Movement of $1/z$ in complex plane

Question

How is moving $\frac{1}{z}$ in complex plane if $z$ is described by a circle which has radius $r$ and center $a+b*i$

I've just started complex algebra and still having some trouble imagining it. How one does even solve this kind of problems, I don't want the complete solution I just want some appropriate ways for working with complex numbers to solve this kind of problems. Thank you.)

Answer 1

Let $c=a+bi$. Then the points $z$ of your circle are characterized by $|z-c|=r$. This translates to $$ (z-c)(\bar{z}-\bar{c})=r^2 $$ You want to find $w=1/z$, so you can substitute $z=1/w$: $$ \left(\frac{1}{w}-c\right)\left(\frac{1}{\bar{w}}-\bar{c}\right)=r^2 $$ and, doing the simplifications, $$ (|c|^2-r^2)w\bar{w}-\bar{c}w-c\bar{w}+1=0 $$ Now write $w=x+iy$: then $\bar{c}w=(a-ib)(x+iy)=ax-ibx+aiy+by$ and $c\bar{w}=ax+ibx-aiy+by$, so the equation becomes $$ (|c|^2-r^2)(x^2+y^2)-2ax-2by+1=0 $$ Can you recognize this locus?

Answer 2

Hint: Compute $\dfrac1{r\bigl(\cos(\theta)+i\sin(\theta)\bigr)}$.

Answer 3

Multiply it by $z^*/z^*$, where $z = r[\cos\theta + i \sin\theta]$ and $z^* = r[\cos\theta - i \sin\theta]$, that should get you

$$ \frac{1}{z} = \frac{1}{z}\frac{z^*}{z^*} = \frac{z^*}{|z|^2} = \frac{r[\cos\theta - i \sin\theta]}{r^2} = \frac{1}{r}[\cos\theta - i\sin\theta] $$ and $$ \theta = arctan(b/a) $$

Can you take it from there?