Unit 5: Part B Solutions (Complex Variables & Transforms)

Solution

Step 1: Identify the function and singularity.
The integrand is f(z) / (z-a) where f(z) = 3z²+7z+1 and a = -1.
The singularity is at z = -1.

Step 2: Locate the singularity relative to the contour.
The contour C is the circle |z| = 1/2, which is a circle centered at the origin (0,0) with a radius of 0.5.
The singularity z = -1 is at a distance of |-1| = 1 from the origin.
Since 1 > 1/2, the singularity z = -1 lies outside the contour C.

Step 3: Apply Cauchy's Integral Theorem.
The function (3z²+7z+1)/(z+1) is analytic everywhere inside the contour C.
By Cauchy's Integral Theorem, if a function is analytic everywhere inside and on a closed contour, the integral around that contour is zero.

Solution

Step 1: Identify f(z), singularity, and contour.
This is in the form ∫ f(z)/(z-a) dz.
f(z) = z
Singularity: a = 2
Contour C: |z-2|=3/2. This is a circle centered at z=2 with a radius of 1.5.

Step 2: Locate the singularity.
The singularity a = 2 is at the center of the contour, so it is inside C.

Step 3: Apply Cauchy's Integral Formula.
Cauchy's Integral Formula states: ∫_C f(z)/(z-a) dz = 2πi · f(a).
Here, a = 2 and f(z) = z, so f(a) = f(2) = 2.
Integral = 2πi · f(2) = 2πi · (2)

Solution

Step 1: Identify f(z), singularities, and contour.
f(z) = sin(πz²) + cos(πz²)
Singularities are at z = 1 and z = 2.
Contour C: |z|=3. This is a circle centered at the origin with radius 3.
Both singularities (z=1 and z=2) are inside the contour.

Step 2: Use Partial Fractions.
1/((z-1)(z-2)) = A/(z-1) + B/(z-2)
1 = A(z-2) + B(z-1)

• If z=1: 1 = A(-1) → A = -1
• If z=2: 1 = B(1) → B = 1

So, the integral becomes:
∫_C [ -f(z)/(z-1) + f(z)/(z-2) ] dz

Step 3: Apply Cauchy's Integral Formula to both parts.
Integral = -∫_C f(z)/(z-1) dz + ∫_C f(z)/(z-2) dz
Integral = -[ 2πi · f(1) ] + [ 2πi · f(2) ] = 2πi · [f(2) - f(1)]
f(1) = sin(π(1)²) + cos(π(1)²) = sin(π) + cos(π) = 0 - 1 = -1
f(2) = sin(π(2)²) + cos(π(2)²) = sin(4π) + cos(4π) = 0 + 1 = 1
Result = 2πi · [1 - (-1)] = 2πi · (2)

Solution

Step 1: Identify singularities and contour.
Singularities are at z = 0, z = 1, and z = 2.
Contour C: |z|=3/2 = 1.5.
Singularities inside C: z = 0, z = 1.
Singularity outside C: z = 2.

Step 2: Reformulate the integral.
We must group the "outside" singularity with f(z).
Let f(z) = (4-3z) / (z-2).
The integral becomes: ∫_C f(z) / (z(z-1)) dz

Step 3: Use Partial Fractions on the remaining denominator.
1/(z(z-1)) = A/z + B/(z-1)
1 = A(z-1) + B(z)

• If z=0: 1 = A(-1) → A = -1
• If z=1: 1 = B(1) → B = 1

The integral is: ∫_C [ -f(z)/z + f(z)/(z-1) ] dz

Step 4: Apply Cauchy's Integral Formula.
Integral = -[ 2πi · f(0) ] + [ 2πi · f(1) ] = 2πi · [f(1) - f(0)]
f(0) = (4 - 3(0)) / (0 - 2) = 4 / -2 = -2
f(1) = (4 - 3(1)) / (1 - 2) = 1 / -1 = -1
Result = 2πi · [-1 - (-2)] = 2πi · (1)

Solution

Step 1: State the Taylor's series formula.
f(z) = f(a) + f'(a)(z-a)/1! + f''(a)(z-a)²/2! + f'''(a)(z-a)³/3! + ...
Here, a = π/4.

