SOLUTION: This is one of the so called examples in my book Simplify in euler form: (1-i)^5 (sqrt3+i)^3 How is it possible to get to this? (sqrt2 e(-ipi/4))^5 (2e(ipi/6)^3. And from the

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: This is one of the so called examples in my book Simplify in euler form: (1-i)^5 (sqrt3+i)^3 How is it possible to get to this? (sqrt2 e(-ipi/4))^5 (2e(ipi/6)^3. And from the      Log On


   



Question 40889: This is one of the so called examples in my book
Simplify in euler form: (1-i)^5 (sqrt3+i)^3
How is it possible to get to this?
(sqrt2 e(-ipi/4))^5 (2e(ipi/6)^3.
And from there is simplifies down to -32-32i, that I am ok with but why is it no one can explain the first step? Smoke and mirrors?
And in all examples they conveniently use (1-i) but what about any other numbers?
Surely someone must know how to explain this?

Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
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...