Question

Find the 2nd of the three cube roots of: $Z=27(\cos(2\pi/3)+i\sin(2\pi/3))$. a. $Z^{\frac{1}{3}}=2(\cos(8\pi/9)+i\sin(8\pi/9))$ b. $Z^{\frac{1}{3}}=3(\cos(8\pi/9)+i\sin(8\pi/9))$ c. $Z^{\frac{1}{3}}=2(\cos(8\pi/9)+i\sin(8\pi/9))$ d. $Z^{\frac{1}{3}}=3(\cos(8\pi/9)+i\sin(8\pi/9))$

Name: find the 2nd of the three cube roots of z27cos2pi3isin2pi3 a zfrac132cos8pi9isin8pi9 b zfrac133cos8pi9isin8pi9 c zfrac132cos8pi9isin8pi9 d zfrac133cos8pi9isin8pi9 11788
Uploaded: 2021-08-22T09:11:16-08:00
Duration: 2 min 27 s
Channel: Gaurav Kalra
Description: find the 2nd of the three cube roots of z27cos2pi3isin2pi3 a zfrac132cos8pi9isin8pi9 b zfrac133cos8pi9isin8pi9 c zfrac132cos8pi9isin8pi9 d zfrac133cos8pi9isin8pi9 11788

          Find the 2nd of the three cube roots of:
$Z=27(\cos(2\pi/3)+i\sin(2\pi/3))$.
a. $Z^{\frac{1}{3}}=2(\cos(8\pi/9)+i\sin(8\pi/9))$
b. $Z^{\frac{1}{3}}=3(\cos(8\pi/9)+i\sin(8\pi/9))$
c. $Z^{\frac{1}{3}}=2(\cos(8\pi/9)+i\sin(8\pi/9))$
d. $Z^{\frac{1}{3}}=3(\cos(8\pi/9)+i\sin(8\pi/9))$

$find the 2nd of the three cube roots of z27cos2pi3isin2pi3 a zfrac132cos8pi9isin8pi9 b zfrac133cos8pi9isin8pi9 c zfrac132cos8pi9isin8pi9 d zfrac133cos8pi9isin8pi9 11788$

Added by Thomas A.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Gaurav Kalra

08/22/2021

Step 1

We can use De Moivre's Theorem for roots of complex numbers. If $Z = r(\cos\theta + i\sin\theta)$, then the $n$-th roots are given by: $Z_k = r^{1/n} \left( \cos\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\left(\frac{\theta + 2k\pi}{n}\right) \right)$ for $k = 0, Show more…

Show all steps

Thanks for your feedback!

Find the 2nd of the three cube roots of: $Z=27(\cos(2\pi/3)+i\sin(2\pi/3))$. a. $Z^{\frac{1}{3}}=2(\cos(8\pi/9)+i\sin(8\pi/9))$ b. $Z^{\frac{1}{3}}=3(\cos(8\pi/9)+i\sin(8\pi/9))$ c. $Z^{\frac{1}{3}}=2(\cos(8\pi/9)+i\sin(8\pi/9))$ d. $Z^{\frac{1}{3}}=3(\cos(8\pi/9)+i\sin(8\pi/9))$