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Okay. I know how to solve for all values of $\sqrt{-1} $ but $\sqrt{-1+\iota} $ confuses me a bit. I got the value of r to be $\sqrt 2 $

I ended up with this: $ z_k = \sqrt[8]{2}\;cos(\frac{2k+1}{4}\pi)+ i\sqrt[8]{2}\;cos(\frac{2k+1}{4}\pi)$

I used wolfram alpha to see the possible values and the first one seems to be $ \sqrt[8]{2}\;e^{3i\pi/16} $

The final result confuses me. How did exponent end up in there?

This is my first post and I hope it doesn't violate any rules.

Saad Aleem
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3 Answers3

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The trick is Euler's formula:

$$e^{i \theta}=\cos(\theta)+i\sin(\theta).$$

Multiplying by a factor of $r>0$ you get the polar form of a general complex number: $r e^{i \theta}=r \cos(\theta) + ir\sin(\theta)$.

This form is really nice for roots, because the rules for exponents of real numbers carry over here, with some caveats. For instance, let's look at $\sqrt[3]{-1+i}$. One way to write $-1+i$ would be $\sqrt{2} e^{i 3 \pi/4}$. If you now take its third root in this form, you use the exponent rules and get $\sqrt[3]{-1+i}=2^{1/6} e^{i \pi/4}$. For consistency with the rest of this answer let me rewrite this as:

$$\sqrt[3]{-1+i}=2^{1/6} e^{i 3 \pi/12}.$$

But there are two more. We can find them by introducing an initially superfluous factor of $1$. For one of them, $-1+i=\sqrt{2} e^{i 3 \pi/4} e^{i 2 \pi}$. This last term doesn't do anything initially, but when we use the exponent rules, we get a different third root, namely

$$\sqrt[3]{-1+i}=2^{1/6} e^{i \pi/4} e^{i 2 \pi/3} = 2^{1/6} e^{i 11 \pi/12}.$$

We can do it one more time: starting from $\sqrt{2} e^{i 3 \pi/4} e^{i 4 \pi}$ we get

$$\sqrt[3]{-1+i}=2^{1/6} e^{i \pi/4} e^{i 4 \pi/3} = 2^{1/6} e^{i 19 \pi/12}.$$

If we do it again, what we get will be the first root multiplied by $e^{i 2 \pi}$...which is exactly the first root again! So there are only three third roots. In general there are $n$ nth roots.

Ian
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  • The question actually takes $ \sqrt[4]{-1 + i} $. Also, could you please explain how you rewrote it in the exponent form? – Saad Aleem Apr 18 '15 at 15:55
  • @AleemSaadullah I know you were asked for the fourth root; I did the third root to leave the fourth root to you. As for rewriting, think about the circle. The vector corresponding to $-1+i$ has length $\sqrt{1^2+1^2}=\sqrt{2}$. So now $(-1+i)/\sqrt{2}=-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$. This point is on the unit circle at the angle $3 \pi/4$ (halfway through the second quadrant). – Ian Apr 18 '15 at 15:58
  • Got it! Although, I'm still getting the last two fourth roots wrong. I think it has something to do with the sign of the real part. Here is the equation I got:
    $ \sqrt[8]{z}[ cos(\frac {3\pi + 8 \pi k}{16}) + isin(\frac {3\pi + 8 \pi k}{16})] $     – Saad Aleem Apr 18 '15 at 22:34
  • @AleemSaadullah You need $2$ instead of $z$ of course. As for the arguments, you should have $3 \pi/16$, $3 \pi/16 + \pi/2$, $3 \pi/16 + \pi$, and $3 \pi/16 + 3 \pi/2$. That is, a quarter of the original argument plus $2 \pi k/4$ for $k=0,1,2,3$. – Ian Apr 19 '15 at 00:09
  • Yes. That was a typo. I'm just putting in the values of $ k $ till $ n-1 $ and it gives me the same values as the method you suggested. I guess I'll just disregard the answers that I'm getting from wolframalpha. Thanks for all the answers. Appreciate it. – Saad Aleem Apr 19 '15 at 00:49
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The key to this question is Euler's formula

$$e^{ix}=cos(x)+i sin(x)$$

It is a remarkable formula and you should look into it further, but now let's see how it applies to the problem. When you apply it to the result that Wolfram alpha gave, the formula will make more sense to you.

Sheepman
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We want to use Euler's identity here. We can rewrite any complex number as $Ae^{i\theta}$ by Euler's identity $$Ae^{i\theta}=A\cos(\theta)+Ai\sin(\theta)$$ We can also go backwards from this formula using the fact that for any complex number of the form $r+ic$ $$A=\sqrt{r^2+c^2}$$ and $$\theta=\arccos(r/A)$$ If we want to take the $n$-th root of a complex number, the simplest root to find will always be $\theta/n$. However, Euler's identity tells us that so long as $n\theta_p\equiv \theta \mod 2\pi$, then $\theta_p$ will be a solution to $\sqrt [n]{e^{i\theta}}$. Hope that helps!

Archaick
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