Transformation of complex numbers on the Argand plane If $P$ represents the complex number $z$ and $Q$ represents $\frac{1}{z}$ prove that if P describes a circle with radius $r$ and a centre at $w$ that $Q$ will also describe a circle which has a centre at $\frac{\overline{w}}{|w|^2-r^2}$ and find its radius.
Is saying that $Q$ has a radius of $\frac{1}{r}$ by letting $P=r[\cos\theta+i\sin\theta]$ then using De moivre's theorem on $z^{-1}$ to get $\frac{1}{r}[\cos(-\theta)+isin(-\theta)]$ correct?
I've also got no idea how to obtain an expression of $Q$ from the expression for $P$ which I found to be $|z-w|=r$. I tried squaring both sides but I can't isolate $z$ from $z\cdot{z}-2z\cdot{w}+w\cdot{w}=r^2$
 A: You can write the requirement of $P$ belonging to the circle you specified as $|P-w|^2=r^2$, that is
$$(P-w)\overline{(P-w)} = r^2$$
We suppose $P \neq 0$ for every value of $P$ belonging to that circle, that is $w\overline{w}\neq r^2$. Then $Q = \frac{1}{P}$ implies $P = \frac{1}{Q}$ and we have the equation
$$\left(\frac{1}{Q}-w\right)\overline{\left(\frac{1}{Q}-w\right)} = \left(\frac{1}{Q}-w\right)\left(\frac{1}{\overline{Q}}-\overline{w}\right)= r^2$$
By multiplying left and right by $Q\overline{Q}$ we get
$$\left(1-wQ\right)\left(1-\overline{wQ}\right) = r^2Q\overline{Q}$$
That is,
$$\left(w\overline{w}-r^2\right) Q\overline{Q} -wQ -\overline{wQ} + 1 = 0$$
Divide by $w\overline{w}-r^2$ either side:
$$Q\overline{Q} - \frac{w}{w\overline{w}-r^2}Q - \frac{\overline{w}}{w\overline{w}-r^2}\overline{Q} + \frac{1}{w\overline{w}-r^2} = 0$$
Factoring out the product and completing the expression gives us
$$\left(Q-\frac{\overline{w}}{w\overline{w}-r^2}\right)\left(\overline{Q}-\frac{w}{w\overline{w}-r^2}\right) - \frac{r^2}{\left(w\overline{w}-r^2\right)^2} = 0$$
Finally, that means
$$\left(Q-\frac{\overline{w}}{w\overline{w}-r^2}\right)\left(\overline{Q}-\frac{w}{w\overline{w}-r^2}\right) = \left(\frac{r}{w\overline{w}-r^2}\right)^2$$
which gives you both the center (which is the one you said) and the radius (which is not $\frac{1}{r}$ in general but depends on $w$) of the circle $Q$ belongs to.
Note that we left out the case in which $w\overline{w} = r^2$. In that case, the nonzero points of the initial circle get sent not to a circle, but to a line:
$$wQ +\overline{wQ} = 1$$
A: We have the initial circle equation:
$$
\left(z-w\right)\left(\bar{z}-\bar{w}\right)=r^{2}
$$
Then we have $q=\frac{1}{z}$. Substitute this into the equation to obtain the following:
$$
\begin{align}
r^{2}&=\left(\frac{1}{q}-w\right)\left(\frac{1}{\bar{q}}-\bar{w}\right)\\
\\
&=\frac{1}{q\bar{q}}-\frac{1}{q}\bar{w}-\frac{1}{\bar{q}}w+w\bar{w}
\end{align}
$$
Multiply by $q\bar{q}$, substitute $w\bar{w}=|w|^{2}$, then do some algebra works to obtain the following equation:
$$
\begin{align}
\left(q-\frac{\bar{w}}{|w|^{2}-r^{2}}\right)\left(\bar{q}-\frac{w}{|w|^{2}-r^{2}}\right)&=\left(\frac{r}{|w|^{2}-r^{2}}\right)^{2}
\end{align}
$$
