Proving a Complex number equality

Question

To Prove: If $\displaystyle p=\operatorname{cis}\theta =\cos\theta+i\sin\theta$ and $\displaystyle q=\operatorname{cis}\phi =\cos\phi+i\sin\phi$, then show that

$\displaystyle \frac{(p+q)(pq-1)}{(p-q)(pq+1)}=\frac{\sin\theta+\sin\phi}{\sin\theta-\sin\phi}$

My Attempt: $\displaystyle p=e^{i\theta} $ and $\displaystyle q=e^{i\phi} $

Then we have

$\displaystyle \frac{(e^{i\theta}+e^{i\phi})(e^{i(\theta+\phi)}-e^{i0})}{(e^{i\theta}-e^{i\phi})(e^{i(\theta+\phi)}+e^{i0})}$

I think I am going in the wrong direction here. But putting the full $\cos$ and $\sin$ in the problem will just blow the size of the expression. Any hints ?

Answer 1

Starting from the expression which you have just derived, we can remain in the complex exponential domain and simplify further in this domain as follows:- $$\displaystyle \frac{(e^{i\theta}+e^{i\phi})(e^{i(\theta+\phi)}-e^{i0})}{(e^{i\theta}-e^{i\phi})(e^{i(\theta+\phi)}+e^{i0})}=\frac{e^{i(2\theta+\phi)}-e^{i\theta}+e^{i(\theta+2\phi)}-e^{i\phi}}{e^{i(2\theta+\phi)}+e^{i\theta}-e^{i(\theta+2\phi)}-e^{i\phi}}\\=\frac{e^{i(\theta+\phi)}(e^{i\theta}-e^{-i\theta}+e^{i\phi}-e^{-i\phi})}{e^{i(\theta+\phi)}(e^{i\theta}-e^{-i\theta}-e^{i\phi}+e^{-i\phi})}\\=\frac{\color{blue}{(e^{i\theta}-e^{-i\theta})}+\color{red}{(e^{i\phi}-e^{-i\phi})}}{\color{blue}{(e^{i\theta}-e^{-i\theta})}-\color{red}{(e^{i\phi}+e^{-i\phi})}}\\=\frac{\color{blue}{2i\sin\theta}+\color{red}{2i\sin\phi}}{\color{blue}{2i\sin\theta}-\color{red}{2i\sin\phi}}\\=\frac{\sin\theta+\sin\phi}{\sin\theta-\sin\phi}$$

Answer 2

Another nice way of seeing this equality is noting that

$$\sin\theta=\frac{1}{2i}(p-p^*)\quad\text{and}\quad\sin\phi=\frac{1}{2i}(q-q^*)$$

So the right-hand side is

$$\frac{p-p^*+q-q^*}{p-p^*-q+q^*}=\frac{(p+q)-(p+q)^*}{(p-q)-(p-q)^*}$$

Now note that $p^*=1/p$ and $q^*=1/q$, so we get

$$\frac{(p+q)-(\frac{1}{p}+\frac{1}{q})}{(p-q)-(\frac{1}{p}-\frac{1}{q})}=\frac{(p+q)-\frac{p+q}{pq}}{(p-q)+\frac{p-q}{pq}}= \frac{(p+q)(1-\frac{1}{pq})}{(p-q)(1+\frac{1}{pq})}=\frac{(p+q)(pq-1)}{(p-q)(pq+1)}$$