Step 2: Find the derivatives and evaluate at a = π/4.

• f(z) = sin z    →  f(π/4) = sin(π/4) = 1/√2
• f'(z) = cos z    →  f'(π/4) = cos(π/4) = 1/√2
• f''(z) = -sin z   →  f''(π/4) = -sin(π/4) = -1/√2
• f'''(z) = -cos z  →  f'''(π/4) = -cos(π/4) = -1/√2
• f⁽⁴⁾(z) = sin z   →  (The pattern repeats)

Step 3: Substitute the values into the formula.
sin z = (1/√2) + (1/√2)(z-π/4)/1! - (1/√2)(z-π/4)²/2! - (1/√2)(z-π/4)³/3! + ...

Solution

Step 1: Use Partial Fractions.
f(z) = 1/((z+2)(z+3)) = A/(z+2) + B/(z+3)
1 = A(z+3) + B(z+2)

• If z = -2:   1 = A(1) → A = 1
• If z = -3:   1 = B(-1) → B = -1

f(z) = 1/(z+2) - 1/(z+3)

Step 2: Rearrange terms for the region |z| < 2.
The region |z| < 2 implies |z/2| < 1 and |z/3| < 1. We must use the (1+r)^-1 form.
Term 1: 1/(z+2) = 1 / (2(1 + z/2)) = (1/2) · (1 + z/2)^-1
Term 2: -1/(z+3) = -1 / (3(1 + z/3)) = (-1/3) · (1 + z/3)^-1

Step 3: Expand using the geometric series (1+r)^-1 = 1 - r + r² - ... = Σ (-1)ⁿrⁿ.
Term 1 = (1/2) Σ_n=0^∞ (-1)ⁿ (z/2)ⁿ = Σ_n=0^∞ (-1)ⁿ zⁿ / 2ⁿ⁺¹
Term 2 = (-1/3) Σ_n=0^∞ (-1)ⁿ (z/3)ⁿ = Σ_n=0^∞ (-1)ⁿ⁺¹ zⁿ / 3ⁿ⁺¹

Step 4: Combine the series.
f(z) = Σ_n=0^∞ [ (-1)ⁿ / 2ⁿ⁺¹ + (-1)ⁿ⁺¹ / 3ⁿ⁺¹ ] zⁿ
f(z) = Σ_n=0^∞ (-1)ⁿ [ 1/2ⁿ⁺¹ - 1/3ⁿ⁺¹ ] zⁿ

Solution

Step 1: Use Long Division and Partial Fractions.
First, divide (z²-1) by (z²+5z+6) (which is (z+2)(z+3)).
(z²-1) / (z²+5z+6) = 1 + (-5z-7) / (z²+5z+6)
Now use partial fractions on the remainder:
(-5z-7) / ((z+2)(z+3)) = A/(z+2) + B/(z+3)
-5z-7 = A(z+3) + B(z+2)

• If z = -2:   10-7 = A(1) → A = 3
• If z = -3:   15-7 = B(-1) → B = -8

f(z) = 1 + 3/(z+2) - 8/(z+3)

Step 2: Rearrange for the region 2 < |z| < 3.
This region means |z| > 2 (so |2/z| < 1) and |z| < 3 (so |z/3| < 1).
Term 1: 1 (this is already a valid series)
Term 2: 3/(z+2). Since |z| > 2, we factor out z:
3 / (z(1 + 2/z)). Use (1+r)^-1 with r = 2/z.
= (3/z) · (1 - 2/z + (2/z)² - ...) = Σ_n=0^∞ 3 · (-1)ⁿ · 2ⁿ / zⁿ⁺¹
Term 3: -8/(z+3). Since |z| < 3, we factor out 3:
-8 / (3(1 + z/3)). Use (1+r)^-1 with r = z/3.
= (-8/3) · (1 - z/3 + (z/3)² - ...) = Σ_n=0^∞ (-8/3) · (-1)ⁿ · (z/3)ⁿ

