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

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Answer 1 · 2016-09-30T10:31:04

As $e^(itheta)=costheta+isintheta$

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

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

and $e^(pi/8i)*e^(pi/2i)=(cos(pi/8)+isin(pi/8))(cos(pi/2)+isin(pi/2))$

= $cos(pi/8)cos(pi/2)+cos(pi/8)xxisin(pi/2)+isin(pi/8)xxcos(pi/2)+isin(pi/8)xxisin(pi/2)$

= $cos(pi/8)cos(pi/2)+icos(pi/8)sin(pi/2)+isin(pi/8)cos(pi/2)+i^2sin(pi/8)sin(pi/2)$

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

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

= $cos(pi/8+pi/2)+isin(pi/8+pi/2)$

= $e^(pi/8+pi/2)$

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