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

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Answer 1 · 2017-10-15T19:42:32

#= e^(((31pi)/8)i)#

# e^(((3pi)/8)i) * e^ (3(pi/2)i#
#e^(((3pi)/8)i) * e^(((7pi)/2)i)#

#e^(((3pi)/8)i)* e^ (((28pi)/8)i)#
#e^ (((3pi + 28pi)/8)(i+i) = e^(((31pi)/8)i)#

Answer 2 · 2017-10-15T20:53:31

#cos(pi/8)-isin(pi/8)# or #-(-1)^(7/8)#

Are you sure its #e^(3pi/2)# not #e^((3pi)/2)# because it makes more sense when you write it in trig form.

Remember this beauty?
#e^(ipi)=-1#
That was Euler's identity. This is the generalized formula:
#e^(ix)=cosx+isinx#

Therefore, we can break down the two terms involved:

#e^(i3/8pi)=cos((3pi)/8)+isin((3pi)/8)#

#e^(i7/2pi)=cos((7pi)/2)+isin((7pi)/2)=0+i*(-1)=-i#

#e^(i3/8pi)=(e^(i3/2pi))^(1/4)=(-i)^(1/4)#
#e^(i3/8pi)*e^(i3/2pi)=(e^(i3/2pi))^(1/4)*e^(i7/2pi)=(e^(i3/2pi))^(5/4)=(-i)^(5/4)=-(-1)^(7/8)#

Or

#e^(i3/8pi)*e^(i7/2pi) = sin((3pi)/8)-icos((3pi)/8)#

#=cos(pi/8)-isin(pi/8)#