How do you multiply #e^(( 3 pi )/ 2 i) * e^( 3 pi/2 i ) # in trigonometric form?

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Answer 1 · 2016-03-27T08:35:51

#e^((3pi)/2i)*e^((3pi)/2i)=-1#

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

#e^((3pi)/2i)=cos((3pi)/2)+isin((3pi)/2)#.

Hence, #e^((3pi)/2i)*e^((3pi)/2i)=#

#{cos((3pi)/2)+isin((3pi)/2)}*{cos((3pi)/2)+isin((3pi)/2)}# or

= #{cos^2((3pi)/2)+i^2sin^2((3pi)/2)+2isin((3pi)/2)cos((3pi)/2)}#

= #{cos^2((3pi)/2)+(-1)sin^2((3pi)/2)+i(2sin((3pi)/2)cos((3pi)/2))}#

= #{(cos^2((3pi)/2)-sin^2((3pi)/2))+i(2sin((3pi)/2)cos((3pi)/2))}#

= #cos(2xx(3pi)/2)+isin(2xx(3pi)/2)#

= #cos3pi+isin3pi#

= #-1+ixx0=-1#