Linear function

1.

(a) $w = \frac{1}{z}$ takes $z_{1}$ by inverting the point $z$ with respect to the unit circle $\vert z\vert = 1$, and then takes symmetrical transformation of $z_{1}$.

Since the two points P and Q are on the unit circle, they appear at points P'and Q'symmetrical with respect to the real axis by $w = \frac{1}{z}$. Also, the point at infinity of a straight line is reflected at the origin. Furthermore, $w = \frac{1}{z}$ is circle to circle, this straight line maps to the circle going through P', Q'.

(b) $w = \frac{1}{z}$ takes $z_{1}$ by inverting the point $z$ with respect to the unit circle $\vert z\vert = 1$, and then takes symmetrical transformation of $z_{1}$.

Since the point P is on the unit circle, it maps to the point P' which is symmetric with respect to the real axis by $w = \frac{1}{z}$. Also, the point at infinity of a straight line is reflected at the origin. Furthermore, $w = \frac{1}{z}$ is circle to circle. This straight line maps to the circle with the radius of the distance from the orign to P'.

(c) $w = \frac{1}{z}$ takes $z_{1}$ by inverting the point $z$ with respect to the unit circle $\vert z\vert = 1$, and then takes symmetrical transformation of $z_{1}$.

A point $a$ maps to $\frac{1}{a}$, a point $i$ maps to $\frac{1}{i}$, a point $-i$ maps to $\frac{1}{-i}$. Furthermore, $w = \frac{1}{z}$ is circle to circle, a cricle going through 3 points $a,i,-i$ maps to a circle going through 3 points $\frac{1}{a},\frac{1}{i},\frac{1}{-i}$.

2.

(a) An invariant point is a point that satisfies $f(z) = z$.

Let $w = \frac{1}{z} = z$. Then $z^2 = 1$. Thus we use the quadratic formula.

$z^{2} = 1 = e^{0}$ implies the roots are $\cos{\frac{0 + 2k\pi}{2}} + i\sin{\frac{0 + 2k\pi}{2}} (k = 0,1)$. In other words, $\pm 1$.

(b) An invariant point is a point that satisfies $f(z) = z$.

Let $w = \frac{az + b}{cz + d} = z$. Then $cz^2 + dz= az + b$. Rewrite this, we have

$$cz^2 + (d -a)z - b = 0$$

Thus,

$$z = \frac{(a - d) \pm \sqrt{(a -d)^2 + 4bc}}{2c}$$