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

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Answer 1 · 2017-09-25T14:18:47

The answer is #=1#

We apply Euler's identity

#e^(itheta)=costheta +i sin theta#

#i^2=-1#

#e^a *e^b=e^(a+b)#

So,

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

#e^(i3pi/2)=cos(3/2pi)+isin(3/2pi)=0-1*i=-i#

Therefore,

#e^ (ipi/2)* e^(i3pi/2)=i*-i=-i^2=1#

We can also perform

#e^(ipi/2)*e^(i3pi/2)=e^((ipi/2+i3pi/2))=e^(i4pi/2)=e^(i2pi)#

#=cos(2pi)+isin(2pi)#

#=1+i*0=1#