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

Question

2086 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-10T13:47:46

#sin((11pi)/6)+isin((11pi)/6)#

A number given a #re^(itheta)# can be written as #r(costheta+isintheta)#

#r_1(cos(theta_1)+isin(theta_1))*r_2(cos(theta_2)+isin(theta_2))=r_1r_2(cos(theta_1+theta_2)+isin(theta_1+theta_2))#

#r_1r_1=1*1=1#
#theta_1+theta_2=pi/3 +(3pi)/2=(11pi)/6#

#sin((11pi)/6)+isin((11pi)/6)#

#e^((11pi)/6i#

Answer 2 · 2018-05-10T14:12:32

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

This is another point of view.

By Euler's relation

#e^(itheta)=costheta+isintheta#

#I^2=-1#

#e^(pi/3i)=cos(pi/3)+isin(pi/3)=1/2+sqrt3/2i#

#e^(3/2pii)=cos(3/2pi)+isin(3/2pi)=0-i#

Therefore,

# e^(pi/3i)*e^(3/2pii) = (1/2+sqrt3/2i)*(-I) #

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

#=sqrt3/2-1/2i#