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Disclaimer: I know what complex numbers are.

Let $x,\space n\in\Bbb R$

What is the complex algebraic solution to $\sqrt[n]{-x}$? Could I have a 'general' formula and a walk through on how to accomplish this.

I know about roots of Unity such that: $$\large\sqrt[n]{\pm 1}=\pm e^{(2\pi ki)/n} $$ And that: $$\sqrt{-x} = i\sqrt{x}$$ So really, it's just that I do not understand when it comes to higher radicals.

rurouniwallace
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Ali Caglayan
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    The answer is simpler when $x\le 0$, so take $x\gt 0$. It is enough to find a single $n$-th root $w$ of $-x$, since then all $n$-th roots are given by $we^{2k\pi i/n}$, where $k$ ranges from $0$ to $n-1$. – André Nicolas Jul 21 '13 at 20:13
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    As a general rule, when $n$ is an arbitrary real number, we prefer $$x^{1/n}$$ to $\sqrt[n]{x}$ because of the word "root" implicit in the latter notation strongly implying $n$ is an integer. Not manditory, but generally preferred. – Thomas Andrews Jul 21 '13 at 20:30
  • @Alizter Do you want $n$ to be any real number, or do you want $n$ to be an integer? – Pedro Jul 21 '13 at 20:47
  • @PeterTamaroff I stated that so that complex (strictly speaking non -real) radicals could be ruled out. – Ali Caglayan Jul 21 '13 at 21:57

4 Answers4

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You have $x\in \Bbb R$ and I presume $n\in\Bbb N$. You are looking for solutions of $$z^n=-x$$

Find $w$ such that $w^n =-1$. Then you want to solve $$w^nz^n=(wz)^n=x$$ that is $$z_0^n=x$$ where $z_0=wz$. Then you can find $z$.

Can you take it from here?

Pedro
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DeMoivre's formula states:

$$(\cos{x}+i\sin{x})^n=\cos(nx)+i\sin(nx)$$

Rewrite $\sqrt[n]{-x}$ as follows:

$$\sqrt[n]{-x}=\sqrt[n]{-1}\cdot\sqrt[n]{x}$$

$$\sqrt[n]{-1}=(-1)^{1/n}=(\cos{\pi}+i\sin{\pi})^{1/n}=(\cos{\frac{\pi}{n}}+i\sin{\frac{\pi}{n}})$$

rurouniwallace
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  • DeMoivre really only works for integer values of $n$, otherwise it only gives you one possible value for the expression. – Thomas Andrews Jul 21 '13 at 20:33
  • @ThomasAndrews That's true. But that's only if you're looking for y's that satisfy $-x=y^n$. However the root function itself only yields a single output for a single input, right? – rurouniwallace Jul 21 '13 at 20:41
  • That's the question - what does $\sqrt[x]{z}$ mean when $x=\sqrt{2}$? This seems to be the heart of the question. Usually, as I mentioned in the comment above, we don't use the $\sqrt[x]{z}$ notation unless $x$ is positive integer, preferring the notation $y^{1/x}$. But $z^w$ is not a single-valued function on the non-zero complex numbers $z$, either, unless you are willing to either leave it undefined or make it discontinous. – Thomas Andrews Jul 21 '13 at 20:48
  • @ThomasAndrews I don't see that $x$ being a real number rather than an integer is at the heart of the question, since it's not stated in the title nor is it mentioned anywhere except the one point where the OP states $x\in\mathbb{R}$. In most of the question as stated the OP seems to be focusing more on the fact that the radicand is negative. But anyway even if it's written $y=x^{1/n}$, a power function still only one output to one input (since it is a function, after all). – rurouniwallace Jul 21 '13 at 21:05
  • Well, in your formulation, it is $n$ which is an arbitrary real number - I write $\sqrt[x]{y}$ in my comment, which might have confused things. – Thomas Andrews Jul 21 '13 at 21:07
  • @Thomas Sorry, that was a typo, I did mean to say :$n$ being a real number rather than an integer" and "the OP states $x\in\mathbb{R}$ (talk about confusing things lol) – rurouniwallace Jul 21 '13 at 21:10
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First, we rarely write $\sqrt[n]{z}$ unless $n$ is a positive integer. We prefer to write $z^{1/n}$ for other $n$, just for clarity.

For general real $y$, the complex function $z\to z^y$ is not a single-valued function(*). If y is an integer, then $z^y$ can be defined to be a single-valued function on all of the complex plane. If $y$ is rational, then for each $z\neq 0$ there are finitely many possible values for $z^y$. When $y$ is irrational, then there are infinitely many possible values for $z^y$ when $z\neq 0$.

The only way to define $z^w$ for $z,w$ complex is to define the natural logarithm of $z$. But there are infinitely many such logarithms for any $z$ - if $e^v = z$, then $e^{v+2\pi k i}=z$ for any integer $k$. So $z^w$ can take any of the values $$z^w=e^{wv}\left(e^{w\cdot 2\pi i}\right)^k$$

When $w$ is an integer, then $e^{w\cdot 2\pi i}=1$, so there is only one possible value. When $w$ is a rational number, then $e^{w\cdot 2\pi i}$ is a root of unity, so there are only finitely many possible values. When $w$ is irrational or complex, then there are infinitely many possible values for $z^w$.

(*) You can, of course, choose a single value for the function, but then you have to give up other nice properties - either the function will not be continuous or the function will not be defined for all $z\neq 0$. You also do not get nice properties like $z_1^wz_2^w=(z_1z_2)^w$, while a multi-valued function will give you this formula, if you take it to mean for any values $z_1^w$ and $z_2^w$, their product is some value for $(z_1z_2)^w$.

It's an imperfect answer, but there is no perfect answer for general exponentiation for complex numbers - the standard rules that you know and love for $z$ a positive real just break down when you try to extend to non-zero complex $z$. You have to compromise some property, and it turns out the most "elegant" compromise is to remove the single-valued condition on exponentiation. But that is not obvious at the beginning.

Thomas Andrews
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You want to find a solution to $-x = z^n$.

Let $z = r e^{i t}$, where $0 \le t < 2 \pi$. $z^n = r^n e^{i t n}$. If $z^n = -x$, since $x$ is real, $r^n = |x|$, so $r = |x|^{1/n} = e^{\ln |x|/n} $ and $itn = \pi i$ or $t = \pi/n$.

marty cohen
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