If $a = (\cos heta + i\sin heta)$, prove that $\dfrac{1+a}{1-a} = \left(\cot\dfrac{ heta}{2} ight)i$.

Question

If $a = (\cos	heta + i\sin	heta)$, prove that $\dfrac{1+a}{1-a} = \left(\cot\dfrac{	heta}{2}ight)i$.

Accepted Answer

>

> [!TLDR]
> Use half-angle identities to factor $1+a$ and $1-a$, then cancel the common factor.
>
> Given $a=\cos	heta+i\sin	heta$.
>
> $1+a=1+\cos	heta+i\sin	heta=2\cos^2\dfrac{	heta}{2}+i\cdot2\sin\dfrac{	heta}{2}\cos\dfrac{	heta}{2}=2\cos\dfrac{	heta}{2}\left[\cos\dfrac{	heta}{2}+i\sin\dfrac{	heta}{2}ight]$.
>
> $1-a=1-\cos	heta-i\sin	heta=2\sin^2\dfrac{	heta}{2}-i\cdot2\sin\dfrac{	heta}{2}\cos\dfrac{	heta}{2}=2\sin\dfrac{	heta}{2}\left[\sin\dfrac{	heta}{2}-i\cos\dfrac{	heta}{2}ight]$.
>
> Note $\sin\dfrac{	heta}{2}-i\cos\dfrac{	heta}{2}=-i\left[\cos\dfrac{	heta}{2}+i\sin\dfrac{	heta}{2}ight]$ (check: $-i\cos	frac{	heta}{2}-i^2\sin	frac{	heta}{2}=-i\cos	frac{	heta}{2}+\sin	frac{	heta}{2}$, matches).
>
> So $1-a=-2i\sin\dfrac{	heta}{2}\left[\cos\dfrac{	heta}{2}+i\sin\dfrac{	heta}{2}ight]$.
> …

Q.If $a = (\cos\theta + i\sin\theta)$ , prove that $\dfrac{1+a}{1-a} = \left(\cot\dfrac{\theta}{2}\right)i$ .

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Q.If $a = (\cos\theta + i\sin\theta)$ , prove that $\dfrac{1+a}{1-a} = \left(\cot\dfrac{\theta}{2}\right)i$ .

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