The polar form of $ z $ is:
$$ z = 2 \sqrt{ 73 } \cdot \left( cos ~ 249^o + i \cdot sin ~ 249^o \right) $$The polar form of the complex number $ z = a + bi$ is:
$$ z = r \cdot \left(cos~\theta + i \cdot sin ~ \theta \right) $$where, $ r $ is the modulus of $ z $ and $~ \theta ~$ is an argument of $ z $.
The moduo for $ a = -6 $ and $ b = -16 $ is:
$$ r = \sqrt{a^2 + b^2} = \cdots =2 \sqrt{ 73 } $$To find argument $ \theta $ we use one of the following formulas:
$$ \begin{aligned} \theta & = \arctan \left( \frac{b}{a} \right) \text{ if } a > 0 \\ \theta & = \arctan \left( \frac{b}{a} \right) + 180^o \text{ if } a < 0 \\ \\ \theta & = 90^o \text{ if } a = 0 \text{ and } b > 0 \\ \\ \theta & = 270^o \text{ if } a = 0 \text{ and } b < 0 \end{aligned} $$In this example: $ a = -6 $ and $ b = -16 $ so:
$$ \theta = arctan \left( \frac{b}{a} \right) + 180^o $$$$ \theta = arctan \left( \frac{ -16 }{ -6 } \right) + 180^o $$$$ \theta = arctan \left( \frac{ 8 }{ 3 } \right) + 180^o $$$$ \theta \approx 249^o $$