How can you use trigonometric functions to simplify # 15 e^( ( 3 pi)/8 i ) # into a non-exponential complex number?

1 Answer
Jan 15, 2016

#15e^((3pi)/8i) = 15/2(sqrt(2-sqrt(2)) + sqrt(2+sqrt(2))i)#

Explanation:

Using the identity #e^(itheta) = cos(theta) + isin(theta)# we have:

#15e^((3pi)/8i) = 15(cos((3pi)/8)+isin((3pi)/8))#

If we do not want trig functions either, we can simplify further using the half angle formulas.

#cos((3pi)/8) = cos(1/2((3pi)/4)) = sqrt((1+cos((3pi)/4))/2) = sqrt(2-sqrt(2))/2#

#sin((3pi)/8) = sin(1/2((3pi)/4)) = sqrt((1-cos((3pi)/4))/2) = sqrt(2+sqrt(2))/2#

Thus

#15e^((3pi)/8i) = 15(sqrt(2-sqrt(2))/2 + sqrt(2+sqrt(2))/2i)#

#= 15/2(sqrt(2-sqrt(2)) + sqrt(2+sqrt(2))i)#