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

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Answer 1 · 2016-09-30T10:22:19

$e^(pi/2i)*e^(pii)=e^(pi/2+pi)$

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

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

$e^(pii)=cospi+isinpi$

and $e^(pi/2i)*e^(pii)=(cos(pi/2)+isin(pi/2))(cospi+isinpi)$

= $cos(pi/2)cospi+cos(pi/2)xxisinpi+isin(pi/2)xxcospi+isinpixxisin(pi/2)$

= $cos(pi/2)cospi+icos(pi/2)sinpi+isin(pi/2)cospi+i^2sinpisin(pi/2)$

= $cos(pi/2)cospi+icos(pi/2)sinpi+isin(pi/2)cospi-sinpisin(pi/2)$

= $[cos(pi/2)cospi-sinpisin(pi/2)]+i[cos(pi/2)sinpi+sin(pi/2)cospi]$

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

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

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