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Let $z = a+bi$, $w = c+di$ and $s = z+w$.

It is very easy and straightforward to calculate $s$, $\arg(s)$ and $|s|$, but what I want to know, is how $\arg(s)$ and $|s|$ are related to $z$ and $w$.

What is $\arg(s)$ in terms of $\arg(z)$ and $\arg(w)$?
And likewise, what is $|s|$ in terms of $|z|$ and $|w|$?

Funny enough, complex addition seems to be more difficult than multiplication, when we think about $\arg(z)$ and $|z|$. Even complex multiplication is easier, a simple formula for that using only $|x|$, $Arg(x)$ and other basic operations, can be written like this: $$zw = |z||w|e^{(Arg(z)+Arg(w))i}$$ And for complex exponentiation it's a bit more complicated and complex (literally): $$z^w = \frac{|z|^{Re(w)}}{e^{Arg(z)Im(w)}}e^{(Arg(z)Re(w)+ln(|z|)Im(w))i}$$

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  • $\begingroup$ Not much can be said about that, at best all you can say is that $|s| \leq |z| + |w|$ $\endgroup$ – Naweed G. Seldon Nov 8 '18 at 19:34
  • $\begingroup$ Looks like a duplicate of math.stackexchange.com/questions/2412297/… $\endgroup$ – Naweed G. Seldon Nov 8 '18 at 19:36
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    $\begingroup$ You can't say much about $\arg(s)$ in terms of $\arg(z)$ and $\arg(w)$, and you can't say much about $|s|$ in terms of $|z|$ and $|w|$. However, in terms of $\arg(z)$ and $\arg(w)$ and $|z|$ and $|w|$, you can say everything you want to, using basic trigonometry. $\endgroup$ – Arthur Nov 8 '18 at 19:38
  • $\begingroup$ Yes that post is similar, but im not just asking about that, im asking how to express arg(s) and |s| in terms of |z|, arg(z), |w| and arg(w). Calculating arg(z+w) is just atan((Im(z)+Im(w))/(Re(z)+Re(w)) + or minus pi depending on wether Re(z+w) and Im(z+w) is positive or not. @NaweedSeldon $\endgroup$ – Nils Phillip Talgö Nov 8 '18 at 19:50
  • $\begingroup$ Maybe expressing arg(s) is easier by using a combination of arg(z) and |z|? When we think about it, we both know the angles of z and w, and we also know their lenghts, doesnt that form a triangle? link $\endgroup$ – Nils Phillip Talgö Nov 8 '18 at 19:58
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There is no shortcut. Convert to Cartesian, add and back to polar.

This yields

$$|z+w|=\sqrt{(|z|\cos(\text{Arg }z)+|w|\cos(\text{Arg }w))^2+(|z|\sin(\text{Arg }z)+|w|\sin(\text{Arg }w))^2}=\sqrt{|z|^2+2|z||w|\cos(\text{Arg }z-\text{Arg }w)+|w|^2}$$

and similar for the argument.

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  • $\begingroup$ If you compare the formula for multiplication and exponentiation to this, addition actually seems more complicated. Who would have thought that in some way, $2+3$ is actually harder than $2*3$ and $2^3$. $\endgroup$ – Nils Phillip Talgö Nov 9 '18 at 9:49
  • $\begingroup$ @NilsPhillipTalgö: have a look at $(2+3i)^{1-2i}$... $\endgroup$ – Yves Daoust Nov 9 '18 at 9:50
  • $\begingroup$ Actually, the polar form is very close to the logarithm. $\log z=\log|z|+i\text{Arg }z$. This explains why the product in polar form is easy, $\log zw=\log z+\log w$. And also why addition doesn't work: $\log(z+w)=???$. $\endgroup$ – Yves Daoust Nov 9 '18 at 9:53

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