1
$\begingroup$

I was trying to find solution to $\arg(z+w)$, where $z$ and $w$ are two complex numbers in terms of $\arg(z)$ and $\arg(w)$. Making a parallelogram out of vector addition of the $2$ complex numbers in Argand plane leads to $$\arg(z+w)=\frac{\arg(z)+\arg(w)}{2}$$

Am I correct or there are some cases to be accounted for? (consider only principal arguments)

$\endgroup$
  • 2
    $\begingroup$ Is Arg $(z+w)$ the same as Arg $(z+2w)$ as implied by your question? $\endgroup$ – Mark Bennet Aug 31 '17 at 14:56
  • $\begingroup$ Try your formula on Arg$ (z+z)$. $\endgroup$ – lulu Aug 31 '17 at 14:56
  • $\begingroup$ Please use MathJax. $\endgroup$ – José Carlos Santos Aug 31 '17 at 14:58
  • $\begingroup$ @lulu ( arg(z) +arg(z) )/ 2= arg(2z) =arg(z). What's wrong? $\endgroup$ – john doe Aug 31 '17 at 15:02
  • $\begingroup$ @johndoe Are you interested in $\operatorname{arg}(z)+\operatorname{arg}(z)$? Or in $\operatorname{arg}(z+z)$? $\endgroup$ – José Carlos Santos Aug 31 '17 at 15:08
1
$\begingroup$

In order to compute $\arg(z+w)$ you will need to know not only $\arg z$ and $\arg w$, but also $|z|$ and $|w|$. Then it is a trigonometry problem.

$\endgroup$
2
$\begingroup$

First of all, your conclusion that $\arg(z+w)=(\arg z+\arg w)/2$ can only be true when $|z|=|w|$. Furthermore, it's not always true even in that case. In some circumstances it is rotated by $\pi$. I think you need a more general expression for this, such as

$$\arg(z+w)=\frac{1}{2i}[\ln(z+w)-\ln(z^*+w^*)]$$

$\endgroup$
  • $\begingroup$ Does * mean conjugate? Any hints on how I can obtain this form? $\endgroup$ – john doe Aug 31 '17 at 15:31
  • $\begingroup$ Yes, * means conjugate. As for the derivation, consider that $z=|z|e^{i\theta}$, $z^*=|z|e^{-i\theta}$. Then $z/z^*=e^{2i\theta}$ and $\theta=(1/2i)\ln(z/z^*)=(1/2i)(\ln z-\ln z^*)$. $\endgroup$ – Cye Waldman Aug 31 '17 at 15:35
1
$\begingroup$

There are at least two problems with your idea.

First, you are drawing the parallelogram spanned by $0$, $z$, $w$, and $z+w$, and assuming that the diagonal ($0$ to $z+w$) will be the angle bisector of the angle formed by $z$, $0$, $w$. That will be okay if the parallelogram is a rhombus, in other words if $|z|=|w|$ (although there is still a second problem, see below). If $|z| \neq |w|$ then this will not work. That is what the comment about $z+w$ versus $z+2w$ is getting at.

Second, even if $|z|=|w|$, you will still have some problems with the restriction to principal arguments. Say for example $z = 1 = 1+0i$ (so $\arg(z)=0$), and $|w|=1$. What happens if $\arg(w)=\frac{3\pi}{4}$? What happens if $\arg(w)=\pi$? What happens if $\arg(w)=\frac{5\pi}{4}$? Out of these three scenarios, I think you will find that your idea works fine in the first one, but not in the other two.

$\endgroup$
1
$\begingroup$

Your question presumes that $\arg(z+w)$ is completely determined by $\arg(z)$ and $\arg(w)$. However, consider $z=1-i$, $w_1=1+i$, and $w_2=100+100i$.

It is clear that $\arg(w_1)=\arg(w_2)=\frac{\pi}{4}$. However, $\arg(z+w_1)=0$, while $\arg(z+w_2)$ is very close to $\frac{\pi}{4}$.

Thus, knowing $\arg(z)$ and $\arg(w)$ is not sufficient to determine $\arg(z+w)$.

$\endgroup$
  • $\begingroup$ So is there no closed form relation between the three arguments? $\endgroup$ – john doe Aug 31 '17 at 15:19
  • $\begingroup$ Indeed, no. It's like trying to solve a triangle knowing only one angle, and no side lengths. $\endgroup$ – G Tony Jacobs Aug 31 '17 at 15:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.