Solution

Part (i): f(z) = z²
Let z = x + iy.
f(z) = (x+iy)² = (x² - y²) + i(2xy)
Here, u(x,y) = x² - y² and v(x,y) = 2xy.
Check Cauchy-Riemann (C-R) equations:
1. ∂u/∂x = 2x
2. ∂v/∂y = 2x
   ⇒ ∂u/∂x = ∂v/∂y

3. ∂u/∂y = -2y
4. ∂v/∂x = 2y
   ⇒ ∂u/∂y = -∂v/∂x
Since the C-R equations are satisfied for all (x,y) and the partial derivatives are continuous everywhere, f(z)=z² is analytic everywhere (entire).

Part (ii): f(z) = e^z
Let z = x + iy.
f(z) = e^x+iy = e^x · e^iy = e^x(cos y + i sin y)
Here, u(x,y) = e^xcos y and v(x,y) = e^xsin y.
Check C-R equations:
1. ∂u/∂x = e^xcos y
2. ∂v/∂y = e^xcos y
   ⇒ ∂u/∂x = ∂v/∂y

3. ∂u/∂y = -e^xsin y
4. ∂v/∂x = e^xsin y
   ⇒ ∂u/∂y = -∂v/∂x
Since the C-R equations are satisfied for all (x,y) and the partial derivatives are continuous everywhere, f(z)=e^z is analytic everywhere (entire).
Derivative: f'(z) = ∂u/∂x + i ∂v/∂x
f'(z) = (e^xcos y) + i(e^xsin y) = e^x(cos y + i sin y) = e^z.

Solution

Part (i): C-R Equations
f(z) = u + iv, where u = x + ay and v = bx + cy.
For f(z) to be analytic, it must satisfy the C-R equations.
1. ∂u/∂x = ∂v/∂y
    ∂/∂x(x + ay) = ∂/∂y(bx + cy)
    1 = c
2. ∂u/∂y = -∂v/∂x
    ∂/∂y(x + ay) = -∂/∂x(bx + cy)
    a = -b

Part (ii): Proof
Let |f(z)|² = u² + v². The operator (∂²/∂x² + ∂²/∂y²) is the Laplacian ∇².
LHS = ∇²(u² + v²)
∂/∂x (u² + v²) = 2u(∂u/∂x) + 2v(∂v/∂x)
∂²/∂x² (u² + v²) = 2 [ (∂u/∂x)² + u(∂²u/∂x²) + (∂v/∂x)² + v(∂²v/∂x²) ]
By symmetry (swapping x and y):
∂²/∂y² (u² + v²) = 2 [ (∂u/∂y)² + u(∂²u/∂y²) + (∂v/∂y)² + v(∂²v/∂y²) ]
Adding these two expressions gives the LHS:
LHS = 2 [ (∂u/∂x)² + (∂u/∂y)² + (∂v/∂x)² + (∂v/∂y)² + u(∇²u) + v(∇²v) ]
Since f(z) is analytic, u and v are harmonic, which means ∇²u = 0 and ∇²v = 0.
LHS = 2 [ (∂u/∂x)² + (∂u/∂y)² + (∂v/∂x)² + (∂v/∂y)² ]
Now apply C-R equations: ∂u/∂y = -∂v/∂x and ∂v/∂y = ∂u/∂x
LHS = 2 [ (∂u/∂x)² + (-∂v/∂x)² + (∂v/∂x)² + (∂u/∂x)² ]
LHS = 2 [ 2(∂u/∂x)² + 2(∂v/∂x)² ] = 4 [ (∂u/∂x)² + (∂v/∂x)² ]
RHS = 4|f'(z)|²
f'(z) = ∂u/∂x + i ∂v/∂x
|f'(z)|² = (∂u/∂x)² + (∂v/∂x)²
RHS = 4 [ (∂u/∂x)² + (∂v/∂x)² ]
Since LHS = RHS, the identity is proved.

