Found the following question:
On the Argand diagram $P$ represents the complex number $z$ and $Q$ the complex number $\frac{1}{z}$. If $P$ lies on the straight line $x=1$ prove that $Q$ will lie on a certain circle and find its centre and radius.
I can visualise why this is true, as $|z| ≥ 1$, hence $\frac{1}{|z|} ≤ 1$. You can also show this graphically.
However, is there an algebraic method that shows this?
On
Algebraic proof
Set $z=r(\cos \theta + i \sin \theta), z\neq 0.$ Then $\frac{1}{z}=\frac{1}{r}\big(\cos (-\theta)+i \sin(-\theta)\big).$
The point $P(z)$ lies on the line $\iff {r\cos \theta =1}.$
The implication to prove is in fact an equivalence.
The point $Q(\frac{1}{z})$ lies on the circle centered at $A({1\over 2})$ with radius $1\over 2$ $$\begin{aligned}\iff &\left|{1\over z}-{1\over 2}\right|={1\over2} \\ \iff & \left|{1\over z}-{1\over 2}\right|^2={1\over4} \\ \iff &{\frac{1}{r^2}-\frac{1}{r}\cos (-\theta)}=0\\ \iff & r\cos(\theta)=1\end{aligned}$$ If you prefer, you can do the proof in the opposite direction.
Proof using circle inversion
With $z \mapsto \frac{1}{z}$ is defined inversion in circle, in particular, in unit circle. Unit circle is a reference circle.
The line $Re( z)=1$ is tangent to the reference circle at the point $T(1+0i).$
Therefore, the inverse of the line is the circle passing through the center of the reference circle, which is origine, and which is tangent to the line (and to the reference circle) at $T.$
If $P = 1 + yi$, then $Q = \frac1{1+yi} = \frac1{1+y^2} + \frac{y}{1+y^2}i$.
The point $(\frac1{1+y^2}, \frac{y}{1+y^2})$ can be checked to be at distance $\frac12$ from the point $(\frac12,0)$.