Question

Which expression is a cube root of -1+i√3?
A. ^3√2(cos120 + i sin 120)
B. ^3√4(cos320 + i sin 320)
C. ^3√2(cos40 + i sin 40)
D. ^3√4(cos280 + i sin 280)

Answer 1

Answer:

The correct option is C.

Step-by-step explanation:

Root Of Complex Numbers

If a complex number is expressed in polar form as

$Z=(r,\theta)$

Then the cubic roots of Z are

$\displaystyle Z_1=\left(\sqrt[3]{r},\frac{\theta}{3}\right)$

$\displaystyle Z_2=\left(\sqrt[3]{r},\frac{\theta}{3}+120^o\right)$

$\displaystyle Z_3=\left(\sqrt[3]{r},\frac{\theta}{3}+240^o\right)$

We are given the complex number in rectangular components

$Z=-1+i\sqrt{3}$

Converting to polar form

$r=\sqrt{(-1)^2+(\sqrt{3})^2}=2$

$\displaystyle tan\theta=\frac{\sqrt{3}}{-1}=-\sqrt{3}$

It's located in the second quadrant, so

$\theta=120^o$

The number if polar form is

$Z=(2,120^o)$

Its cubic roots are

$\displaystyle Z_1=\left(\sqrt[3]{2},\frac{120^o}{3}\right)=\left(\sqrt[3]{2},40^o\right)$

$\displaystyle Z_2=\left(\sqrt[3]{2},40^o+120^o\right)=\left(\sqrt[3]{2},160^o\right)$

$\displaystyle Z_3=\left(\sqrt[3]{2},40^o+240^o\right)=\left(\sqrt[3]{2},280^o\right)$

Converting the first solution to rectangular coordinates

$z_1=\sqrt[3]{2}(\ cos40^o+i\ sin40^o)$

The correct option is C.