How do you multiply e^(( pi )/ 8 i) * e^( pi/2 i ) eπ8ieπ2i in trigonometric form?

1 Answer
Sep 30, 2016

As e^(itheta)=costheta+isinthetaeiθ=cosθ+isinθ

e^(pi/8i)=cos(pi/8)+isin(pi/8)eπ8i=cos(π8)+isin(π8) and

e^(pi/2i)=cos(pi/2)+isin(pi/2)eπ2i=cos(π2)+isin(π2)

and e^(pi/8i)*e^(pi/2i)=(cos(pi/8)+isin(pi/8))(cos(pi/2)+isin(pi/2))eπ8ieπ2i=(cos(π8)+isin(π8))(cos(π2)+isin(π2))

= cos(pi/8)cos(pi/2)+cos(pi/8)xxisin(pi/2)+isin(pi/8)xxcos(pi/2)+isin(pi/8)xxisin(pi/2)cos(π8)cos(π2)+cos(π8)×isin(π2)+isin(π8)×cos(π2)+isin(π8)×isin(π2)

= cos(pi/8)cos(pi/2)+icos(pi/8)sin(pi/2)+isin(pi/8)cos(pi/2)+i^2sin(pi/8)sin(pi/2)cos(π8)cos(π2)+icos(π8)sin(π2)+isin(π8)cos(π2)+i2sin(π8)sin(π2)

= cos(pi/8)cos(pi/2)+icos(pi/8)sin(pi/2)+isin(pi/8)cos(pi/2)-sin(pi/8)sin(pi/2)cos(π8)cos(π2)+icos(π8)sin(π2)+isin(π8)cos(π2)sin(π8)sin(π2)

= [cos(pi/8)cos(pi/2)-sin(pi/8)sin(pi/2)]+i[cos(pi/2)sin(pi/8)+sin(pi/2)cos(pi/8)][cos(π8)cos(π2)sin(π8)sin(π2)]+i[cos(π2)sin(π8)+sin(π2)cos(π8)]

= cos(pi/8+pi/2)+isin(pi/8+pi/2)cos(π8+π2)+isin(π8+π2)

= e^(pi/8+pi/2)eπ8+π2