Question

How do you write the complex number in trigonometric form `3+sqrt3i`?

Answer 1

`=2sqrt3(cos(pi/6)+sin(pi/6))`

Explanation:

let `z=3+isqrt3`
We have to calculate the modulus `∣z∣=sqrt(3^2+(sqrt3)^2)=sqrt(9+3)=sqrt12=2sqrt3`
Now we have to divide z by ∣z∣
`z/(∣z∣)`=`3/(2sqrt3)+isqrt3/(2sqrt3)=sqrt3/2+i/2`
So `z=2sqrt3(sqrt3/2+i/2)`
Now you compare this to
`z=r(costheta+isintheta )`
You can see that
`costheta=sqrt3/2`
and `sintheta=1/2`
As the signs are positive, the results is in the first quadrant
`theta=30º orpi/6`
So the anwer is `z=2sqrt3(cos(pi/6)+sin(pi/6))`