How do you evaluate # e^( (3 pi)/2 i) - e^( (4 pi)/3 i)# using trigonometric functions?

1 Answer
Feb 24, 2016

#e^((3π)/2)i−e^((4π)/3)i =1/2+i(-1+sqrt(3)/2)#

Explanation:

Let's recall the Euler's formula
#e^(ix)=cos(x)+isin(x)#.

Based on this formula,
#e^((3π)/2)i−e^((4π)/3)i =#
#= cos((3π)/2)+isin((3π)/2)-cos((4π)/3)-isin((4π)/3)=#
#= cos(π+π/2)+isin(π+π/2)-cos(π+π/3)-isin(π+π/3)#

Recall that
#cos(π+x) = -cos(x)# and
#sin(π+x) = -sin(x)#

Using these identities, the above expression looks like
# -cos(π/2)-isin(π/2)+cos(π/3)+isin(π/3) =#

Now notice that
#cos(π/2)=0#
#sin(π/2)=1#
#cos(π/3)=1/2#
#sin(π/3)=sqrt(3)/2#

Finally, our expression equals to
#0-i+1/2+isqrt(3)/2 = 1/2+i(-1+sqrt(3)/2)#