How do you multiply #e^(( 4 pi )/ 3 i) * e^( 3 pi/2 i ) # in trigonometric form?

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Answer 1 · 2018-04-22T13:19:47

The answer is #=-sqrt3/2+1/2i#

Apply Euler's identity

#e^(itheta)=costheta+isintheta#

#I^2=-1#

#cos(4/3pi)=-1/2#

#sin(4/3pi)=-sqrt3/2#

#sin(3/2pi)=-1#

#cos(3/2pi)=0#

Therefore,

#e^(4/3ipi).e^(3/2ipi)#

#=(cos(4/3pi)+isin(4/3pi))(cos(3/2pi)+isin(3/2pi))#

#=(-1/2-isqrt3/2)(0-i)#

#=-sqrt3/2+1/2i#