How do you evaluate # e^( ( 7 pi)/4 i) - e^( ( 2 pi)/3 i)# using trigonometric functions?
1 Answer
Jan 29, 2018
# e^((7pi)/4i) - e^((2pi)/3i) = 1/2+sqrt(2)/2+i{-sqrt(2)/2 - sqrt(3)/2} #
Explanation:
For convenience, Let us denote the sum by:
# S =e^((7pi)/4i) - e^((2pi)/3i) #
We know from Euler's Formula , that:
# e^(i theta) = costheta+isintheta #
And so we can represent the sum in trigonometric form by:
# S = (cos((7pi)/4)+isin((7pi)/4)) - (cos((2pi)/3)+isin((2pi)/3)) #
# \ \ = cos((7pi)/4)-cos((2pi)/3)+i{sin((7pi)/4) - sin((2pi)/3)} #
# \ \ = sqrt(2)/2-(-1/2)+i{-sqrt(2)/2 - sqrt(3)/2} #
# \ \ = 1/2+sqrt(2)/2+i{-sqrt(2)/2 - sqrt(3)/2} #
