Answer to Question #117369 in Algebra for jay

Question

Simplify the following expressions:
(a) (cos π/4 + i sin π/4)(cos 3π/4 + isin 3π/4)

(b) (cos π/4 + i sin π/4)^2 / (cos π/6 + isin π/6)

Answer 1

$a)(cos(π/4) + i sin(π/4))(cos(3π/4) + isin (3π/4))= сos(\pi/4)*cos(3\pi/4) + cos(\pi/4)*isin(3\pi/4) + isin(\pi/4)*cos(3\pi/4)+isin(\pi/4)*isin(3\pi/4)= 0.5*(cos(-\pi/2)+cos(\pi)) + cos(\pi/4)*isin(3\pi/4) + isin(\pi/4)*cos(3\pi/4) -0.5(cos(-\pi/2)-cos(\pi))= 0.5(-1) + i*0.5(sin(\pi) +sin(\pi/2)) +i*0.5(sin(\pi) +sin(-\pi/2)) - 0.5*1= = -1 + i*0.5 - i*0.5= -1 ;$

$b) (cos(π/4) + i sin(π/4))^2 / (cos(π/6) + isin(π/6))=(cos^2 (\pi/4)+isin(\pi/2)+ I^2sin^2 (\pi/4))/(cos(π/6) + isin(π/6))= (0.5 + i - 0.5)/(cos(π/6) + isin(π/6))= i *(cos(π/6) - isin(π/6))/(cos(π/6) - isin(π/6))*(cos(π/6) + isin(π/6))= i* (cos(π/6) - isin(π/6))/(cos^2 (\pi/6) - i^2sin^2(\pi/6)= i*(cos(π/6) - isin(π/6))/(0.75 +0.25)= i*(cos(π/6) - isin(π/6))=0.5+ (\sqrt3/2)i$