4
$\begingroup$

Basically, is there a formula for $\sqrt[i]{z}$.

I was thinking about complex numbers and trying to come up with formulas like $log_i(x)$ and $x^i$. I then thought about this:

$$\sqrt[i]{z} = f(z)$$

So I started trying. We know that $x^i = \cos(\ln(x))+i\sin(\ln(x))$, from Euler's formula. I then tried to reverse it:

$$\ln(x) \to e^x$$ $$\cos(x) \to \arccos(x)$$ $$...e^{\arccos(x)}$$

But, obviously, it will give you the wrong answer for $i\sin(\ln x)$.

I'm also unsure if $\arccos$ is even defined for values > 1.

  • 0
    $\arccos$ has a complex definition btw. Mainly, look up hyperbolic trig functions. – 2017-01-31
  • 0
    @Olly Britton do the provided answers answer your question? If so, you should consider accepting one of them by clicking the checkmark. – 2017-01-31

3 Answers 3

0

Note that $1/i=-i$. So, we may in fact define $$ f(z)= z^{-i} $$ Keep in mind, however, that $z^w$ is a multi-valued function for any non-real complex number $w$.

1

It's a multi-valued function (see here for an explanation of branch cuts). Specifying a branch and taking an appropriate subset of $\mathbb{C}$ let's you define it on that subset.

To see how to define it, notice that $\frac{1}{i}=-i$ and so $z^{\frac{1}{i}}=z^{-i}=\frac{1}{z^i}$ and see if you can derive it from there via Euler's formula.

  • 1
    Writing it as $z^{-i}$ is probably better for use though, but +1 anyways for covering everything. :-) – 2017-01-31
0

Hint:

$$\sqrt[i]{z}=z^{1/i}=z^{-i}$$

And then apply Euler's formula.

  • 0
    So it's $$\cos(-\ln x) + i\sin(-\ln x)$$ – 2017-01-31
  • 0
    @OllyBritton and take the negatives out of the trig functions for simplicity. – 2017-01-31
  • 0
    Wait... you can do that? – 2017-01-31
  • 1
    @OllyBritton $$\cos(-x)=\cos(x)\\\sin(-x)=-\sin(x)$$ – 2017-01-31