz− e^−iz = −2

⇔ (e^iz)²+ 2(e^iz) − 1 = 0 Substitute w = e^iz to obtain the equation w²+ 2w − 1 = 0.

Then

w = −2 ± √

4 + 4 2

= −1 ± 2√ 2 2

= −1 ±√ 2 so that

e^iz = −1 ±√ 2

⇔ iz = ln 

−1 +√ 2

+ i2nπ or

iz = ln  

−1 −√ 2

+ i(2n + 1)π

⇔ z = −i ln(√

2 − 1) + 2nπ or z = −i ln(√

2 + 1) + (2n + 1)π Then since

√ 1

2 − 1 =√ 2 + 1 we can write the answer as

z = i ln(√

2 + 1) + 2nπ or z = −i ln(√

2 + 1) + (2n + 1)π 2.2 For z = x + iy where x, y ∈ R we have

e^iz = e^i(x+iy) = e^−y+ix = e^−ye^ix = e^−y(cos x + i sin x) Hence

Re e^iz = e^−ycos x and

Re e^iz = e^−ycos x = 0

⇔ cos x = 0 and y ∈ R

⇔ x = ±π

2 + 2nπ, y ∈ R Hence

z = (π

2 + nπ) + iy, n ∈ Z, y ∈ R

From f (z) = (2x³+ y³− 7y − x) + i(x³− 2y³+ 5y + 4x), we have

u = 2x³+ y³− 7y − x, v = x³− 2y³+ 5y + 4x Then

u_x = 6x²− 1 v_x= 3x²+ 4 u_y = 3y²− 7 v_y == −6y²+ 5

These first partial derivatives are all continuous everywhere, so g will therefore be differentiable wherever the Cauchy-Riemann equations are satisfied.

(See the section on sufficient conditions for differentiability) .Now u_x = v_y ⇔ 6x²− 1 = −6y²+ 5

⇔ x²+ y² = 1 Similarly

uy = −vx⇔ 3y²− 7 = 3x²+ 4

⇔ x²+ y² = 1

f (z) is differentiable on the circle x²+ y² = 1 and if (x0, y0) satisfies x²₀+ y²₀ = 1 we get f⁰(x₀+ iy₀) = u_x(x₀+ iy₀) + iv_x(x₀+ iy₀)

= 6x²₀− 1 + i(3x²₀+ 4) which is also

= 6x²₀− 1 + i(3(1 − y²₀) + 4)

= 6x²₀− 1 + i(−3y₀²+ 7) or

f⁰(x₀+ iy₀) = u_y(x₀+ iy₀) + iv_y(x₀+ iy₀)

= 3y₀²− 7 + i(−6y₀²+ 5 ) which is also

= 3(1 − x²₀) − 7 + i(−6y²₀+ 5 )

= −3x²₀− 4 + i(−6y₀²+ 5 ) 3.2

We get

∂

∂x(e^usin v) = e^uu_xsin v + e^uv_xcos v and

∂²

∂x²(e^usin v) = e^u(u_x)²sin v + e^uu_xxsin v + e^uu_xv_xcos v + e^uu_xv_xcos v + e^uv_xxcos v − e^u(v_x)²sin v

MAT3705/103

Similarly

∂

∂y(e^usin v) = e^uu_ysin v + e^uv_ycos v

∂²

∂y²(e^usin v) = e^u(u_y)²sin v + e^uu_yysin v + e^uu_yv_ycos v + e^uu_yv_ycos v + e^uv_yycos v − e^u(v_y)²sin v Hence

∂²

∂x²(e^usin v) + ∂²

∂y²(e^usin v) = e^usin v((u_x)²− (v_y)²) + e^usin v(u_xx+ u_yy) + e^ucos v(v_xx+ v_yy) + 2e^ucos v(u_xv_x+ u_yv_y) + e^usin v((u_y)²− (v_x)²)

Since u + iv is analytic, it follows that u and v are harmonic and we get u_xx+ u_yy = 0,

vxx+ vyy = 0,

u_x = v_y and u_y = −v_x Hence also

(u_x)² = (v_y)² (u_y)² = (v_x)² and

u_xv_x = −u_yv_y Hence

∂²

∂x²(e^usin v) + ∂²

∂y²(e^usin v) = 0 and so e^usin v is harmonic.

QUESTION 4 4.1 We have that

|z|² = zz = (√

2)² = 2 Hence in the circle |z| =√

2 we get

z = 2 z hence

(z)² = 4

z² on the circle C : |z| =√ 2 Hence

Z

C

4 z²dz =

Z

C

z²dz

C

f (z)

(z − z0)ⁿ⁺¹dz = 2πi

n! f⁽ⁿ⁾(z₀) where C is the circle |z| = 3.

Hence z₀ = −1 and n = 3.

Clearly z₀ = −1 lies inside the circle |z| = 3.

Also

f (z) = e^2z, f⁰(z) = 2e^2z, f⁰⁰(z) = 4e^2z and f⁰⁰⁰(z) = 8e^2z Hence f⁰⁰⁰(−1) = 8e⁻² and so

Z

|z|=3

e^2z

(z + 1)⁴dz = 2πi 3! (8

e²) = 8πi 3e² 4.3

We have that f (z) is analytic in the domain |z| < 1 and |f (0)| = |1 + i| =√ 2.

