1A (Vectors and Matrices)

\frac{a z + b}{z + 1} = z ⟺ z^{2} + (1 - a) z - b = 0.

A unique solution at

z = i

(z - i)^{2} = z^{2} - 2 i z - 1.

Comparing coefficients gives

a = 1 + 2 i, b = 1 .

ii

3 i

and

1 + i

(z - 3 i) (z - 1 - i) = z^{2} - (1 + 4 i) z + (3 - 3 i) .

Comparing this with

z^{2} + (1 - a) z - b

gives

a = 2 + 4 i, b = - 3 + 3 i .

Write

z = x + i y

. Rationalizing gives

\frac{z - 2}{z} = \frac{x ^{2} - 2 x + y ^{2} + 2 i y}{x ^{2} + y ^{2}} .

An argument of

π /4

requires equal positive real and imaginary parts. Hence

x^{2} - 2 x + y^{2} = 2 y, y > 0.

Completing squares yields

(x - 1)^{2} + (y - 1)^{2} = 2, y > 0 .

Thus the constant-argument locus of a Möbius transformation is the part above the real axis of the circle with centre

(1, 1)

2

. Its endpoints

z = 0

and

z = 2

are excluded because the quotient is respectively undefined and zero.

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