Question

How do I write [4(cos(7π/9)+isin(7π/9))]^3 in standard form for the complex number a+bi.

Answer 1

\(\large [4(cos(7π/9)+isin(7π/9))]^3 \)

Answer 2

hint 1:-
(cos theta +i sin theta ) ^n=(cos n theta +i sin n theta )
hint 2:-
e^itheta =(cos theta +i sin theta )

Answer 3

First use Demoivre's theorem to get rid of the exponent:
\[\large \left(re^{i\theta}\right)^n=r^ne^{in\theta}\]
So,
\[\bigg[4\left(\cos\frac{7\pi}{9}+i\sin\frac{7\pi}{9}\right)\bigg]^3=64\left(\cos\frac{7\pi}{3}+i\sin\frac{7\pi}{3}\right)\]
Find the trig ratios and simplify.

Answer 4

OK thanks