# What does $\sqrt[n]{z}$ mean, when $z$ is an arbitrary complex number?

What does $\sqrt[n]{z}$ mean, when $z$ is an arbitrary complex number? Is it a single complex number, or the set of $n$-th roots of $z$?

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When I see that symbol, I expect it to mean a single $n^{th}$ root of $z$. The problem is, without additional context, I don't know which one is meant! The symbol is not well-defined, since there are indeed $n$ distinct $n^{th}$ roots of $z$ if $z$ is not $0$. Usually we define a function sending complex numbers to one of their $n^{th}$ roots, for example by sending $re^{i\theta}$ to $\root n \of re^{i\theta/n}$, where $\theta \in [0,2\pi)$. But notice that this function is not continuous on the positive real line. (The positive real axis is said to be a branch cut for this function. We could in fact use any ray from the origin as a branch cut to define an unambiguous $n^{th}$ root function.) After explaining how the function is defined, one is then free to (ab)use the notation $\root n \of z$.

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In case my notation is unfamiliar: $re^{i\theta}$ is another way of writing $r(\cos\theta+i\sin\theta)$. –  Brett Frankel Mar 9 '12 at 5:35
The problem is that the author of a textbook at hand uses it without any preceding definition; he writes, "if $\sqrt{i} = x + iy$, where $x,y\in\mathbb{R}$, then $x=y=\pm 1/\sqrt{2}$". –  Pteromys Mar 9 '12 at 9:10
Another abuse of notation. He's being ambiguous by saying that "the squareroot" could be either of the two values. You should read this as "$x=y=1/\sqrt2$ and $x=y=-1/\sqrt2$ are the two complex numbers that, when squared, give you $i$. –  Brett Frankel Mar 9 '12 at 18:53

This symbol is for the set of all $n-th$ root of $z$.

$$z= r(\cos\theta+i\sin\theta)$$ For $k= 0,1,,,n-1$, we have: $$z^\frac{1}{n}=\{ r^{\frac{1}{n}}[\cos(2\pi k +\theta)+i\sin(2\pi k+\theta)]^\frac{1}{n}| k=0,1,..n-1\}$$ that is $$z^\frac{1}{n}=\{ r^{\frac{1}{n}}[\cos\frac{(2\pi k +\theta)}{n}+i\sin\frac{(2\pi k+\theta)}{n}]| k=0,1,..n-1\}$$

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If I may respectfully disagree, when I see the symbol I don't think the set of roots, I expect it to mean a specific $n^{th}$ root (and since the symbol is not defind, I search for context to see which root is meant). I think mathematicians usually expect that symbol to be a function, and use other notation to indicate the set of roots $\{w:w^n=z\}$. –  Brett Frankel Mar 9 '12 at 5:09
@BrettFrankel, yeah i too agree with you... but question was what does $z^\frac{1}{n}$ mean... As you said this doesn't make sense....only to make sense of this symbol we can say that this is a set.... +1 for ur answer as i think ur answer is more appropriate. –  zapkm Mar 9 '12 at 5:14
Yes, the symbol is not well-defined. But mathematicians still abuse the notation and use it to denote functions, they just explain which root they mean. –  Brett Frankel Mar 9 '12 at 5:17