What is (−2√3−2i)^{4} equivalent to?

Express the answer in polar form.

(−2√3−2i)^{4} = __ ___{1}* cis _*__ _{2}

Answer options:

256 ; 128 ; 512

(5π/3) ; (2π/3) ; (7π/6) ; (π6)

I've tried going on Youtube to see if I could get some kind of direction to solve this problem, but I have no idea. If someone could help me understand how to do this, I would appreciate that lots:)

auxiarc May 16, 2020

#1**+1 **

First: you will need to place the complex number -2·sqrt(3) - 2i into r·cis(theta) form.

For the complex number a + bi r = sqrt( a^{2} + b^{2} ) and theta = tan^{-1}( b/a)

For your problem: a = -2·sqrt(3) and b = -2:

r = sqrt( ( -2·sqrt(3) )^{2} + (-2)^{2} ) = sqrt( 12 + 4 ) = sqrt( 16 ) = 4

theta = tan^{-1}( -2 / (-2·sqrt(3) ) = tan^{-1}( sqrt(3)/3 ) = 7pi/6

r·cis(theta) = 4·cis( 7pi/6 )

General rule: [ r·cis(theta) ]^{n} = r^{n}·cis( n·theta )

Now, to find the 4^{th} power, take the r-value to the 4^{th} power, and multiply the angle by 4.

[ 4·cis( 7pi/6 ) ]^{4} = 4^{4}·cis( 4·7pi/6 ) = 256·cis( 14pi/3 )

(You'll have to reduce the 14pi/3 into an angle smaller than 2pi.)

geno3141 May 17, 2020

#3**+1 **

14/3·pi

Keep subtracting 2pi from (14/3)pi until you have a value smaller than 2pi

(14/3)pi - 2pi = (8/3)pi

(8/3)pi - 2pi = (2/3)pi

The reason that you subtract 2pi is that every 2pi is one revolution around the circle.

Going around the circle for one whole revolution (or 2pi) results in the same values for sin, cos, and tan.

geno3141 May 17, 2020