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

2 Answers
Apr 11, 2018

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

Explanation:

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)

Apr 11, 2018

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

Explanation:

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