Solution

Step 1: Relate the expression to a known complex function.
Consider the function g(z) = tan(z).
tan(z) = sin(z)/cos(z) = sin(x+iy) / cos(x+iy)
tan(z) = (sin(x)cosh(y) + i cos(x)sinh(y)) / (cos(x)cosh(y) - i sin(x)sinh(y))

Step 2: Rationalize the denominator.
Multiply numerator and denominator by the conjugate of the denominator, (cos(x)cosh(y) + i sin(x)sinh(y)).
Denominator = cos²(x)cosh²(y) + sin²(x)sinh²(y)
   = cos²(x)cosh²(y) + (1-cos²(x))(cosh²(y)-1)
   = cos²(x)cosh²(y) + cosh²(y) - 1 - cos²(x)cosh²(y) + cos²(x)
   = cosh²(y) + cos²(x) - 1 = ( (1+cosh(2y))/2 ) + ( (1+cos(2x))/2 ) - 1
   = (1/2) [ cosh(2y) + cos(2x) ]
Numerator (Real Part) = (sin x cosh y)(cos x cosh y) - (cos x sinh y)(sin x sinh y)
   = sin(x)cos(x) [ cosh²(y) - sinh²(y) ]
   = (1/2)sin(2x) · (1) = (1/2)sin(2x)

Step 3: Find the real part, u.
u = Real(tan(z)) = (Numerator Real Part) / Denominator
u = ( (1/2)sin(2x) ) / ( (1/2)(cosh(2y) + cos(2x)) )
u = sin(2x) / (cosh(2y) + cos(2x))

Step 4: Conclude the function.
This perfectly matches the given 'u'.
Therefore, the analytic function is f(z) = tan(z) + C (where C is a real constant, or Ci for an imaginary constant).

Solution

Part (i): Harmonic Check and Milne-Thomson
1. Check if v is harmonic (v_xx + v_yy = 0):
∂v/∂x = e^-2y(-2y sin(2x) + sin(2x) + 2x cos(2x))
∂²v/∂x² = e^-2y(-4y cos(2x) + 2cos(2x) + 2cos(2x) - 4x sin(2x))
∂²v/∂x² = e^-2y(4cos(2x) - 4y cos(2x) - 4x sin(2x))

∂v/∂y = e^-2y(cos(2x)) - 2e^-2y(y cos(2x) + x sin(2x))
∂v/∂y = e^-2y(cos(2x) - 2y cos(2x) - 2x sin(2x))
∂²v/∂y² = e^-2y(-2cos(2x)) - 2e^-2y(cos(2x) - 2y cos(2x) - 2x sin(2x))
∂²v/∂y² = e^-2y(-4cos(2x) + 4y cos(2x) + 4x sin(2x))
Adding them: v_xx + v_yy = e^-2y [ (4cos 2x - 4y cos 2x - 4x sin 2x) + (-4cos 2x + 4y cos 2x + 4x sin 2x) ] = 0.
Yes, v is harmonic.
2. Find f(z) using Milne-Thomson:
f'(z) = ∂u/∂x + i ∂v/∂x = ∂v/∂y + i ∂v/∂x
Replace x=z, y=0:
∂v/∂y(z,0) = e⁰(cos(2z) - 0 - 2z sin(2z)) = cos(2z) - 2z sin(2z)
∂v/∂x(z,0) = e⁰(0 + sin(2z) + 2z cos(2z)) = sin(2z) + 2z cos(2z)
f'(z) = (cos(2z) - 2z sin(2z)) + i(sin(2z) + 2z cos(2z))
f'(z) = (cos(2z) + i sin(2z)) + 2iz cos(2z) - 2z sin(2z)
f'(z) = e^i2z + 2iz (cos(2z) + i sin(2z))    (since -1 = i²)
f'(z) = e^i2z + 2iz (e^i2z) = (1 + 2iz)e^i2z
This is the derivative of (z · e^i2z) by the product rule.
f(z) = ∫ (1 + 2iz)e^i2z dz = z e^i2z + C

Part (ii): f(z) from u
u = (1/2)log(x²+y²).
In polar coordinates, z = re^iθ, x²+y² = r².
u = (1/2)log(r²) = log(r).
We know the analytic function f(z) = log(z) is defined as:
log(z) = log(re^iθ) = log(r) + log(e^iθ) = log(r) + iθ
The real part (u) of log(z) is log(r), which matches the given u.

