A Simple Complex Number Equation

**Moon** · June 19th 2010, 09:39 AM

Hello, I'm new here.

I have a problem with understanding this little equation:
$ z^{4} = i $

According to WolframAlpha, the solutions are:
$ 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:

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

Apparently, the solutions are:
$ z_{3} = - \sqrt{- \sqrt{i}} $
and:
$ z_{4} = \sqrt{- \sqrt{i}} $

But how do I go from that to this?:
$ z_{3} = - (-1)^{\frac{5}{8}}$
$ 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:

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

You may delete this topic, sorry for the fuss.
Thanks,
Moon

**tonio** · June 19th 2010, 04:01 PM

Originally Posted by Moon

Hello, I'm new here.

I have a problem with understanding this little equation:
$ z^{4} = i $

According to WolframAlpha, the solutions are:
$ 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:

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

Apparently, the solutions are:
$ z_{3} = - \sqrt{- \sqrt{i}} $
and:
$ z_{4} = \sqrt{- \sqrt{i}} $

But how do I go from that to this?:
$ z_{3} = - (-1)^{\frac{5}{8}}$
$ 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:

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

You may delete this topic, sorry for the fuss.
Thanks,
Moon

That's a weird way to write down complex numbers...I'd do the following:

$z^4=i=e^{\pi i/2+2k\pi i}=e^{\frac{\pi i}{2}(1+4k)}\,,\,\,k=0,1,2,3$ , so now taking the fourth rooth on both sides and using De Moivre's theorem we get:

$z_k=e^{\frac{\pi i}{8}(1+4k)}$ , and the four different roots are obtained by taking the four different values of k.

Tonio

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