How do you evaluate e^( ( 7 pi)/4 i) - e^( ( 2 pi)/3 i)e7π4i−e2π3i 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} e7π4i−e2π3i=12+√22+i{−√22−√32}
Explanation:
For convenience, Let us denote the sum by:
S =e^((7pi)/4i) - e^((2pi)/3i) S=e7π4i−e2π3i
We know from Euler's Formula , that:
e^(i theta) = costheta+isintheta eiθ=cosθ+isinθ
And so we can represent the sum in trigonometric form by:
S = (cos((7pi)/4)+isin((7pi)/4)) - (cos((2pi)/3)+isin((2pi)/3)) S=(cos(7π4)+isin(7π4))−(cos(2π3)+isin(2π3))
\ \ = 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}
