I have to get $\Im$, $\Re$, the absolut value as well as the argument $\phi$ of the complex number

$$z = \left(-\frac{1}{\sqrt2}+\sqrt\frac{3}{2}i\right)^8$$

I do this by transforming $z' = -\frac{1}{\sqrt2}+\sqrt\frac{3}{2}i$ into polar-coordinates:

$|z'| = \sqrt{\left(\frac{1}{\sqrt2}\right)^2+\left(\sqrt{\frac{3}{2}}\right)^2 }= \sqrt2$

$arg~z' = \phi = arctan\frac{\Im~z}{\Re~z} = arctan\frac{\sqrt\frac{3}{2}}{-\frac{1}{\sqrt2}} = arctan (-\sqrt3) = -\frac{\pi}{3}$

So that gives me:

$re^{i\phi} = |z|e^{i~arg~z} \rightarrow z' =\sqrt2 e^{-\frac{i\pi}{3}}$

Now I don't have to continue any further because the rest should be clear to me. My problem is that the solution states that $z' = \sqrt2 e^{\frac{2\pi i}{3}}$ and thus the argument is $\frac{2\pi}{3}$ (I suppose). But why is that?

Thank you for advice!



I suppose that some $i$'s are missing in your post.

The answer is effectively $\frac{2\pi}{3}$ because $x<0$ and $y>0$. This kind of ambiguity is a classical problem. Just have a look at here.

  • $\begingroup$ Thanks - I got that and also fixed the $i's$ x) ... ouch $\endgroup$ – FunkyPeanut Jan 25 '14 at 14:05

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