Argument of sum of two complex numbers 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)
 A: 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
$$\arg(z+w)=\frac{1}{2i}[\ln(z+w)-\ln(z^*+w^*)]$$
A: 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.
A: 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.
A: 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 $\arg(z+w)$.
A: 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 $w=w_R+i w_I$.
A: 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.
A: 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.
