Question

How do you convert `-9 - 9sqrt2 i` to polar form?

Answer 1

`-9-9sqrt2i=9sqrt3(costheta+isi ntheta)`, where `theta=tan^(-1)sqrt2` and `theta` is in `Q3`.

Explanation:

If a complex number `a+bi=rcostheta+irsintheta`, `a=rcostheta` and `b=rsintheta` and squaring and adding `a^2+b^2=r^2` or `r=sqrt(a^2+b^2)`.

Hence, as we have `-9-9sqrt2i`, `r=sqrt((-9)^2+(-9sqrt2)^2)=sqrt(81+162)=sqrt243=9sqrt3`.

Hence `costheta=-1/sqrt3` and `sintheta=-sqrt(2/3)` i.e. `theta` is in third quadrant and `tantheta=sqrt2`

and `-9-9sqrt2i=9sqrt3(costheta+isi ntheta)`, where `theta=tan^(-1)sqrt2` and `theta` is in `Q3`.