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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:)

 May 16, 2020
 #1
avatar+20906 
+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( a2 + b2 )     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  =  rn·cis( n·theta )

 

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

 

[ 4·cis( 7pi/6 ) ]4  =  44·cis( 4·7pi/6 )  =  256·cis( 14pi/3​ ) 

 

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

 May 17, 2020
 #2
avatar+182 
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I get everything up until the last part, in the parenthesis. What do you mean 14pi/3 has to be reduced into an angle smaller than 2pi? How would I do that? 

auxiarc  May 17, 2020
 #3
avatar+20906 
+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.

 May 17, 2020

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