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

Question

1249 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-18T14:47:19

#e^((4pi)/3i)*e^((3pi)/2i)=cos((17pi)/2)+isin((17pi)/2)#

We can write exponential fom of complex number #e^(itheta)# in trigonometric form as #costheta+isintheta#

Hence #e^((4pi)/3i)=cos((4pi)/3)+isin((4pi)/3)#

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

Hence #e^((4pi)/3i)*e^((3pi)/2i)#

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

= #cos((4pi)/3)cos((3pi)/2)+i^2sin((4pi)/3)sin((3pi)/2)+i{sin((4pi)/3)cos((3pi)/2)+cos((4pi)/3)sin((3pi)/2)}#

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

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

= #cos((17pi)/2)+isin((17pi)/2)#

Observe that this is #e^((17pi)/2)# which is nothing but #e^((4pi)/3+(3pi)/2)#.