✅ Após resolver os cálculos, concluímos que a forma polar do referido ponto dado é:

              [tex]\Large\displaystyle\text{$\begin{gathered}\boxed{\boxed{\:\:\:\bf P_{P}=(2\sqrt{6},\:135^{\circ})\:\:\:}}\end{gathered}$}[/tex]

Seja o ponto em sua forma cartesiana:

                   [tex]\Large\displaystyle\text{$\begin{gathered} P_{C} = (-2\sqrt{3},\,2\sqrt{3})\end{gathered}$}[/tex]

Para escrevermos o ponto "P" em coordenadas polares, fazemos:

     [tex]\Large \text {$\begin{aligned}P_{P} & = (\tau,\,\theta)\\& = \left[\sqrt{x_{C}^{2} + y_{C}^{2}},\:\arctan\left(\frac{y_{C}}{x_{C}}\right)\right]\\& = \left[\sqrt{(-2\sqrt{3})^{2} + (2\sqrt{3})^{2}},\:\arctan\left(-\frac{2\sqrt{3}}{2\sqrt{3}}\right)\right]\\& = \left[\sqrt{(-2)^{2}\cdot(\sqrt{3})^{2} + 2^{2}\cdot(\sqrt{3})^{2}},\arctan(-1)\right]\\& =\left[\sqrt{4\cdot3 + 4\cdot3},\:135^{\circ}\right]\\& = \left[2\sqrt{6},\:135^{\circ}\right]\end{aligned} $}[/tex]

✅ Portanto, o ponto na forma polar é:

     [tex]\Large\displaystyle\text{$\begin{gathered} P_{P} = (2\sqrt{6},\:135^{\circ})\end{gathered}$}[/tex]

[tex]\LARGE\displaystyle\text{$\begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Bons \:estudos!!\:\:\:Boa\: sorte!!\:\:\:}}}\end{gathered}$}[/tex]

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