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: 43

Answer

$z=5(\displaystyle \cos[\tan^{-1}(\frac{3}{4})]+i\sin[\tan^{-1}(\frac{3}{4})])$

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 $z$. ------- $z=4+3i$ lies in Q.I. $r=|z|=\sqrt{(4)^{2}+(3)^{2}}=\sqrt{25}=5$ $\tan\theta =\displaystyle \frac{3}{4}$ In Q.I, $\displaystyle \theta=\tan^{-1}(\frac{3}{4}) \approx 0.6435$ So, approximately, $z=5(\cos 0.6435+i\sin 0.6435)$ but more precisely, $z=5(\displaystyle \cos[\tan^{-1}(\frac{3}{4})]+i\sin[\tan^{-1}(\frac{3}{4})])$
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