A complex number a is unimodular if |a|=1 . Show that a is…

A complex number $a$ is unimodular if $|a|=1$. Show that $a$ is unimodular if and only if it can be written as the quotient of two complex conjugate numbers.

Direction 1 · If $a=z/\bar z$ for some $z\neq0$, then $|a|=1$

Using $|z_1/z_2|=|z_1|/|z_2|$ and $|\bar z|=|z|$:

$$ |a|=\left|\frac{z}{\bar z}\right|=\frac{|z|}{|\bar z|}=\frac{|z|}{|z|}=1 $$

Direction 2 · If $|a|=1$, then $a=z/\bar z$ for some $z\neq0$

Since $|a|=1$, write $a=e^{i\theta}$ for some real $\theta$. Choose:

$$ z=e^{i\theta/2} \quad\Longrightarrow\quad \bar z=e^{-i\theta/2} $$

Then:

$$ \frac{z}{\bar z}=\frac{e^{i\theta/2}}{e^{-i\theta/2}}=e^{i\theta}=a $$

$$ \boxed{|a|=1 \iff a=\frac{z}{\bar z} \text{ for some } z\in\mathbb{C}^*} $$

Step 3 · Geometric meaning

Writing $z=re^{i\varphi}$ ($r>0$), $\ \dfrac{z}{\bar z}=e^{2i\varphi}$: the map $z\mapsto z/\bar z$ sends every nonzero $z$ to the point on the unit circle at twice the argument of $z$, regardless of $|z|$. So every unimodular number is reached (twice over — by $z$ and by $-z$, and more generally by any $z$ with the right argument mod $\pi$).

Step 4 · Verification examples