How do you multiply #e^((11pi )/ 12 ) * e^( pi/4 i ) # in trigonometric form?

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Answer 1 · 2018-05-30T08:23:43

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

Apply Euler's relation

#e^(itheta)=costheta+isintheta#

#e^(11/12ipi)=cos(11/12pi)+isin(11/12pi)#

#e^(1/4ipi)=cos(1/4pi)+isin(1/4pi)#

#i^2=-1#

Therefore,

# e^(11/12ipi)*e^(1/4ipi)=(cos(11/12pi)+isin(11/12pi))*(cos(1/4pi)+isin(1/4pi))#

#=cos(11/12pi)cos(1/4pi)-sin(11/12pi)sin(1/4pi)+i(cos(11/12pi)sin(1/4pi)+sin(11/12pi)cos(1/4pi)))#

#=cos(11/12pi+1/4pi)+isin(11/12pi+1/4pi)#

#=cos(14/12pi)+isin(14/12pi)#

#=cos(7/6pi)+isin(7/6pi)#

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