Complex and Trigonometric Identities

Question

How can I get this result:

$$\frac{1+cis\theta}{1-cis\theta}=-\frac{1}{i\tan(\theta/2)}$$

I've tried to expand $1-cis\theta$ as $(1+cis(\theta/2))(1-cis(\theta/2))$, but it doesn't help.

Answer 1

Note that $\text{cis}(\theta) = e^{i\theta}$. I find this change of notation really helps. Anyway, divide the top and bottom by $e^{i\theta/2}$ to get $$ \frac{1+e^{i\theta}}{1-e^{i\theta}} = -\frac{e^{i\theta/2}+e^{-i\theta/2}}{e^{i\theta/2}-e^{-i\theta/2}} $$ and substitute $e^{\pm i\theta/2} = \cos(\theta/2)\pm i\sin(\theta/2) $.

Answer 2

Using Double-Angle Formulas

$$\frac{1+\text{cis}2\phi}{1-\text{cis}2\phi}=\frac{2\cos^2\phi+2i\sin\phi\cos\phi}{2\sin^2\phi-2i\sin\phi\cos\phi}$$

$$=\frac{2\cos\phi(\cos\phi+i\sin\phi)}{2i\sin\phi(\cos\phi+i\sin\phi)}$$

We can safely cancel out $(\cos\phi+i\sin\phi)$ assuming it to be non-zero, which is always for real $\phi$