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Find the complex number, lying in the second quadrant, and having the smallest possible real part, which satisfies the equation

$w^8=15-15i$

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    Rewrite $w^8$ in the [polar form](http://en.wikipedia.org/wiki/Complex_number#Polar_form): $r(\cos\theta+i\sin\theta)$. Then, see the [de Moivre's formula](http://en.wikipedia.org/wiki/De_Moivre%27s_formula).2012-08-21

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$z=15-15i\Longrightarrow |z|=15\,\sqrt 2\,\exp({7\pi i}/{4}+2k\pi i),\,k\in\Bbb Z$

$\Longrightarrow w^8=z\Longrightarrow w=z^{1/8}=15^{1/8}\,2^{1/16}\,\exp({7\pi i}/32+{k\pi i}/{4})$

Now just observe that as $\,k\,$ runs from $\,0\,$ to $\,7\,$, we get all the possible (eight) values on the right-hand side above...

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    Wow, seco$n$d typo discovered $n$ow. Of course, it shall be corrected now. Thanks @AndréNicolas2012-08-21
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Hint: use the polar form of complex numbers.