Further show that under this transformation the unit circle w-plane is mapped onto a straight line in the z plane write the name of this line.

Let the transformation be $w = \frac{az + b}{cz + d}$

Now putting the value of z and w we get

1) w = 0, z = 1 0 = $\frac{a(1) + b}{c(1) + d} \therefore a+ b = 0$ i.e. a = -b

2) w = 1, z = I 1 = $\frac{a(i) + b}{c(i) + d} \therefore ci + d = ai + b \therefore (a-c)i = d - b$

3) w = $\infty$, z = -1 $\therefore \infty = \frac{a(-1) + b}{c(-1) + d} \therefore -c + d = 0$ i.e. c = d

Hence we have a = -b (1)

(a - c)i = d - b (2)

c = d (3)

$\therefore$ substitute (1) and (3) in equation (2)

(a - c)i = c + a

$\therefore$ ai - a = c + ci

a(i - 1) = c(i + 1)

$\frac{a}{c} = \frac{i + 1}{i - 1} * \frac{i + 1}{i + 1} \\ = \frac{(i + 1)^2}{-1 -1} \\ = \frac{i^2 + 2i + 1}{-2}$

$\therefore$ a = -ci

i.e.

$c = \frac{-1}{i} a \\ \therefore c = ai \\ \therefore c - d = ai$

$w = \frac{az + b}{cz + d} \\ w = \frac{az - a}{(ai)z + ai}$

Now

$w = \frac{a(z - 1)}{ai(z + 1)} \\ \therefore w = -i \frac{z - 1}{z + 1}$

Now the second part:

From the given question unit circle in w plane which is mathematically written as

$|w| = 1$ i.e. $\bigg|\frac{zi - i}{-1 -z} \bigg| = 1$

$\therefore |zi - i| = |-1 - z|$

Where z = x + iy

$\therefore |(x + iy)i - i| = |-1 - x - iy|$

Seperating into real and imaginary parts

$\therefore |-y + i(x - 1)| = |-(x + 1) - iy|$

Now taking the magnitude i.e. $\sqrt{real^2 + imag^2}$

$\therefore \sqrt{(-y)^2 + (x - 1)^2} = \sqrt{[-(x + 1)]^2 + (-y)^2}$

$\therefore y^2 + (x - 1)^2 = (x + 1)^2 + y^2 \\ \therefore (x - 1)^2 = (x + 1)^2 \\ \therefore x^2 - 2x + 1 = x^2 + 2x + 1 \\ \therefore -2x = 2x \\ \therefore 4x = 0 \\ \therefore x = 0$

Hence the map is y-axis