How do you multiply #e^(( 19pi )/ 12 ) * e^( pi/2 i ) # in trigonometric form?

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Answer 1 · 2018-01-26T11:10:51

The answer is #=(sqrt6+sqrt2)/4+i(sqrt6-sqrt2)/4#

Apply Euler's Identity

#e^(itheta)=costheta+isintheta#

#i^2=-1#

Therefore,

#e^(ipi/2)=cos(pi/2)+isin(pi/2)=0+i*1=i#

#e^(i19/12pi)=cos(19/12pi)+isin(19/12pi)#

#cos(19/12pi)=cos(5/4pi+1/3pi)#

#=cos(5/4pi)cos(1/3pi)-sin(5/4pi)sin(1/3pi)#

#=(-sqrt2/2)*(1/2)-(-sqrt2/2)*(sqrt3/2)#

#=-sqrt2/4+sqrt6/4#

#=(sqrt6-sqrt2)/4#

#sin(19/12pi)=sin(5/4pi+1/3pi)#

#=sin(5/4pi)cos(1/3pi)+cos(5/4pi)sin(1/3pi)#

#=(-sqrt2/2)*(1/2)+(-sqrt2/2)*(sqrt3/2)#

#=-sqrt2/4-sqrt6/4#

#=-(sqrt6+sqrt2)/4#

Finally,

#e^(i19/12pi)*e^(ipi/2)=((sqrt6-sqrt2)/4-i(sqrt6+sqrt2)/4)*(i)#

#=(sqrt6+sqrt2)/4+i(sqrt6-sqrt2)/4#