Show that $\frac 1z$ maps circle $|z-1|=1$ onto vertical line $x=\frac 12$

Question

I would like to show that

The complex function $f(z)=1/z$ maps the circle $|z-1|=1$ onto the vertical line $x=1/2$.

So far, I know that if $z=x+iy$, then $f(z)$ is equivalent to

$$w=u(x,y)+iv(x,y)=\frac{x}{x^2+y^2}+i\frac{-y}{x^2+y^2},$$

but I'm pretty new to complex analysis and I don't know how to follow.

Answer 1

Note that, for each $z\in\mathbb C$,\begin{align}\lvert z-1\rvert=1&\iff\lvert z-1\rvert^2=1\\&\iff\lvert z\rvert^2-2\operatorname{Re}(z)+1=1\\&\iff\lvert z\rvert^2=2\operatorname{Re}(z).\end{align}So, if $z\neq0$\begin{align}\operatorname{Re}\left(\frac1z\right)=\frac12&\iff\frac{\operatorname{Re}(z)}{\lvert z\rvert^2}=\frac12\\&\iff\lvert z\rvert^2=2\operatorname{Re}(z)\\&\iff\lvert z-1\rvert=1.\end{align}

Answer 2

WLOG $z-1=\cos2t+i\sin2t$

$$\dfrac1z=\dfrac1{2\cos t(\cos t+i\sin t)}=\dfrac{\cos t-i\sin t}{2\cos t}$$

Answer 3

$z=0$ is mapped to $f(0) = \infty$. For $z \ne 0$ we have with $w = \frac 1z$ $$ |z-1| = 1 \iff |\frac 1w - 1| = 1 \iff |w-1|^2 = |w|^2 \\ \iff |w|^2 - 2 \operatorname{Re}w + 1 = |w|^2 \iff \operatorname{Re}w = \frac 12 \, , $$ where we have used the formula $$ |z_1 + z_2|^2 = |z_1|^2 + 2 \operatorname{Re}(z_1 \bar z_2) + |z_2|^2 \, . $$

Therefore the image is the (extended) line $$ \{ w \mid \operatorname{Re}w = \frac 12 \} \cup \{ \infty \} \, .$$

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If you are familiar with Möbius transformations then you can argue that the image of the circle

• must be a line (because Möbius transformations maps circles to circles or lines, and $f(0) = \infty$),
• pass through $w = \frac 12$ (because $f(2) = \frac 12$),
• be orthogonal to the real axis (because Möbius transformations preserve angles and $f$ maps the real axis onto itself),

and that leaves only the (extended) line $x= \frac 12$ plus the point at infinity.

Answer 4

Let $z-1 = e^{ia}$, then

$$\frac 1z = \frac 1{1+e^{ia}}=\frac {1+e^{-ia}}{(1+e^{ia})(1+e^{-ia})}=\frac {1+\cos a - i\sin a}{2+2\cos a}=\frac 12 -\frac12\tan\frac a2 i $$

Thus, $\frac 1z$ is of a vertical line $x=\frac 12$.