How do you simplify #z^6 = a + bi# given #z = sqrt 3 + i#?

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Answer 1 · 2016-07-08T19:08:12

#a = -64, b=0#

if you are familiar with Euler's rule #z = e^(ix) = cos x + i sin x# we can express z as

#z = 2( sqrt 3 /2 + i/2) = 2 (cos pi/6 + i sin pi/6 ) = 2 e^( i pi/6) #

thus #z^6 =( 2 e^( i pi/6))^6#

#= 2^6 * (e^( i pi/6))^6#

#= 64 * e^( i pi)#

and then reversing back into original form

#= 64 (cos pi + i sin pi)#

#= 64 (-1 + i (0))#

# = -64#