How do you multiply #e^(( 5 pi )/ 8 i) * e^( 3 pi/2 i ) # in trigonometric form?

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Answer 1 · 2018-03-13T16:30:01

The answer is #=1/2((sqrt(2+sqrt2))-i(sqrt(2-sqrt2)))#

Apply Euler's Identity

#e^(itheta)=costheta+isintheta#

#e^(5/8pii)=cos(5/8pi)+isin(5/8pi)#

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

#cos2theta=2cos^2theta-1=1-2sin^2theta#

#costheta=sqrt((1+cos2theta)/2)#

#sintheta=sqrt((1-sin2theta)/2)#

#cos(5/8pi)=sqrt((1+cos(10/8pi)/2))#

#cos(10/8pi)=cos(5/4pi)=cos(pi+1/4pi)#

#=cospicos(1/4pi)-sin(pi)sin(1/4pi)#

#=-1*sqrt2/2-0#

#=-sqrt2/2#

#cos(5/8pi)=sqrt((1-sqrt2/2)/2)=(sqrt(2-sqrt2))/(2)#

#sin(5/8pi)=sqrt((1-sin(10/8pi)/2)#

#sin(10/8pi)=sin(5/4pi)=2*-sqrt2/2=-sqrt2#

#sin(5/8pi)=sqrt((1+sqrt2/2)/2)=1/2sqrt(2+sqrt2)#

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

And finally,

#e^(5/8pii)*e^(3/2pii)=((sqrt(2-sqrt2))/(2)+i1/2sqrt(2+sqrt2))*(-i)#

#=1/2(sqrt(2+sqrt2))-i1/2(sqrt(2-sqrt2))#