polar from of `-sqrt7+i sqrt21`

Explanation

`z=x+iy,=-sqrt(7)`
`y=sqrt(21)`
`r=sqrt(x^(2)+y^(2))=sqrt(7+21)=sqrt(28)=2sqrt(7)`
`tan theta, (y)/(x)=(sqrt(21))/(-sqrt(7))=-sqrt(3)`
Since the given point lies in the second quadrant we look for a solution of
`tan theta=-sqrt(3)` that lies in `[(pi)/(2),pi]`, we find that `tan theta =(2pi)/(3)` is such solution.
`:.- sqrt(7)+isqrt(21)=2sqrt(7)"cos" (2pi)/(3)`
( or) `2sqrt(7)("cos"(2pi)/(3)+i+"sin"(2pi)/(3))`