How do you multiply #e^(( 7 pi )/ 12 ) * e^( pi/2 i ) # in trigonometric form?

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Answer 1 · 2017-11-22T14:43:55

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

Apply Euler's relation

#e^(itheta)=costheta+isintheta#

Therefore,

#e^(i7/12pi)=cos(7/12pi)+isin(7/12pi)#

#cos(7/12pi)=cos(1/4pi+1/3pi)=cos(1/4pi)cos(1/3pi)-sin(1/4pi)sin(1/3pi)#

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

#=(sqrt2-sqrt6)/4#

#sin(7/12pi)=sin(1/4pi+1/3pi)=sin(1/4pi)cos(1/3pi)+cos(1/4pi)sin(1/3pi)#

#=(sqrt2/2).(1/2)+(sqrt2/2)(sqrt3/2)#

#=(sqrt2+sqrt6)/4#

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

#i^2=-1#

Therefore,

#e^(i7/12pi).e^(ipi/2)=(((sqrt2-sqrt6)+(sqrt2+sqrt6)i)/4)(i)#

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