1. ## Complex Numbers

If $\displaystyle z = cis\theta$, where $\displaystyle \theta < \frac{\pi}{4}$ show that:

$\displaystyle \frac{1-z^2}{z+1} = 2\sin(\frac{\theta}{2})cis(\pi-\frac{3\theta}{2})$

Any help would be much appreciated as all my attempts at solving this haven't gone too well.

2. Originally Posted by DekuTrombonist
If $\displaystyle z = cis\theta$, where $\displaystyle \theta < \frac{\pi}{4}$ show that:

$\displaystyle \frac{1-z^2}{z+1} = 2\sin(\frac{\theta}{2})cis(\pi-\frac{3\theta}{2})$

Any help would be much appreciated as all my attempts at solving this haven't gone too well.
Either the answer is wrong or I'm wrong. You decide which:

$\displaystyle \frac{1-z^2}{1+z} = \frac{(1+z)(1-z)}{1+z} = 1-z = 1-\cos\theta - i\sin\theta$.

Next, $\displaystyle 1-\cos\theta = 2\sin^2\tfrac\theta2$ and $\displaystyle \sin\theta = 2\sin\tfrac\theta2\cos\tfrac\theta2$. Also, $\displaystyle -i = \text{cis}\bigl(-\tfrac\pi2\bigr)$. Putting those things together, I get

\displaystyle \begin{aligned}1-z &= 2\sin\tfrac\theta2\bigl(\sin\tfrac\theta2 - i\cos\tfrac\theta2\bigr) \\ &= 2\sin\tfrac\theta2(-i)\bigl(\cos\tfrac\theta2 + i\sin\tfrac\theta2\bigr) \\ &= 2\sin\tfrac\theta2\text{cis}\bigl(-\tfrac\pi2\bigr)\text{cis}\,\tfrac\theta2 \\ &= 2\sin\tfrac\theta2\text{cis}\bigl(\tfrac\theta2-\tfrac\pi2\bigr)\end{aligned}.

3. Thanks for your help on this one. Obviosuly the answer in the question is wrong which probably explains why I couldn't do it.