Question #6abef

1 Answer
Jan 27, 2018

#cos((2015pi)/3)+isin((2015pi)/3)=1/2-isqrt3/2#

Explanation:

#(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#