Question

How do you convert `3 sqrt 2 - 3 sqrt2i` to polar form?

Answer 1

In polar form `3sqrt2-3sqrt2i` is `6(cos(-pi/4)+isin(-pi/4))`

Explanation:

A complex number `a+bi` can be written in polar form as `r(costheta+isintheta)` i.e. `a=rcostheta` and `b=rsintheta`.

Squaring and adding the last two, we get `a^2+b^2=r^2cos^2theta+r^2sin^2theta`

= `r^2(cos^2theta+sin^2theta)=r^2xx1=r^2`

and `b/a=(rsintheta)/(rcostheta)=tantheta`

Here in the complex number `3sqrt2-3sqrt2i`, we have `a=3sqrt2` and `b=-3sqrt2` and hence `r=sqrt(a^2+b^2)=sqrt((3sqrt2)^2+(-3sqrt2)^2)`

= `sqrt(18+18)=sqrt36=6` and

`costheta=(3sqrt2)/6=1/sqrt2` and `costheta=(-3sqrt2)/6=-1/sqrt2`

i.e. `theta=(-pi/4)`

Hence in polar form `3sqrt2-3sqrt2i` is `6(cos(-pi/4)+isin(-pi/4))`