Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 8 - Section 8.3 - Polar Form of Complex Numbers; De Moivre's Theorem - 8.3 Exercises - Page 610: 36


$z=6(\displaystyle \cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$

Work Step by Step

See p. 604, A complex number $z=a+bi$ has the polar (or trigonometric) form $z=r(\cos\theta+i\sin\theta)$ where $r=|z|=\sqrt{a^{2}+b^{2}}$ and $\tan\theta=b/a$. The number $r$ is the modulus of $z$, and $\theta$ is an argument of $z$. ------- $z=3+3i\sqrt{3}$ $r=|z|=\sqrt{(3)^{2}+(3\sqrt{3})^{2}}=\sqrt{9+9\cdot 3}=\sqrt{36}=6$ $\displaystyle \tan\theta=\frac{3\sqrt{3}}{3}=\sqrt{3}$, In quadrant I, $\displaystyle \tan\frac{\pi}{3}=\sqrt{3}$. By symmetry (on the unit circle), $\theta$ can also be $\displaystyle \frac{4\pi}{3}$ (quadrant III). $3+3i\sqrt{3}$ lies in quadrant I, so we select the appropriate argument: $z=6(\displaystyle \cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.