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Answer 1 · 2018-01-26T06:46:35

$(\sqrt{85}, {tan}^{- 1} (\frac{9}{3}))$ $(sqrt85, tan^-1 (9/3))$

Given the complex number,

$2 + 9 i$ $2+9i$

we know that the modulus is $r = \sqrt{x^{2} + y^{2}}$ $r=sqrt(x^2+y^2)$ and the argument is $θ$ $theta$ = ${tan}^{- 1} (\frac{y}{x})$ $tan^-1(y/x)$

Here

$y = 9$ $y = 9$ and $x = 2$ $x = 2$

so

$r = \sqrt{2^{2} + 9^{2}} = \sqrt{85}$ $r=sqrt(2^2+9^2) = sqrt(85)$

$θ = {tan}^{- 1} (\frac{9}{2})$ $theta=tan^-1 (9/2)$

The polar form of the complex number is $(\sqrt{85}, {tan}^{- 1} (\frac{9}{3}))$ $(sqrt85, tan^-1 (9/3))$