If $|z|=1$, $z\neq-1$, show that $z$ may be expressed in the form $ z=\frac{1+it}{1-it}$ where $t\in \mathbb{R}$.

HINT:

$$\frac{1 + i \tan \frac{\theta}{2}}{ 1 - i \tan \frac{\theta}{2}} = \cdots$$

(a picture can make it obvious)

Without appealing to polar form, using only complex algebra...

Solving for $z-1$ gives $z-1= it(1+z)$, and as $z \neq -1$ we can divide to get $$ \frac{z-1}{z+1} = it. $$

Write $z=a+bi$ and start doing the algebra on the left hand side. It's not hard (conjugate by the denominator). When you realize that $a^2+b^2=1$ you end up with $$ \frac{z-1}{z+1} = \frac{2bi}{2a+2} = it. $$ Hence $t = \frac{2b}{2a+2} = \frac{b}{a+1}$ should work, and you can check it.

Like Randall, I would begin by solving that $$ it=\frac{z-1}{z+1}. $$ But from this point on I would use the fact that $1=|z|^2=z\overline{z}$. This allows a rewrite $$ it=\frac{z-1}{z+1}=\frac{z-z\overline{z}}{z+z\overline{z}}=\frac{1-\overline{z}}{1+\overline{z}}=-\overline{\left(\frac{z-1}{z+1}\right)}=-\overline{it}. $$ This implies that $t$ must be real.