How do you multiply $e^((11pi )/ 12 ) * e^( pi/4 i )$ in trigonometric form?

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Answer 1 · 2018-05-30T08:23:43

The answer is $=-sqrt3/2-1/2i$

Apply Euler's relation

$e^(itheta)=costheta+isintheta$

$e^(11/12ipi)=cos(11/12pi)+isin(11/12pi)$

$e^(1/4ipi)=cos(1/4pi)+isin(1/4pi)$

$i^2=-1$

Therefore,

$e^(11/12ipi)*e^(1/4ipi)=(cos(11/12pi)+isin(11/12pi))*(cos(1/4pi)+isin(1/4pi))$

$=cos(11/12pi)cos(1/4pi)-sin(11/12pi)sin(1/4pi)+i(cos(11/12pi)sin(1/4pi)+sin(11/12pi)cos(1/4pi)))$

$=cos(11/12pi+1/4pi)+isin(11/12pi+1/4pi)$

$=cos(14/12pi)+isin(14/12pi)$

$=cos(7/6pi)+isin(7/6pi)$

$=-sqrt3/2-1/2i$

How do you multiply e^((11pi )/ 12 ) * e^( pi/4 i ) e^((11pi )/ 12 ) * e^( pi/4 i ) in trigonometric form?