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Find (2+2i)^8? Write in standard form.

 Mar 12, 2018
 #1
avatar+94618 
+1

(2 + 2i)^8  =

 

r  = √ [ 2^2 + 2^2 ]  =  2√2

 

tan  θ  =  2/2   =  1 

 θ  = pi/4

 

(2 + 2i)^8  =  (2√2)^8* e^(i * 8pi /4 )  =   (2√2)^8 *e^( i * 2pi)  =

 

4096 ( cos 2pi + i * sin (2pi) )  =

 

4096 ( 1 + 0i )  =

 

4096

 

 

cool cool cool

 Mar 12, 2018

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