consider the mobius transformation $f(z) =\frac{1}{z} $ , $z \in \mathbb{C}$, $z \neq 0$.choose the correct options

Question

consider the mobius transformation $f(z) =\frac{1}{z} $ , $z \in \mathbb{C}$, $z \neq 0$. If C denotes a circle with positive radius passing through the origin, then $f$ map $C \setminus \{0\}$ to

choose the correct options

$1.$Circle

$2.$ a line

$3.$ a line passing through the origin

$4.$ a line not passing through the origin

My attempt : i take $z = e^{i\theta}$,$f(z) = 1/z = 1/e^{i\theta}=e^{-i\theta}$

that $f(z) = \cos\theta - i\sin\theta$

After that im not able to proceed further pliz help me

any hints/solution will be appreciated thanks u

Answer 1

What's the point of taking $z=e^{i\theta}$ that gives you a circle not passing through the origin? If you do take a circle passing through the origin, then what you will get is a line not passing through the origin. That is, the third option is the right one.

Answer 2

You can consider the above Möbius transformation $f$ on the extended complex plane $\hat{\mathbb C} =\mathbb C \cup\{ \infty\}$ with $f(0)= \infty$ and $f(\infty)=0$.

Since $0 \in C$ we have $ \infty \in f(\hat{\mathbb C})$. And since $f(\infty)=0$, we see that the third option is correct.

Answer 3

It is easy to show that $A(x^2+y^2)+Bx+Cy+D=0$ under $\frac1z$ goes to $D(u^2+v^2)+Bu-Cv+A=0$

So as the circle does not pass through the origin, $A$ and either $B$ or $C$ are non-zero and $D=0$

Hence, options 2 and 4 true.