Question

What is the polar form of the complex number 4−6i? Keep z in its simplest radical form and round θ to the nearest degree.

A
25√(cos(304°)+isin(304°))

B
25√(cos(56°)+isin(56°))

C
213−−√(cos(56°)+isin(56°))

D
213−−√(cos(304°)+isin(304°))

Answer 1

Answer:

D

Step-by-step explanation:

$z=r(cos \theta+\iota sin \theta)=4-6 \iota\\r cos \theta=4\\r sin \theta=-6\\square~and~add\\r^2(cos^2 \theta+sin ^2\theta)=4^2+(-6)2=16+36\\r=\sqrt{52} =2\sqrt{13} \\cos \theta >0~and~sin ~\theta<0\\so ~\theta~lies~in ~4th~quadrant.\\divide\\tan \theta==\frac{-6}{4} =\frac{-3}{2} \\\theta=tan^{-1} (\frac{-3}{2} ) \approx -56 ^\circ ~or~304 ^\circ\\so~z=2\sqrt{13} (cos(304) ^\circ+\iota sin (304)^\circ)$