# Find $w$ such that $w^8=15-15i$

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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dang, im sry, got the syntax wrong, its suppose to be w^8=15-15*I(imaginary number); –  ryantata Aug 21 '12 at 2:37
Rewrite $w^8$ in the polar form: $r(\cos\theta+i\sin\theta)$. Then, see the de Moivre's formula. –  FrenzY DT. Aug 21 '12 at 2:47

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

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

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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Thanks, @minopret . –  DonAntonio Aug 21 '12 at 3:48
Wow, second typo discovered now. Of course, it shall be corrected now. Thanks @AndréNicolas –  DonAntonio Aug 21 '12 at 3:53

Hint: use the polar form of complex numbers.

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