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SOVA2 [1]
3 years ago
10

Which expression is a cube root of -1+i√3?

Mathematics
1 answer:
Tpy6a [65]3 years ago
8 0

Answer:

<em>The correct option is C.</em>

Step-by-step explanation:

<u>Root Of Complex Numbers</u>

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.

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