Solution

Part (i): Analytic Function from u+v
1. Simplify the given expression V = u+v.
e^2y+e^-2y = 2cosh(2y)
V = u+v = 2sin(2x) / (2cosh(2y) - 2cos(2x)) = sin(2x) / (cosh(2y) - cos(2x))
2. Relate V to a known function.
Consider g(z) = cot(z) = cos(z)/sin(z)
cot(z) = (cos(x+iy)) / (sin(x+iy)) = (cos x cosh y - i sin x sinh y) / (sin x cosh y + i cos x sinh y)
Multiply by the conjugate (sin x cosh y - i cos x sinh y):
Denominator = sin²x cosh²y + cos²x sinh²y = (1/2)(cosh(2y) - cos(2x))
Numerator (Real part) = (cos x cosh y)(sin x cosh y) + (sin x sinh y)(cos x sinh y)
   = (sin x cos x)(cosh²y + sinh²y) = (1/2)sin(2x)(cosh(2y))
Numerator (Imaginary part) = ... = -(1/2)sinh(2y)
So, cot(z) = ( (1/2)sin 2x cosh 2y ) / ( (1/2)(cosh 2y - cos 2x) ) + i · ( -(1/2)sinh 2y ) / ( (1/2)(cosh 2y - cos 2x) )
The real part of cot(z) is u_cot = sin(2x)cosh(2y) / (cosh(2y) - cos(2x)).
The imaginary part of cot(z) is v_cot = -sinh(2y) / (cosh(2y) - cos(2x)).
Our given V = u+v = sin(2x) / (cosh(2y) - cos(2x)). This is not u_cot or v_cot.
3. Use the F(z) = (1+i)f(z) method.
Let F(z) = U + iV = (1+i)f(z) = (1+i)(u+iv) = (u-v) + i(u+v).
We are given V = u+v. So, V = sin(2x) / (cosh(2y) - cos(2x)).
This V is exactly the real part of cot(z) (u_cot), *except* for the cosh(2y) term in the numerator.

There is a high probability of a typo in the question. The standard problem V = u+v = (sin 2x - sinh 2y) / (cosh 2y - cos 2x) leads to f(z) = (1/2)(1+i)cot(z). Assuming the question intended V = sin(2x) / (cosh(2y) - cos(2x)), which is u_cot.

Assuming V = u_cot:
F(z) = U + iV = U + i(u_cot).
We want F(z) to be analytic. F'(z) = U_x + iV_x = V_y - iU_y.
C-R equations for F: U_x = V_y = u_cot,y and U_y = -V_x = -u_cot,x.
Consider g(z) = cot(z) = u_cot + iv_cot. C-R for g: u_cot,x = v_cot,y and u_cot,y = -v_cot,x.
So, U_x = -v_cot,x and U_y = -v_cot,y. This implies U = -v_cot.
F(z) = U + iV = -v_cot + i u_cot = i(u_cot + i v_cot) = i · cot(z).
(1+i)f(z) = i · cot(z)
f(z) = [ i / (1+i) ] cot(z) = [ i(1-i) / 2 ] cot(z) = (1+i)/2 · cot(z).

