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

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Answer 1 · 2017-11-15T07:19:45

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

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

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

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

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

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

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

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

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

= $-1/2+i1/2$

Simplify 1/sqrt2(cos45^@-isin45^@)^51/sqrt2(cos45^@-isin45^@)^5 in the form a+iba+ib using De Moivre's theorem?