Since |f (z)| ≤√

2 = |f (0)| for |z| < 1 and 0 lies in the domain |z| < 1, it follows from the Maximum Modulus Principle that f is constant in the domain |z| < 1.

Hence f (z) = 1 + i for |z| < 1 and so f⁰(z) = 0 for |z| < 1 and f⁰(0) = 0.

4.4

Since |g(z)| > |f (z)| ≥ 0 it follows that |g(z)| > 0 and so g(z) 6= 0 for every z ∈ C.

Since f and g are analytic on C, ^{f (z)}_g(z) is also analytic on C.

But   

f (z) g(z)

< 1 on C so by Louisville’s Theorem ^{f (z)}_g(z) is constant on C.

QUESTION 5 Partial fractions:

1

(z − 1)(z − 2) = A

z − 1 + B z − 2 i.e.

1 = A(z − 2) + B(z − 1) If z = 1 then A = −1; if z = 2 then B = 1.

Hence

1

(z − 1)(z − 2) = −1

z − 1 + 1 z − 2 Since 1 < |z| < 2 we get

¹

z

< 1 and ^z

2

 < 1.

Hence

−1

z − 1 = − 1

z(1 − ¹_z) = −1 z

∞

X

n=0

(1

z)ⁿ= −

∞

X

n=0

1

zⁿ⁺¹ valid for |z| > 1 Also

1

z − 2 = 1

−2(1 − ^z₂) = −1 2

∞

X

n=0

(z 2)ⁿ

= −

∞

X

n=0

zⁿ

2ⁿ⁺¹valid for |z| < 2

MAT3705/103

Hence

1

(z − 1)(z − 2) = −

∞

X

n=0

1 zⁿ⁺¹ −

∞

X

n=0

zⁿ

2ⁿ⁺¹ valid for 1 < |z| < 2

= ... − 1 z⁴ − 1

z³ − 1 z² − 1

z − 1 2− z

2² −z² 2³ − ...

QUESTION 6

6.1 z²− 1

z²− 5iz − 4 = z²− 1 (z − i)(z − 4i) Hence we have isolated singularities at z = i and z = 4i.

Let p(z) = z²− 1 and q(z) = z²− 5iz − 4.

Since p(i) 6= 0, p(4i) 6= 0 , q⁰(i) = 2i − 5i 6= 0, q⁰(4i) = 8i − 5i 6= 0, both of these singularities are simple poles.

Hence

Res_z=ip(z)

q(z) = p(i)

q⁰(i) = i²− 1

2i − 5i = −2

−3i = −2i

3 and

Res_z=4ip(z)

q(z) = p(4i)

q⁰(4i) = (4i)²− 1

2(4i) − 5i = −17

3i = 17i 3 and 6.2.1

Both i and 4i lie inside the circle |z| = 6.

Hence

Z

|z|=6

z²− 1

z²− 5iz − 4dz = 2πi(17i 3 − 2i

3) = 2πi(15i

3 ) = −10π 6.2.2

Since |4i − i| = 3 > 1 and |i − i| = 0 < 1 only the singularity at z = i lies inside the circle |z − i| = 1.

Hence

Z

|z−i|=1

z²− 1

z²− 5iz − 4dz = 2πi(−2i

3) = 4π 3 QUESTION 7.

7.1 We use the parametric representation z = e^iθ(0 ≤ θ ≤ 2π) to describe C . Furthermore ^dz_dθ = ie^iθ = iz i.e. dθ = ^dz_iz

and from cos θ = ¹₂(e^iθ+ e^−iθ),we obtain cos θ = ¹₂(z + z⁻¹)

Z 2π 0

dθ

3 − 2 cos θ = Z

|z|=1

1

3 − 2 ¹₂(z + z⁻¹)
1 izdz

= Z

|z|=1

dz iz(3 − z − ¹_z)

= Z

|z|=1

dz i(3z − z²− 1)

= Z

|z|=1

idz z²− 3z + 1

z = 3 ± 9 − 4

2 = 3 ± 5

2 Note that ³⁺

√ 5

2 > 1 so this singularity lies outside |z| = 1) Then

Res 3−√ 5 2

i

z²− 3z + 1 = = i 2(³⁻

√ 5

2 ) − 3 = − i

√5 Therefore by the residue theorem we get

Z 2π 0

dθ

3 − 2 cos θ = Z

|z|=1

idz z²− 3z + 1

= 2πi (− i

√5)

= 2π

√5

7.2 Question: Show that f (z) − z⁴ has exactly four solutions inside the unit circle under the condition that |f (z)| < 1 for |z| = 1.

Let

g(z) = −z⁴

Then g(z) is analytic inside and on |z| = 1 and has 4 zeros (counting multiplicities) inside |z| = 1.

Also

|g(z)| = −z⁴

 = 1 > |f (z)| on |z| = 1.

Thus by Rouche’s theorem g(z) and f (z) + g(z) = f (z) − z⁴ have the same number of zeros inside

|z| = 1

i.e. f (z) − z⁴ has exactly four solutions inside the unit circle.