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

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Answer 1 · 2018-05-15T13:36:35

#cos((7pi)/6)+isin((7pi)/6)=e^((7pi)/6i)#

#e^(itheta)=cos(theta)+isin(theta)#

#e^(itheta_1)*e^(itheta_2)==cos(theta_1+theta_2)+isin(theta_1+theta_2)#

#theta_1+theta_2=(2pi)/3+pi/2=(7pi)/6#

#cos((7pi)/6)+isin((7pi)/6)=e^((7pi)/6i)#

Answer 2 · 2018-05-15T13:56:16

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

Another method .

#i^2=-1#

Euler's relation

#e^(itheta)=costheta+isintheta#

Therefore,

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

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

#=1/2i-sqrt3/2#

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