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Is there a way of simplifying the expression below? I've tried by making the square root of $i$ equal to the square root of the square root of $-1$ though I'm not sure that its correct.

$$z= \pm \sqrt{i\sqrt{3}}$$

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  • $\begingroup$ A tidbit that comes in handy at times is that $\sqrt{i}=\pm \cfrac{1+i}{\sqrt{2}}\;$ (which can be easily derived from DeMoivre's formula, or directly from $\require{cancel}(1+i)^2=\bcancel{1} + 2i + \bcancel{i^2}=2i\,$). Then: $$z^2 = i\sqrt{3}=\left(\frac{\sqrt[4]{3}}{\sqrt{2}}(1+i)\right)^2 \quad\iff\quad z = \pm \frac{\sqrt[4]{3}}{\sqrt{2}}(1+i)$$ $\endgroup$ – dxiv Mar 30 '17 at 3:17
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Actually, your "solutions" come from the equation $z^2=i\sqrt{3}$. The absolute value of this complex number is $\sqrt{3}$ and its argument is $90$ degrees. One may resort to DeMoivre, where the new absolute value then becomes $\sqrt{\sqrt3}$ (that's the fourth root of 3, if you wish) and the new arguments $45$ and $225$ degrees respectively. So apply $z=r(cos\theta+isin\theta)$ with the given values to arrive at the two solutions. Angles are standard for the sine and cosine, so please give it a try from here.

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  • $\begingroup$ Nit: Did you mean $z^2 = i \sqrt{3}$? $\endgroup$ – WB-man Mar 30 '17 at 2:25
  • $\begingroup$ @WB-man Thanks, I edited $\endgroup$ – imranfat Mar 30 '17 at 2:33
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    $\begingroup$ @imranfat I'm not sure what was going through my head when I was answering the question as I know how to find the roots. Thanks for reminding me. $\endgroup$ – randb Mar 30 '17 at 2:38
  • $\begingroup$ @imranfat Is there a way to simplify the solution $$3^{\frac{1}{4}}({\frac{\sqrt{2} }{2}}+i{\frac{\sqrt{2} }{2}})$$ to something neat? $\endgroup$ – randb Mar 30 '17 at 2:56
  • $\begingroup$ I am afraid not so, in part because of that fourth root... $\endgroup$ – imranfat Mar 30 '17 at 3:05

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