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

1 Answer
Shwetank Mauria
May 19, 2016

#e^((2pi)/3i)*e^((3pi)/2i)=sqrt3/2+1/2i#

Explanation:

#a+ib# can be written in trigonometric form #re^(itheta)=rcostheta+irsintheta=r(costheta+isintheta)#,
where #r=sqrt(a^2+b^2)#, but here in given question #r=1#

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

= #e^(((4pi)/6+(9pi)/6)i)=e^(13pi/6i)#

= #(cos(13pi/6)+isin(13pi/6))#

= #(cos(2pi+pi/6)+isin(2pi+pi/6))#

= #(cos(pi/6)+isin(pi/6))#

= #sqrt3/2+1/2i#

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