Question 40889
Well since you can express
e^iq = cos q + i*sin q
you can express
(sqrt(2)*e(-ipi/4)) as
(sqrt(2))(cos (-pi/4) + i*sin(-pi/4)) which becomes, after evaluating the trig values,
1 - i
And then, by the same method 2e(ipi/6) becomes sqrt(3)+i...
Your book does not seem to be too instructive on how to go backwards however...