Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


share|cite|improve this question
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 Li Aug 21 '12 at 2:47
up vote 3 down vote accepted

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

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

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

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.