ANSWER:
[tex]10(\cos(233.13\operatorname{\degree})+\imaginaryI\sin(233.13\operatorname{\degree}))[/tex]
STEP-BY-STEP EXPLANATION:
We have the following expression of a complex number:
[tex]-6-8i[/tex]
We must convert it to its polar form, for this we must calculate its magnitude and angle, like this:
[tex]\begin{gathered} \text{ Magnitude: }||-6-8i||=\sqrt{(-6)^2+(-8)^2}=\sqrt{36+64}=\sqrt{100}=10 \\ \\ \text{ Angle: }\theta=\arctan\left(\frac{-8}{-6}\right)=53.13\degree+180\degree=233.13\degree \\ \\ \text{ Therefore:} \\ \\ 10(\cos(233.13\degree)+i\sin(233.13\degree)) \end{gathered}[/tex]