Part (ii): Bilinear Map
z₁=0, z₂=-1, z₃=i
w₁=1, w₂=0, w₃=∞
Use the cross-ratio formula: (w-w₁)(w₂-w₃) / ((w-w₃)(w₂-w₁)) = (z-z₁)(z₂-z₃) / ((z-z₃)(z₂-z₁))
Since w₃=∞, the (w₂-w₃) / (w-w₃) part simplifies to -1.
LHS = (w - w₁) / (-1 · (w₂-w₁)) = (w - 1) / (-(0 - 1)) = (w - 1) / 1 = w - 1.
RHS = (z - 0)(-1 - i) / ((z - i)(-1 - 0)) = z(-1-i) / (-(z-i)) = z(1+i) / (z-i)
Equating: w - 1 = z(1+i) / (z-i)
w = 1 + z(1+i) / (z-i) = ( (z-i) + z(1+i) ) / (z-i)
w = (z - i + z + zi) / (z-i) = (2z + zi - i) / (z-i)

Solution

Part (i): Cauchy's Formula
Singularities of z²+2z+5=0 are z = [-2 ± √(4-20)]/2 = -1 ± 2i.
Let a = -1+2i and b = -1-2i.
Contour 1: C₁ is |z+1+i|=2. (Center: -1-i, Radius: 2)
Distance to a: |-1+2i + 1+i| = |3i| = 3. (Outside C₁)
Distance to b: |-1-2i + 1+i| = |-i| = 1. (Inside C₁)
Rewrite integral: ∫_C1 [ (z+4)/(z-a) ] / (z-b) dz
Let f(z) = (z+4)/(z-a) = (z+4)/(z+1-2i).
By CIF: 2πi · f(b) = 2πi · f(-1-2i)
= 2πi · ((-1-2i)+4) / ((-1-2i) - (-1+2i))
= 2πi · (3-2i) / (-4i) = (-π/2) · (3-2i) = -3π/2 + πi
Contour 2: C₂ is |z+1|=1. (Center: -1, Radius: 1)
Distance to a: |-1+2i + 1| = |2i| = 2. (Outside C₂)
Distance to b: |-1-2i + 1| = |-2i| = 2. (Outside C₂)
Since both singularities are outside C₂, the integrand is analytic inside.
By Cauchy's Theorem, the integral is 0.

Part (ii): Cauchy's Formula for Derivatives
Singularities of (z²+2z+4)²=0 are z = [-2 ± √(4-16)]/2 = -1 ± i√3.
Let a = -1+i√3 and b = -1-i√3.
Contour C: |z+1+i|=2. (Center: -1-i, Radius: 2)
Distance to a: |-1+i√3 + 1+i| = |i(1+√3)| ≈ 2.73. (Outside C)
Distance to b: |-1-i√3 + 1+i| = |i(1-√3)| = √3-1 ≈ 0.73. (Inside C)
Rewrite integral: ∫_C [ (z+1)/(z-a)² ] / (z-b)² dz
Let f(z) = (z+1)/(z-a)². The integral is ∫ f(z)/(z-b)² dz.
This matches the formula ∫ f(z)/(z-a)ⁿ⁺¹ dz = 2πi/n! · f⁽ⁿ⁾(a), with n=1.
Result = 2πi/1! · f'(b).
f'(z) = d/dz [ (z+1) / (z-a)² ]
f'(z) = [ 1(z-a)² - (z+1)2(z-a) ] / (z-a)⁴
f'(z) = [ (z-a) - 2(z+1) ] / (z-a)³ = (-z - a - 2) / (z-a)³
Evaluate f'(b):
Numerator: -b - a - 2 = -(-1-i√3) - (-1+i√3) - 2 = (1+i√3 + 1-i√3) - 2 = 2 - 2 = 0.
Since the numerator is 0, f'(b) = 0.

Solution

Step 1: Use Partial Fractions.
f(z) = (2z-1)/((z+2)(z+3)) = A/(z+2) + B/(z+3)
2z-1 = A(z+3) + B(z+2)

• If z = -2:   2(-2)-1 = A(1) → -5 = A
• If z = -3:   2(-3)-1 = B(-1) → -7 = -B → B = 7

f(z) = 7/(z+3) - 5/(z+2)

Step 2: Rearrange for the region |z| < 2.
The region |z| < 2 implies |z/2| < 1 and |z/3| < 1.
Term 1: 7/(z+3) = 7 / (3(1 + z/3)) = (7/3) · (1 + z/3)^-1
Term 2: -5/(z+2) = -5 / (2(1 + z/2)) = (-5/2) · (1 + z/2)^-1

