Prove that $\arg(\frac{z_1}{z_2})=\arg(z_1)-\arg(z_2)$

Question

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?

Answer 1

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, $\arg(\frac{z}{w})=\arg(z)-\arg(w)$

Answer 2

Hint using Euler's $e^{ix} = \cos{x} + i\sin{x}$, then for any numbers $n\in\mathbb{Z}$ one has that $\arg (re^{i(\theta + 2n\pi)}) = \theta$ assuming that $\theta\in[0,2\pi)$. If $\theta\in[-2\pi,0)$ then we normally have $\arg{(e^{i\theta})} = \theta + 2\pi$. Now write $z = |z|e^{i\theta}$ and $w = |w|e^{i\phi}$ where $|z|, |w|\not= 0$ and $\theta,\phi\in[0,2\pi)$. (This representation is unique) now calculate the LHS and RHS separately.

Answer 3

you need some standard formulas in trigonometry:

$\cos(x+y) =\cos x\cos y - \sin x \sin y$ $\cos(x-y) =\cos x\cos y + \sin x \sin y$ $\sin(x+y) =\sin x \cos y + \cos x \sin y$ $\sin(x-y) =\sin x \cos y - \cos x \sin y$

Answer 4

$z = x+iy = |z|(cos\theta + i\sin\theta)\\ w = a+ib = |w|(cos\phi + i\sin\phi)\\ |zw| = |z||w|$

If you have not proven the last fact, I leave that as an exercise.

$\frac {x+iy}{a+ib}\\ \frac {(x+iy)(a-ib)}{(a+ib)(a-ib)}\\ \frac {(xa+yb)+i(ya-xb)}{a^2+b^2}$

We have multiplied numerator and denominator by the conjugate of $w$

$\frac {|z||w|((cos\theta\cos\phi-\sin\theta\sin\phi) + i(\sin\theta\cos\phi + \cos\theta\sin\phi)}{|w|^2}\\ \frac {|z|}{|w|}(\cos(\theta-\phi) + i \sin(\theta-\phi))$