Question

How do you convert `1+jsqrt3` to polar form?

Answer 1

`1+jsqrt3=2(cos(pi/3)+jsin(pi/3))`

Explanation:

To write the complex number, say `a+jb`, where `j^2=-1` in polar coordinates, we write them as

`z=r(costheta+jsintheta)`.

Now as `rcostheta=a` and `rsintheta=b`.

`r=sqrt(a^2+b^2)`, `theta=tan^(-1)(b/a)`

Hence here for complex number `1+jsqrt3`

`r=sqrt(1^2+(sqrt5)^2)=sqrt4=2` and `theta=tan^(-1)(sqrt3/1)=tan^(-1)sqrt3=pi/3`

Hence `1+jsqrt3=2(cos(pi/3)+jsin(pi/3))`