Question

How do you convert #-1+i to polar form?

Answer 1

`(r,theta)=(sqrt2,3pi/4).`

Explanation:

To convert a complex no. `z=x+iy` into polar form, we need to find `r` & `theta` such that `x=rcostheta, y=rsintheta`, where `theta in (-pi,pi]` and `r=sqrt(x^2+y^2).`

In our case, `x=-1, y=1,` so, clearly, `r=sqrt2.`

`x=rcostheta rArr -1=sqrt2costheta rArr costheta=-1/sqrt2, (-ve)`
`y=rsintheta rArr 1=sqrt2sintheta rArr sintheta=1/sqrt2, (+ve)`

We conclude that `theta` lies in the `II^(nd)` Quadrant, `theta=3pi/4.`

Thus `(r,theta)=(sqrt2,3pi/4)`is the desired polar form of the complex no. `z=-1+i.`