How do you multiply #e^(( 5 pi )/ 12 i) * e^( 3 pi/2 i ) # in trigonometric form?

1 Answer
Narad T.
Jul 8, 2017

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

Explanation:

We apply Euler's formula

#e^(itheta)=costheta+isintheta#

#e^(i5/12pi)=cos(5/12pi)+isin(5/12pi)#

#cos(5/12pi)=cos(3/12pi+2/12pi)#

#=cos(1/4pi+1/6pi)#

#=cos(1/4pi)cos(1/6pi)-sin(1/4pi)sin(1/6pi)#

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

#=(sqrt6-sqrt2)/4#

#sin(5/12pi)=sin(1/4pi+1/6pi)#

#=sin(1/4pi)cos(1/6pi)+cos(1/4pi)sin(1/6pi)#

#=sqrt2/2*sqrt3/2+sqrt2/2*1/2#

#=(sqrt6+sqrt2)/4#

Therefore,

#e^(i5/12pi)=(sqrt6-sqrt2)/4+i(sqrt6+sqrt2)/4#

#e^(i3/2pi)=cos(3/2pi)+isin(3/2pi)#

#=0-i#

So,

#e^(i5/12pi).e^(i3/2pi)=((sqrt6-sqrt2)/4+i(sqrt6+sqrt2)/4)*-i#

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

As #i^2=-1#

Verification

#e^(i5/12pi)*e^(i3/2pi)=e^(i(5/12+3/2)pi)#

#=e^(i23/12pi)#

#=cos(-1/12pi)+isin(-1/12pi)#

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

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

#=1/2*sqrt2/2+sqrt3/2*sqrt2/2#

#=(sqrt6+sqrt2)/4#

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

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

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

#=-(sqrt6-sqrt2)/4#

Therefore,

#e^(i23/12pi)=(sqrt6+sqrt2)/4-i(sqrt6-sqrt2)/4#

The result is the same.

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