nth roots of negative numbers

Question

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.

Answer 1

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?

Answer 2

DeMoivre's formula states:

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

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

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

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

Answer 3

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 $t = \pi/n$.

Answer 4

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.