Originally Posted by

**Moon** Hello, I'm new here.

I have a problem with understanding this little equation:

$\displaystyle

z^{4} = i

$

According to WolframAlpha, the solutions are:

$\displaystyle

z_{1} = - \sqrt[8]{-1},

z_{2} = \sqrt[8]{-1},

z_{3} = - (-1)^{\frac{5}{8}},

z_{4} = (-1)^{\frac{5}{8}}

$

I can understand the first two solutions, since this equation can be written like this:

$\displaystyle

(z - \sqrt{\sqrt{i}})(z + \sqrt{\sqrt{i}})(z^2 + \sqrt{i}) = 0

$

And I just don't know how to deal with the following equation:

$\displaystyle

(z^2 + \sqrt{i}) = 0

$

Apparently, the solutions are:

$\displaystyle

z_{3} = - \sqrt{- \sqrt{i}}

$

and:

$\displaystyle

z_{4} = \sqrt{- \sqrt{i}}

$

But how do I go from that to this?:

$\displaystyle

z_{3} = - (-1)^{\frac{5}{8}} $

$\displaystyle

z_{4} = (-1)^{\frac{5}{8}}

$

Please give me a step by step solution, if you can.

Uhmm... I just got it, somehow seeing it on screen made it simpler, I'm sorry to bother you.

It is of course:

$\displaystyle

- \sqrt{-\sqrt{i}} = - \sqrt{-\sqrt{\sqrt{-1}}} = $ $\displaystyle = - \sqrt{-1 * \sqrt{\sqrt{-1}}} = - \sqrt{-1} * \sqrt{\sqrt{i}} =$ $\displaystyle = - i * i^{\frac{1}{4}} = - i^{\frac{5}{4}} = - (- 1)^{\frac{5}{8}}

$

You may delete this topic, sorry for the fuss.

Thanks,

Moon