Prove that $\arg(\frac{z}{w})=\arg(z)-\arg(w)$
My attempt: Let $z=x+iy$, and $y=a+ib$. The polar form is:
$$z=r(\cos(\theta)+i\sin(\theta)) \quad\text{and} \quad w=p(\cos(\beta)+i\sin(\beta))$$
Then,
\begin{align*}
\arg(z)-\arg(w)
\quad\implies\quad&
r(\cos(\theta-\beta)+i\sin(\theta-\beta))\\
&=r(\cos\theta\cos\beta+\sin\theta\sin\beta+i(\sin\theta\cos\beta-\cos\theta\sin\theta))
\end{align*}
I'm stuck here. Some help please?
Best Answer
Continue with what you have,
$$z=r(\cos\theta+i\sin\theta),\>\>\>\>\>w=p(\cos\beta+i\sin\beta)$$
to express
$$\frac zw = \frac rp \frac {\cos\theta+i\sin\theta}{\cos\beta+i\sin\beta}$$ $$=\frac rp \frac {(\cos\theta+i\sin\theta)(\cos\beta-i\sin\beta)} {(\cos\beta+i\sin\beta)(\cos\beta-i\sin\beta)}$$
$$=\frac rp \frac {(\cos\theta\cos\beta+\sin\theta\sin\beta)+i(\sin\theta\cos\beta-\cos\theta\sin\beta)} {\cos^2\beta+\sin^2\beta}$$ $$=\frac rp [ \cos(\theta-\beta)+i\sin(\theta-\beta)] $$
Thus, $\arg(\frac{z}{w})=\arg(z)-\arg(w)$