Answer to Question #117185 in Algebra for Edward

Question

Simplify the following expressions:
(a). (cos π/4 + isin π/4)(cos 3π/4 + isin 3π/4), (b). (cos π/4 + isin π/4)2
(cos π/6 + isin π/6) .

Answer 1

(a). (cos π/4 + isin π/4)(cos 3π/4 + isin 3π/4)=( $\sqrt{2}$/2+i$\sqrt{2}$/2)(- $\sqrt{2}$/2+i$\sqrt{2}$/2)= -2/4+i*2/4-i*2/4+i²*2/4=-1/2+i²*1/2= -1/2 -1/2=-2/2=-1

(b). (cos π/4 + isin π/4)2(cos π/6 + isin π/6)= ( $\sqrt{2}$/2+ i$\sqrt{2}$/2)*2*($\sqrt{3}$/2 + i/2)=($\sqrt{2}$/2+ i$\sqrt{2}$/2)($\sqrt{3}$+i)= $\sqrt{6}$/2 + i*$\sqrt{2}$/2 + i*$\sqrt{6}$/2 + i²*$\sqrt{2}$/2=$\sqrt{6}$/2 - $\sqrt{2}$/2 + i*$\sqrt{2}$/2 + i*$\sqrt{6}$/2= ($\sqrt{2}$($\sqrt{3}$ -1) + $\sqrt{2}$(1+$\sqrt{3}$ ))/2= ($\sqrt{2}$ (3-1))/2= 2$\sqrt{2}$/2 = $\sqrt{2}$

Answer: (a). (cos π/4 + isin π/4)(cos 3π/4 + isin 3π/4)=-1, (b). (cos π/4 + isin π/4)2(cos π/6 + isin π/6)=$\sqrt{2}$