How do you multiply #e^(( 7 pi )/ 4 i) * e^( 3 pi/2 i ) # in trigonometric form?

Question

1231 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-02T10:14:33

#e^(7pi/4i)=(cos(7pi/4)+isin(7pi/4))=1/sqrt2-i(1/sqrt2)#

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

Trigonometric form of #e^(7pi/4i)# can be written as

#(cos(7pi/4)+isin(7pi/4))#

As #cos(7pi/4)=cos(-pi/4)=cos(pi/4)=1/sqrt2# and #sin(7pi/4)=sin(-pi/4)=sin(pi/4)=-1/sqrt2#

#e^(7pi/4i)# can be written as #1/sqrt2-i(1/sqrt2)#

and that of #e^(3pi/2i)# can be written as

#(cos(3pi/2)+isin(3pi/2))# and as #cos(3pi/2)=0# and #sin(pi/2)=-1#,

#e^(3pi/2i)# can be written as #-i#