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

Question

1508 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-04T06:19:17

#e^(pi/4i)*e^(pi/2i)=-1/sqrt2+1/sqrt2i#

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

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

#e^(pi/4i)*e^(pi/2i)#

= #(cos(pi/4)+isin(pi/4))*(cos(pi/2)+isin(pi/2))#

= #cos(pi/2)cos(pi/4)+icos(pi/2)sin(pi/4)+isin(pi/2)cos(pi/4)+i^2sin(pi/2)sin(pi/4)#

= #cos(pi/2)cos(pi/4)+i{cos(pi/2)sin(pi/4)}+sin(pi/2)cos(pi/4)-sin(pi/2)sin(pi/4)#

= #{cos(pi/2)cos(pi/4)-sin(pi/2)sin(pi/4)}+i{cos(pi/2)sin(pi/4)}+sin(pi/2)cos(pi/4)#

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

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

= #-1/sqrt2+1/sqrt2i#