# Thread: Complex fourth root of -1?

1. ## Complex fourth root of -1?

hi, I have a problem which asks me for the 4th complex root of -1...but I have never done this (it was in a course which I have not taken).

Any help would be very much appreciated, thanks!

2. We know that $\displaystyle -1=\exp(\pi i)$. Then let $\displaystyle \sigma = \exp \left( {\frac{{\pi i}}{4}} \right)$.
We see that $\displaystyle \sigma^4=-1$ so we have one fourth root.

If $\displaystyle \xi = \exp \left( {\frac{{\pi i}}{2}} \right)$ then the four roots are $\displaystyle \sigma\xi^k,~k=0,1,2,3$.

3. Thanks a lot!

I have one question though...I understand everything but the last line, where I get a bit confused. The result would be equivalent to (-1)^(1/4)*(-1)^(1/2)*k, but why the second bit? ((-1)^(1/2)*k)

Thanks for your help, and sorry for bothering you again :-/

4. Originally Posted by juanma101285
The result would be equivalent to (-1)^(1/4)*(-1)^(1/2)*k, but why the second bit? ((-1)^(1/2)*k)
That is not all what that means.
The roots are:
$\displaystyle \frac{{\sqrt 2 }}{2} + i\frac{{\sqrt 2 }}{2},~\frac{{-\sqrt 2 }}{2} + i\frac{{\sqrt 2 }}{2},~-\frac{{\sqrt 2 }}{2} - i\frac{{\sqrt 2 }}{2},~\frac{{\sqrt 2 }}{2} - i\frac{{\sqrt 2 }}{2}$.

Because $\displaystyle \frac{{\sqrt 2 }}{2} + i\frac{{\sqrt 2 }}{2}=\exp\left(\frac{i\pi}{4}\right)$

5. You can also use this formula:

$\displaystyle \displaystyle w_k=r^{\frac{1}{n}}*\left(cos\left(\frac{\theta+2\ pi k}{n}\right)+isin\left(\frac{\theta+2\pi k}{n}\right)\right)$

Where r is the modulus and theta the angle.

For your example, the modulus of $\displaystyle -1$ is $\displaystyle \displaystyle\sqrt{a^2+b^2}=\sqrt{(-1)^2+0^2}=1$.

And the angle is $\displaystyle \pi$ since the coordinates are (-1,0).

N is the number of roots and k=0,1,2,3.

Therefore, you would solve

$\displaystyle \displaystyle w_0=1^{\frac{1}{4}}*\left(cos\left(\frac{\pi+2\pi* 0}{4}\right)+isin\left(\frac{\pi+2\pi*0}{4}\right) \right)$ This is the Principle root of $\displaystyle \displaystyle z^4=-1$

$\displaystyle \displaystyle w_1=1^{\frac{1}{4}}*\left(cos\left(\frac{\pi+2\pi* 1}{4}\right)+isin\left(\frac{\pi+2\pi*1}{4}\right) \right)$

and do the same for w_2 and w_3. Also, once you solve w_0, you can divide $\displaystyle 2\pi$ by 4 to figure out the other roots with out using the formula.