Question

How is `e^(ipi)= -1` ? and what is `e^(ipi/4)`?

Answer 1

see below

Explanation:

we use Euler's relationship

`e^(itheta)=costheta+isintheta`

`:e^(ipi)=cospi+isinpi`

`=-1+0i=-1`

`e^(pi/4)=cos(pi/4)+isin(pi/4)`

`=sqrt2/2+isqrt2/2=sqrt2/2(1+i)`