#(1+cos((2pi)/3)+isin((2pi)/3))^2015#
First we need to include the 1 in the complex number's polar form so:
#1+cos((2pi)/3)+isin((2pi)/3)#
#=1-1/2+isqrt[3]/2#
#1/2+isqrt(3)/2#
So now transforming back to polar form:
The modulus of this complex number is:
#sqrt((1/2)^2+(sqrt(3)/2)^2)=sqrt(1/4+3/4)=sqrt(1)=1#
And the argument will be:
#tan^(-1)(((sqrt(3)/2))/((1/2)))=tan^(-1)(sqrt(3))=pi/3#
Hence: #1/2+isqrt(3)/2=cos(pi/3)+isin(pi/3)#
So it follows that:
#(1+cos((2pi)/3)+isin((2pi)/3))^2015=(cos(pi/3)+isin(pi/3))^2015#
In modulus argument form, raising a complex number to some power is given by:
#(r(cos(x)+isin(x))^n=r^n(cos(nx)+isin(nx))#
(In our case the modulus #r# is #1#). Therefore:
#(cos(pi/3)+isin(pi/3))^2015#
#=cos((2015pi)/3)+isin((2015pi)/3)#
#=1/2-isqrt3/2#