Step 3: Expand using the geometric series Σ (-1)ⁿrⁿ.
Term 1 = (7/3) Σ_n=0^∞ (-1)ⁿ (z/3)ⁿ = Σ_n=0^∞ 7 · (-1)ⁿ · zⁿ / 3ⁿ⁺¹
Term 2 = (-5/2) Σ_n=0^∞ (-1)ⁿ (z/2)ⁿ = Σ_n=0^∞ -5 · (-1)ⁿ · zⁿ / 2ⁿ⁺¹

Step 4: Combine the series.
f(z) = Σ_n=0^∞ [ (7 · (-1)ⁿ / 3ⁿ⁺¹) - (5 · (-1)ⁿ / 2ⁿ⁺¹) ] zⁿ
f(z) = Σ_n=0^∞ (-1)ⁿ [ 7/3ⁿ⁺¹ - 5/2ⁿ⁺¹ ] zⁿ

Solution

Part (i): f(z) = 1/((z-1)(z-2))
Partial Fractions: 1/((z-1)(z-2)) = 1/(z-2) - 1/(z-1)
(a) Region 1 < |z| < 2:
This means |z| > 1 (so |1/z| < 1) and |z| < 2 (so |z/2| < 1).
Term 1: 1/(z-2) = 1 / (-2(1 - z/2)) = (-1/2) · (1 - z/2)^-1 = (-1/2) Σ_n=0^∞ (z/2)ⁿ
Term 2: -1/(z-1) = -1 / (z(1 - 1/z)) = (-1/z) · (1 - 1/z)^-1 = (-1/z) Σ_n=0^∞ (1/z)ⁿ
f(z) = - Σ_n=0^∞ zⁿ/2ⁿ⁺¹ - Σ_n=0^∞ z^-(n+1)
(b) Region 0 < |z-1| < 1 (Annulus centered at z=1):
Let w = z-1. Then z = w+1. The region is 0 < |w| < 1.
f(z) = 1/((w+1)-2) - 1/w = 1/(w-1) - 1/w
Term 1: -1/w (This is already a valid series in w)
Term 2: 1/(w-1) = 1 / (-(1-w)) = -1 · (1-w)^-1 = - Σ_n=0^∞ wⁿ
f(z) = -1/(z-1) - Σ_n=0^∞ (z-1)ⁿ

Part (ii): f(z) = 1/((z+1)(z+3))
Partial Fractions: f(z) = (1/2)/(z+1) - (1/2)/(z+3)
(a) Region 1 < |z| < 3:
This means |z| > 1 (so |1/z| < 1) and |z| < 3 (so |z/3| < 1).
Term 1: (1/2)/(z+1) = (1/2) / (z(1 + 1/z)) = (1/2z) Σ_n=0^∞ (-1)ⁿ(1/z)ⁿ
Term 2: -(1/2)/(z+3) = (-1/2) / (3(1 + z/3)) = (-1/6) Σ_n=0^∞ (-1)ⁿ(z/3)ⁿ
f(z) = Σ_n=0^∞ ((-1)ⁿ/2)z^-(n+1) - Σ_n=0^∞ ((-1)ⁿ/6)(z/3)ⁿ
(b) Region |z| > 3:
This means |z| > 3 (so |3/z| < 1) and |z| > 1 (so |1/z| < 1).
Term 1: (1/2)/(z+1) = (1/2) / (z(1 + 1/z)) = (1/2z) Σ_n=0^∞ (-1)ⁿ(1/z)ⁿ
Term 2: -(1/2)/(z+3) = (-1/2) / (z(1 + 3/z)) = (-1/2z) Σ_n=0^∞ (-1)ⁿ(3/z)ⁿ
f(z) = (1/2z) [ Σ (-1)ⁿ(1/z)ⁿ - Σ (-1)ⁿ(3/z)ⁿ ]
f(z) = Σ_n=0^∞ (1/2)(-1)ⁿ(1 - 3ⁿ)z^-(n+1)