Argument of the sum of two complex numbers

Let $r$, $s$ be positive real numbers and $\theta$, $\phi$ real numbers with $|\theta -\phi|<\pi$. Then an argument of $re^{i\theta}+se^{i\phi}$ lies between $\theta$ and $\phi$.

Can someone give a short, clean proof of the statement above that doesn't rely on geometric intuition? Of course the proof may use the not so trivial fact that every nonzero complex number can be written in polar form as well as trigonometrical functions.

I have tried, but I have no idea how to start, probably because it seems so obvious.

• You can visualise this using the parallelogram diagram of complex addition. The argument of the resulting sum can easily be seen to lie between the arguments of the summands. I suppose this depends on your choice of principal argument though. – Antinous Dec 14 '16 at 11:59

You want to show that if you write $re^{i\theta} + se^{i\phi}$ as $te^{i\chi}$, then $\chi$ lies between $\theta$ and $\phi$. Without loss of generality, assume $\theta\leq\phi$. Factoring out $e^{i\theta}$ you get $re^{i\theta}+se^{i\phi} = e^{i\theta}(r + se^{i(\phi-\theta)})$. Since multiplying by $e^{i\theta}$ is just a rotation by an angle $\theta$, it is enough to consider the case where $\theta=0$ and $0\lt \phi\lt \pi$.

In that case, you have $r + se^{i\phi} = r + s(\cos(\phi)+i\sin(\phi))$, and you want to express it in the form $t(\cos(\chi) + i \sin(\chi))$. Looking at real and complex parts, you see that $t\sin(\chi)=s\sin(\phi)$, and $r+s\cos(\phi) = t\cos(\chi)$.

Assume first that $0\lt \phi\lt \frac{\pi}{2}$. Then $\chi$ must also lie in the first quadrant, since we need both $\sin(\chi)$ and $\cos(\chi)$ to be positive (since $t$, $s\sin(\phi)$, and $r+s\cos(\phi)$ are all positive). If, on the other hand, $\frac{\pi}{2}\lt \phi\lt \pi$, then $\cos(\phi)$ is negative. If $r+s\cos(\phi)\gt 0$, then we need $\cos(\chi)\gt 0$, so $\chi$ is in the first quadrant and automatically smaller than $\phi$ and we are done. If $r+s\cos(\phi)$ is negative then we need $\chi$ in the second quadrant.

In the former case, $0\lt \chi,\phi\lt \frac{\pi}{2}$, then from $\sin(\chi)=\frac{s}{t}\sin(\phi)$ and since $\sin(x)$ is increasing on $0\leq x\leq\frac{\pi}{2}$, then $0\lt\chi\lt\phi$ if and only if $\frac{s}{t}\lt 1$, if and only if $s\lt t$.

Now note that $t^2 = ||r+se^{i\phi}||^2 = r^2 + s^2 + 2rs\cos(\phi) \gt s^2$, since all of $r$, $s$, and $\cos(\phi)$ are positive. Since $s$ and $t$ are both positive, $s\lt t$, which shows that $0\lt\chi\lt\phi$, as desired.

In the other case, where $\frac{\pi}{2}\lt\phi\lt \pi$ and $r+s\cos(\phi)\lt 0$, then we know $\chi$ is also in the second quadrant where $\sin(x)$ is decreasing, so from $\sin(\chi)=\frac{s}{t}\sin(\phi)$ we get that $\frac{\pi}{2}\lt \chi\lt \phi$ if and only if $\frac{s}{t}\gt 1$, if and only if $s\gt t$. Here, since $\cos(\theta)\lt 0$, then you have again $t^2 = ||r+se^{i\phi}||^2 = s^2 + r(r + 2s\cos(\phi)) \lt s^2$ (since $r+s\cos(\phi)\lt 0$ in this situation), so you get $t\lt s$ and hence $\frac{\pi}{2}\lt\chi\lt\phi$, as desired

• @Isaac: I had inadvertedly assumed the sum would be in the first quadrant when it could be (as in your example) in the second. I was in the process of fixing that assumption when you posted your comment. – Arturo Magidin Oct 25 '10 at 20:47
• Yes, totally fixed. +1 – Isaac Oct 25 '10 at 20:49

It's enough to do the case $\theta=0$ (just factor out $e^{i\theta}$). Assume $0<\phi<\pi/2$. Then $$0 < \tan\arg(r+s e^{i\phi}) = \frac{s\sin\phi}{r+s\cos\phi} < \frac{s\sin\phi}{s\cos\phi} = \tan\phi,$$ so the argument of the sum is between $0$ and $\phi$ as desired. Something similar should work when $\phi$ is obtuse.