Question

HELP ASAP 20 POINTS CAUSE IM DESPERATE ToT

Answer 1

1. By de Moivre's theorem,

$\left(4\left(\cos\left(\dfrac{7\pi}9\right) + i \sin\left(\dfrac{7\pi}9\right)\right)\right)^3 = 4^3 \left(\cos\left(\dfrac{21\pi}9\right) + i \sin\left(\dfrac{21\pi}9\right)\right) \\\\ ~~~~~~~~ = 64 \left(\cos\left(\dfrac{7\pi}3\right) + i \sin\left(\dfrac{7\pi}3\right)\right) \\\\ ~~~~~~~~ = 64 \left(\cos\left(\dfrac\pi3\right) + i \sin\left(\dfrac\pi3\right)\right) \\\\ ~~~~~~~~ = 64 \left(\dfrac12 + i\dfrac{\sqrt3}2\right) \\\\ ~~~~~~~~ = \boxed{32 + 32\sqrt3\,i}$

2. First write the given number in exponential/trigonometric form.

$z = 5-5\sqrt3\,i$

has modulus

$|z| = \sqrt{5^2 + \left(-5\sqrt3\right)^2} = \sqrt{100} = 10$

and since it lies in the second quadrant of the complex plane, its argument is

$\arg(z) = \pi + \tan^{-1}\left(-\dfrac{5\sqrt3}5\right) = \pi + \tan^{-1}\left(-\sqrt3\right) = \pi - \dfrac\pi3 = \dfrac{2\pi}3$

So, we have

$z = 5 - 5\sqrt3\,i = 10 e^{i2\pi/3} = 10 \left(\cos\left(\dfrac{2\pi}3\right) + i \sin\left(\dfrac{2\pi}3\right)\right)$

Now we apply de Moivre's theorem again, and make sure to account for the multivalued-ness of the exponential function. For $k\in\{0,1,2,3,4\}$, the fifth roots of $z$ are

$z^{1/5} = 10^{1/5} e^{i(2\pi/3 + 2\pi k)/5}$

$k=0 \implies z^{1/5} = 10^{1/5} e^{i2\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{2\pi}{15}\right) + i \sin\left(\dfrac{2\pi}{15}\right)\right)}$

$k=1 \implies z^{1/5} = 10^{1/5} e^{i8\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{8\pi}{15}\right) + i \sin\left(\dfrac{8\pi}{15}\right)\right)}$

$k=2 \implies z^{1/5} = 10^{1/5} e^{i14\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{14\pi}{15}\right) + i \sin\left(\dfrac{14\pi}{15}\right)\right)}$

$k=3 \implies z^{1/5} = 10^{1/5} e^{i20\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{4\pi}3\right) + i \sin\left(\dfrac{4\pi}3\right)\right)}$ $k=4 \implies z^{1/5} = 10^{1/5} e^{i26\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{26\pi}{15}\right) + i \sin\left(\dfrac{26\pi}{15}\right)\right)}$