Question

Let $f(z)=j\,\dfrac{1-z}{1+z}$, where $z$ is complex and $j=\sqrt{-1}$. The inverse function $f^{-1}(z)$ maps the {real axis} to the ______.

• A. unit circle with centre at the origin
• B. unit circle with centre not at the origin
• C. imaginary axis
• D. real axis

Hint:

Cayley transforms like $j\frac{1-z}{1+z}$ map the unit circle $\leftrightarrow$ real axis. Set $z=e^{j\theta}$ to verify quickly.

Answer 1

The Correct Option is A

Let $w=f(z)=j\dfrac{1-z}{1+z}$. The inverse image of the real axis is the set of $z$ for which $w\in\mathbb{R}$. If $w$ is real, then \[ \frac{1-z}{1+z}=\frac{w}{j}=-jw \Rightarrow \text{purely imaginary}. \] Write $z=e^{j\theta}$ (points on the unit circle). Then \[ \frac{1-z}{1+z}=\frac{1-e^{j\theta}}{1+e^{j\theta}} = \frac{e^{j\theta/2}\!\left(e^{-j\theta/2}-e^{j\theta/2}\right)} {e^{j\theta/2}\!\left(e^{-j\theta/2}+e^{j\theta/2}\right)} = \frac{-2j\sin(\theta/2)}{2\cos(\theta/2)} = -j\tan\!\left(\frac{\theta}{2}\right), \] which is purely imaginary, hence $w=j\!\left(\frac{1-z}{1+z}\right)$ is real. Conversely, if $w$ is real, the above ratio is purely imaginary, which implies $|z|=1$. Therefore $f^{-1}$ maps the real axis to the unit circle (centre 0).