How do, I write this in the form a+bi?

Question

#2e^(i12pi/2)#

6585 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-27T07:22:40

#2e^(i12pi/2)=2+0i#

#ae^(itheta)=a(costheta+isintheta)#

Hence #2e^(i12pi/2)=2(cos(12pi/2)+isin(12pi/2))#

= #2(cos6pi+isin6pi)#

= #2(1+ixx0)#

= #2+0i#

Answer 2 · 2017-05-27T07:28:36

See explanation.

The complex number is given in the trigonometric form. Its module is #2#, the argument (angle) is #(12pi)/2=6pi#.

The formula transforming algebraic form to trigonometric form is:

In the example we have:

#2e^(i*6pi)=2*(cos6pi+isin6pi)=2*(1+0i)=2#