Unit 5: Part A Solutions (Complex Variables)

Let f(z) = u + iv be an analytic function.

Given: The imaginary part is constant. Let v(x, y) = c, where c is a constant.

1. ∂v/∂x = 0

2. ∂v/∂y = 0

By the Cauchy-Riemann (C-R) equations, since f(z) is analytic:

∂u/∂x = ∂v/∂y = 0

∂u/∂y = -∂v/∂x = 0

Since both partial derivatives of u are zero, u(x, y) must also be a constant, say k.

Therefore, f(z) = u + iv = k + ic, which is a constant. (Proved)

We check the Cauchy-Riemann (C-R) equations.

For u = x² - y²:

∂u/∂x = 2x

∂u/∂y = -2y

For v = -y / (x²+y²):

∂v/∂y = [(-1)(x²+y²) - (-y)(2y)] / (x²+y²)² = (-x²-y²+2y²) / (x²+y²)² = (y²-x²) / (x²+y²)²

∂v/∂x = (-(-y) · 2x) / (x²+y²)² = 2xy / (x²+y²)²

Check C-R 1: ∂u/∂x = ∂v/∂y ?

2x ≠ (y²-x²) / (x²+y²)²

Check C-R 2: ∂u/∂y = -∂v/∂x ?

-2y ≠ -[ 2xy / (x²+y²)² ]

Since the C-R equations are not satisfied, the function is not analytic.

To be harmonic, φ must satisfy Laplace's equation: ∂²φ/∂x² + ∂²φ/∂y² = 0.

Derivatives w.r.t. x:

∂φ/∂x = 3x² - 3y² + 6x

∂²φ/∂x² = 6x + 6

Derivatives w.r.t. y:

∂φ/∂y = -6xy - 6y

∂²φ/∂y² = -6x - 6

Sum:

∂²φ/∂x² + ∂²φ/∂y² = (6x + 6) + (-6x - 6) = 0

Since Laplace's equation is satisfied, the function is harmonic.

Let φ = 3x²y - y³. We check Laplace's equation.

Derivatives w.r.t. x:

∂φ/∂x = 6xy

∂²φ/∂x² = 6y

Derivatives w.r.t. y:

∂φ/∂y = 3x² - 3y²

∂²φ/∂y² = -6y

Sum:

∂²φ/∂x² + ∂²φ/∂y² = (6y) + (-6y) = 0

Since Laplace's equation is satisfied, the function is harmonic.

We check if u_xx + u_yy = 0.

Derivatives w.r.t. x:

∂u/∂x = e^x(cos y - sin y)

∂²u/∂x² = e^x(cos y - sin y)

Derivatives w.r.t. y:

∂u/∂y = e^x(-sin y - cos y)

∂²u/∂y² = e^x(-cos y + sin y) = -e^x(cos y - sin y)

Sum:

∂²u/∂x² + ∂²u/∂y² = e^x(cos y - sin y) - e^x(cos y - sin y) = 0

Since Laplace's equation is satisfied, the function is harmonic.

We need to find the conjugate function v using the C-R equations.

1. ∂u/∂x = e^xsin y

2. ∂u/∂y = e^xcos y

From C-R 1: ∂v/∂y = ∂u/∂x = e^xsin y

Integrate w.r.t. y to find v:

v = ∫ e^xsin y dy = -e^xcos y + g(x)    (where g(x) is a function of x only)

From C-R 2: ∂v/∂x = -∂u/∂y = -e^xcos y

Differentiate our v w.r.t. x:

∂v/∂x = ∂/∂x(-e^xcos y + g(x)) = -e^xcos y + g'(x)

Compare the two expressions for ∂v/∂x:

-e^xcos y + g'(x) = -e^xcos y  &implies  g'(x) = 0  &implies  g(x) = C (a constant)

So, v = -e^xcos y + C.

w = u + iv = e^xsin y + i(-e^xcos y + C)

w = -i e^x(cos y + i sin y) + iC = -i e^xe^iy + iC = -i e^z + iC

(where K is a complex constant, iC)

The transformation w = z + c is a simple translation (shift).

The original curve |z| = 2 is a circle centered at the origin (0, 0) with a radius of 2.

The transformation shifts the center by the vector (3 + 2i).

New center = (0 + 3) + i(0 + 2) = 3 + 2i.

The radius remains unchanged.

Alternatively:

Let z = w - (3 + 2i). Substitute this into |z| = 2:

|w - (3 + 2i)| = 2    or    |w - 3 - 2i| = 2

This is the equation of a circle.

Invariant (or fixed) points are found by setting w = z.

z = (2z + 6) / (z + 7)

z(z + 7) = 2z + 6

z² + 7z = 2z + 6

z² + 5z - 6 = 0

Factor the quadratic equation:

(z + 6)(z - 1) = 0

The invariant points are z = 1 and z = -6.

We use the cross-ratio formula:

( (w-w₁)(w₂-w₃) ) / ( (w-w₃)(w₂-w₁) ) = ( (z-z₁)(z₂-z₃) ) / ( (z-z₃)(z₂-z₁) )

Since z₃ = ∞, the term (z₂-z₃) / (z-z₃) simplifies to -1.

RHS: ( (z-0) / ( (z₂-z₁) ) ) · (-1) = ( z / (1-0) ) · (-1) = -z

LHS: ( (w-i)(1 - (-i)) ) / ( (w - (-i))(1 - i) )

= ( (w-i)(1+i) ) / ( (w+i)(1-i) )

= [ (w-i) / (w+i) ] · [ (1+i) / (1-i) ]

Simplify the constant term: (1+i)/(1-i) = (1+i)² / ((1-i)(1+i)) = (1 + 2i - 1) / (1 - (-1)) = 2i / 2 = i

Equate LHS and RHS:

i · ( (w-i) / (w+i) ) = -z

(w-i) / (w+i) = -z / i = iz

w - i = iz(w + i)

w - i = izw + i²z

w - i = izw - z

w - izw = i - z

w(1 - iz) = i - z