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Could anyone give me some explanation how can I compute sum of multibranched functions? For example, if I have $z=re^{i\varphi}$ then $$\sqrt{z}=\sqrt{r}(\cos(\varphi/2+k\pi)+i\sin (\varphi/2+k\pi))$$ for some integer $k$. Therefore, is $$\sqrt{z}+\sqrt{z}=2\sqrt{r}(\cos(\varphi/2+k\pi)+i\sin (\varphi/2+k\pi))$$ for some integer $k$ or $$\sqrt{z}+\sqrt{z}=\sqrt{r}(\cos(\varphi/2+k_1\pi)+i\sin (\varphi/2+k_1\pi))+\sqrt{r}(\cos(\varphi/2+k_2\pi)+i\sin (\varphi/2+k_2\pi))$$ for some integers $k_1,k_2$?

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I don't understand what you intended to say with $\sqrt{z}+\sqrt{z}$. Did you mean to add the two possible square roots of $z$? In that case, you'll want to remember the Vieta formulae... – J. M. Oct 22 '11 at 10:31
What in the world is the point of adding these superfluous $k$'s when you can just use what you have in the first line? If you're trying to sum multiple roots from different branches, you have to specify which branches you want each root in first. @J.M. I think Vieta's formulas are overdoing it when we have the square root function. The only two roots are $\pm\sqrt{z}$ so... – anon Oct 22 '11 at 10:32
@anon: I was thinking in general; that is, assuming he wanted to add up all the possible cube roots of a number... something like that. – J. M. Oct 22 '11 at 10:36
I read the formula in some formula collection of mine, and I was wondering how to use the result as an intermediate result of some calculation. For example, I think that in third degree equations one meet these intermediate results. – student Oct 22 '11 at 10:38
@student: I read your last comment because I happened to come back to this page. If you want people to be notified of your responses, you need to put an '@' in front of their name. – joriki Oct 22 '11 at 11:31
up vote 1 down vote accepted

The notation $\sqrt z$ usually refers to the principal square root. If your equations are interpreted with this convention, then the first of your two options is correct. However, there may be situations that require you to sum different roots of the same value. In that case, you would need a different notational device to represent the left-hand side, e.g., you could write


and in that case you might need something like your second option to compute this sum.

Your first equation is incorrect, since you get two different values on the right-hand side for different values of $k$, so you need some form of notation on the left-hand side that depends on $k$. For instance, you could say that the square roots of $z$ are given by

$$\omega_k=\sqrt{r}\left(\cos(\varphi/2+k\pi)+\mathrm i\sin (\varphi/2+k\pi)\right)\;.$$

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The sum of the $n$ complex roots of any polynomial $P_n(x)=x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n$ is $(-a_1)$ by Vieta's formulæ. The $n$-th roots of $z$ are the $n$ roots of the polynomial $P_n(x)=x^n-z$ by definition hence their sum is zero for every $z$ and every $n\geqslant2$.

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Oh Vieta... you really do solve a lot of things like these. :) – J. M. Oct 22 '11 at 10:37
Ah, ok. What about $\sqrt{-1}+\sqrt[3]{-1}$? Do we meet this kind of situations in math where one has to compute different multibanched functions? – student Oct 22 '11 at 10:43
@student: did you intend to say $\pm\sqrt{-1}+(-1)^{1/3}$, where "$(-1)^{1/3}$" is any of the three possible cube roots? – J. M. Oct 22 '11 at 10:51
By joriki's answer, I realized that the book I read uses bad notation. For example, if I type 2x^3+4x^2-1=0 to Wolfran alpha, it contains sum of roots of complex numbers. I was wondering how to read those answers. I guess there might be some equation where WA gives $\sqrt{-1}+\sqrt[3]{-1}$ as exact result and some approximation of it. – student Oct 22 '11 at 11:00
student: now that the dust settled, let me mention that I still do not know what was your question. I read it carefully and I thought I knew unambiguously what it was when I posted my answer but now, you have added comments which do not fit what I thought, nor, in my opinion, the text of your question. In the end, this is to suggest you try to reach more precise formulations of your future questions, if any. – Did Oct 23 '11 at 5:50

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