Question

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

Answer 1

`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)`