Question

Simplify `1/sqrt2(cos45^@-isin45^@)^5` in the form `a+ib` using De Moivre's theorem?

Answer 1

`1/sqrt2(cos45^@-isin45^@)^5=-1/2+i1/2`

Explanation:

According to De Moivre's theorem `(costheta+isintheta)^n=cosntheta+isinntheta`

Hence `1/sqrt2(cos45^@-isin45^@)^5`

= `1/sqrt2(cos(-45^@)+isin(-45^@))^5`

= `1/sqrt2(cos(-5xx45^@)+isin(-5xx45^@))`

= `1/sqrt2(cos(-225^@)+isin(-225^@))`

= `1/sqrt2(cos(360^@-225^@)+isin(360^@-225^@))`

= `1/sqrt2(cos135^@+isin135^@)`

= `1/sqrt2(-1/sqrt2+i1/sqrt2)`

= `-1/2+i1/2`