Proving arg(z/w)=arg(z)-arg(w)

I need to prove that $$arg\left(\frac{z}{w}\right)=arg(z)-arg(w)$$ However, I am a little stuck as to how to go about this.

I know the proof for $arg(zw)=arg(z)+arg(w)$ happens by letting $z=r(cos\theta+isin\theta)$ and $w=s(cos\phi+isin\phi)$ and then multiplying them together and expanding out the brackets, combining the arguments using the double angle formulae.

I have been told that the proof of this is analogous, but I have absolutely no idea how I would simplify $$\frac{r(cos\theta+isin\theta)}{s(cos\phi+isin\phi)}$$

Is there some identity I don't know about?

EDIT: We have not covered $e^{ix}=cos(x)+isin(x)$ yet, so the proof will not need to use that fact.

• $e^{ix} = \cos x + i\sin x$ – Axoren Jan 23 '15 at 2:29
• To be pedantic, ${\rm arg}(zw) \equiv {\rm arg}(z) + {\rm arg}(w) \pmod{2\pi}$ – DanielV Jan 23 '15 at 2:50
• In $arg(zw)=arg(z) + arg(w)$, replace $w$ by $w^{-1}$. – Mick Jan 23 '15 at 2:50
• @DanielV That observation certainly isn't pedantry---for one, it leads to the conflict that complex exponents $z^w$ in general depend on the choice of argument function (equivalently, branch of the logarithm function). – Travis Jan 23 '15 at 2:54

You could multiply by $\frac{\cos\phi-i\sin\phi}{\cos\phi-i\sin\phi}$ and use: $$\sin(x-y) = \sin x \cos y - \cos x \sin y$$ $$\cos(x-y) = \cos x \cos y + \sin x \sin y$$

Use the fact that

$$\arg\left(z^n\right) = n\arg\left(z\right)$$

$$\arg\left(\frac{z}{w}\right) = \arg\left(zw^{-1}\right) = \arg\left(z\right) + \arg\left(w^{-1}\right) = \arg\left(z\right) - \arg\left(w\right)$$

If we interpret $\arg$ as the imaginary part of a branch of the logarithm function, then this is not true: If we choose the branch so that $\arg \zeta \in [0, 2 \pi)$, then $$\arg\left(\frac{1}{-1}\right) = \arg(-1) = \pi$$ but $$\arg 1 - \arg (-1) = 0 - \pi = - \pi.$$

That said, the identity can hold for certain choices of branch and restricted values of $z, w$ (for example if we choose the branch so that $\arg \zeta \in [-\pi, \pi)$ and restrict to $z, w \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right)$).

It's anyway more natural (though often not more convenient) to view the argument as a function that takes values modulo $2\pi$ (equivalently, as elements of $\mathbb{S^1}$), which corresponds to the fact that the above argument function $\arg$ of any branch of $\log$ (with domain $\mathbb{C} - \{0\}$ anyway) satisfies. $$\arg \frac{z}{w} \equiv (\arg z - \arg w) \bmod 2\pi.$$

What you need to use is not an identity but (it can be called at most) a method: You multiply both the numerator and the denominator by the complex conjugate of the latter one:

$\dfrac{a+b\imath}{c+d\imath}=\dfrac{(a+b\imath)(c-d\imath)}{c^2+d^2}$

Using this you will get the cosine/sine of the difference of angles formulae.

Observe that this is actually the way to find a simpler form of the (multiplicative) inverse of a nonzero complex number.