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 02:37
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2Rewrite $w^8$ in the polar form: $r(\cos\theta+i\sin\theta)$. Then, see the de Moivre's formula. – Frenzy Li Aug 21 '12 at 02:47
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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...
DonAntonio
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Wow, second typo discovered now. Of course, it shall be corrected now. Thanks @AndréNicolas – DonAntonio Aug 21 '12 at 03:53
