Question

How do you evaluate `e^( ( pi)/4 i) - e^( ( 7 pi)/6 i)` using trigonometric functions?

Answer 1

`e^((pi i)/4)` = `cos (pi/4) + i sin (pi/4)= (sqrt2)/2 +i(sqrt2)/2`

`e^((7pi i)/6)`= `cos (7pi/6)+ isin (7pi/6)= -cos (pi/6)-isin (pi/6)`
`=-sqrt3/2 -i1/2`
(In the third quadrant both sin and cos are negative)

Thus, `e^((pi i)/4)` -`e^((7pi i)/6)`= `(sqrt2 +sqrt3)/2 +i(sqrt2 +1)/2`