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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)

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    $\begingroup$ Is Arg $(z+w)$ the same as Arg $(z+2w)$ as implied by your question? $\endgroup$ Aug 31, 2017 at 14:56
  • $\begingroup$ Try your formula on Arg$ (z+z)$. $\endgroup$
    – lulu
    Aug 31, 2017 at 14:56
  • $\begingroup$ Please use MathJax. $\endgroup$ Aug 31, 2017 at 14:58
  • $\begingroup$ @lulu ( arg(z) +arg(z) )/ 2= arg(2z) =arg(z). What's wrong? $\endgroup$
    – john doe
    Aug 31, 2017 at 15:02
  • $\begingroup$ @johndoe Are you interested in $\operatorname{arg}(z)+\operatorname{arg}(z)$? Or in $\operatorname{arg}(z+z)$? $\endgroup$ Aug 31, 2017 at 15:08

7 Answers 7

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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.

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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

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

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  • $\begingroup$ Does * mean conjugate? Any hints on how I can obtain this form? $\endgroup$
    – john doe
    Aug 31, 2017 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 $\theta=(1/2i)\ln(z/z^*)=(1/2i)(\ln z-\ln z^*)$. $\endgroup$ Aug 31, 2017 at 15:35
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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.

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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 $\arg(z+w)$.

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  • $\begingroup$ So is there no closed form relation between the three arguments? $\endgroup$
    – john doe
    Aug 31, 2017 at 15:19
  • $\begingroup$ Indeed, no. It's like trying to solve a triangle knowing only one angle, and no side lengths. $\endgroup$ Aug 31, 2017 at 15:21
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Answering the question in the title, you have simply $$\arg(z+w)=\arctan\frac{z_I+w_I}{z_R+w_R},$$ where $z=z_R+i z_I$ and $w=w_R+i w_I$.

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If the arg of $z+w$ is determined by the args of $z,w$, then $\operatorname{arg}(i+0)=\operatorname{arg}(i+1)$ which is not true.

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Appears answered, but I will add something anyways: if the numbers have the same magnitude, then you can use a variant of the Dirichlet Kernel to write this as:

\begin{equation} \begin{split} \exp{j \theta_{1}} + \exp{ \theta_{2}} &= \exp{ j \theta_{1}} \left(1 + \exp{ j (\theta_{2} - \theta_{1}) } \right) \\ &= \exp{ j \theta_{1}} \left( \displaystyle \sum_{k=0}^{1} \exp{ j k (\theta_{2} - \theta_{1})} \right) \\ &= \exp{ j \theta_{1}} \exp{j \frac{1}{2} (\theta_{2} - \theta_{1}) }\cfrac{\sin ( \theta_{2} - \theta_{1} )}{\sin ( \frac{1}{2}(\theta_{2} - \theta_{1}) ) } \\ &= \exp{j \frac{1}{2} (\theta_{2} + \theta_{1}) } \cfrac{\sin ( \theta_{2} - \theta_{1} )}{\sin ( \frac{1}{2}(\theta_{2} - \theta_{1}) ) } \end{split} \end{equation} The argument is the term in the $\exp\left( \right)$ function, while the amplitude is the ratio of sines. If the numerator and denominator differ in sign, the phase includes an additional phase term of $\pm\pi$, and I did not keep track of that here. Feel free to call me out on any other phase related oversights that are sourced by divide by zeros here.

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