How do you multiply #e^(( 9 pi )/ 4 i) * e^( pi/2 i ) # in trigonometric form?

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Answer 1 · 2016-09-26T10:23:14

As #e^(itheta)=costheta+isintheta#, we have

#e^((9pi)/4i)=cos((9pi)/4)+isin((9pi)/4)# and

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

Hence #e^((9pi)/4i)*e^(pi/2i)=(cos((9pi)/4)+isin((9pi)/4))(cos(pi/2)+isin(pi/2))#

= #cos((9pi)/4)(cos(pi/2)+isin(pi/2))+isin((9pi)/4))(cos(pi/2)+isin(pi/2))#

= #cos((9pi)/4)cos(pi/2)+icos((9pi)/4)sin(pi/2))+isin((9pi)/4)cos(pi/2)+i^2sin((9pi)/4)sin(pi/2))#

= #cos((9pi)/4)cos(pi/2)+icos((9pi)/4)sin(pi/2))+isin((9pi)/4)cos(pi/2)-sin((9pi)/4)sin(pi/2))#

= #(cos((9pi)/4)cos(pi/2)-sin((9pi)/4)sin(pi/2))+i(sin((9pi)/4)cos(pi/2)+cos((9pi)/4)sin(pi/2))#

= #cos(((9pi)/4)+(pi/2))+isin(((9pi)/4)+(pi/2))#

= #cos((11pi)/4)+isin((11pi)/4)#

= #e^((11pi)/4)#