How do you multiply #e^(( 5 pi )/ 4 i) * e^( 3 pi/2 i ) # in trigonometric form?

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Answer 1 · 2018-04-11T09:00:52

Using the formula #e^(itheta) = cos(theta) + isin(theta)# and some trigonometric identities.

The formula:

#e^((5pi)/4i) = cos((5pi)/4) + isin((5pi)/4)#

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

Multiplication:

#e^((5pi)/4i) xxe^((3pi)/2i) = (cos((5pi)/4) + isin((5pi)/4))(cos((3pi)/2) + isin((3pi)/2))#

# = cos((5pi)/4)cos((3pi)/2)+icos((5pi)/4)sin((3pi)/2) + isin((5pi)/4)cos((3pi)/2) -sin((5pi)/4)sin((3pi)/2) #

# = cos((5pi)/4)cos((3pi)/2)-sin((5pi)/4)sin((3pi)/2) + i(cos((5pi)/4)sin((3pi)/2) + sin((5pi)/4)cos((3pi)/2)) #

#= cos((5pi)/4 + (3pi)/2) + isin((5pi)/4 + (3pi)/2)#

# = cos((11pi)/4) + isin((11pi)/4)#

#= e^((11pi)/4i)#

Its much easier if you do this:

#e^((5pi)/4i) xxe^((3pi)/2i) = e^((5pi)/4i + (3pi)/2i)#

#= e^((11pi)/4i)#

Answer 2 · 2018-04-11T09:08:58

The answer is #=-sqrt2/2+sqrt2/2i#

Apply Eulers' Identity

#e^(itheta)=costheta+isintheta#

Therefore,

#e^(5/4pii)*e^(3/2pii)=(cos(5/4pi)+isin(5/4pi))(cos(3/2pi)+isin(3/2pi))#

#cos(5/4pi)=cos(1/4pi+pi)=cos(1/4pi)cos(pi)-sin(1/4pi)sin(pi)#

#=-sqrt2/2#

#sin(5/4pi)=sin(1/4pi+pi)=sin(1/4pi)cos(pi)+cos(1/4pi)sin(pi)#

#=-sqrt2/2#

#cos(3/2pi)=cos(1/2pi+pi)=cos(1/2pi)cos(pi)-sin(1/2pi)sin(pi)#

#=0#

#sin(3/2pi)=sin(1/2pi+pi)=sin(1/2pi)cos(pi)+cos(1/2pi)sin(pi)#

#=-1#

Finally,

#e^(5/4pii)*e^(3/2pii)=(-sqrt2/2-sqrt2/2i)(-i)#

#=-sqrt2/2+sqrt2/2i#