DeMoivre’s theorem.

Evaluate:  ( 5 + 5·sqrt(3)·i )⁷

 

1)  Write  5 + 5·sqrt(3)·i  in r·cis( theta )  form:

      If you have  a + bi  use  r  =  sqrt(a² + b²)  and  theta  =  tan^-1( b/a ):

      r  =  sqrt( ( 5 )² + ( 5·sqrt(3) )²  =  sqrt( 25 + 75  =  sqrt( 100 )  =  10

      theta  =  tan^-1( 5·sqrt(3) / 5 )  =  60^o   [If you are to use radians, use pi/3.]

          [You need to check which quadrant you should use; since both 5 and 5·sqrt(3) are

           positive, the angle is in the first quadrant; so 60^o is the correct angle.]

 

2)  You now have:  5 + 5·sqrt(3)·i   =   10·cis( 60^o )

     To raise to a power, raise the constant to that power and multiply the angle by the power.

     [ 10·cis( 60^o ) ]⁷  =  10⁷ · cis( 60^o · 7 )  =  10⁷ · cis( 420^o )  =  10⁷ · cis( 60^o )

 

3)  To write this in rectangular form, use:  x  =  r·cos( theta )     and     y  =  r·sin( theta )

      x  =  10⁷·cos( 60^o )  =  10⁷· (1/2)  =  5·10⁶

      y  =  10⁷·sin 60^o )  =  10⁷· (sqrt(3)/2)  =  5·sqrt(3)·10⁶

      --->     5·10⁶  +  5·sqrt(3)·10⁶·i