Convert the complex number `(-16)/(1+isqrt(3))`into polar form.

Explanation

`z = (-16)/(1+isqrt3) = (-16)/(1+isqrt3)**(1-isqrt3)/(1-isqrt3)`

`=(-16+16sqrt3i)/(1+3) = -4+4sqrt3i`

`:.-4+4sqrt3i = r(costheta+isintheta) `

`rcostheta = -4 and rsintheta = 4sqrt3`

`r^2(cos^2theta+sin^2theta) =16+48`

`r^2 = 64=> r = 8`

`:.costheta = -4/8 = -1/2 and sintheta = 4sqrt3/8 = sqrt3/2`

So, `theta = (2pi)/3`

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