Question

Find the cube root of 4 - 4√3i

that graphs in the second
quadrant.
[?] (cos[]° + i sin[__ _]°)
Use degree measure.
Enter

Answer 1

The answer is

`2 \cos(100) + i \sin(100)`

Explanation:

This is a complex number,

`a + bi`

First, convert this to de movire form.

`r( \cos( \alpha ) + i \sin( \alpha )`

where

`r = \sqrt{ {a}^(2) + {b}^(2) }`

and

`\alpha = \tan {}^( - 1) ( (b)/(a) )`

`a = 4`

`b = - 4 √( 3) i`

`r = \sqrt{ {4}^(2) + ( - 4 √(3)) {}^(2) }`

`r = √(16 + 48)`

`r = √(64) = 8`

and

`\alpha = \tan {}^( - 1) ( ( - 4 √(3) )/(4) )`

`\alpha = \tan {}^( - 1) ( - √(3) )`

Here, our a is positive and b is negative so our angle in degrees must lie in the fourth quadrant, that angle is 300 degrees.

So

`\alpha = 300`

So our initially form is

`8( \cos(300) + i \sin(300) )`

Now, we use the roots of unity formula. To do this, we first take the cube root of the modulus, 8,

`\sqrt[3]{8} = 2`

Next, since cos and sin have a period of 360 we add 360 to each degree then we divide it by 3.

`\sqrt[3]{8} ( \cos( (300 + 360n)/(3) ) + \sin( (300 + 360n)/(3) )`

`2 \cos(100 + 120n) + i \sin(100 + 120n)`

Since 100 is in the second quadrant, we let n=0,

`2 \cos(100) + i \sin(100)`