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)#