Resposta:
[tex]\textsf{Leia abaixo}[/tex]
Explicação passo a passo:
[tex]\mathsf{z = -5 + 5i}[/tex]
[tex]\mathsf{\rho = \sqrt{a^2 + b^2}}[/tex]
[tex]\mathsf{\rho = \sqrt{(-5)^2 + 5^2}}[/tex]
[tex]\mathsf{\rho = \sqrt{25 + 25}}[/tex]
[tex]\mathsf{\rho = \sqrt{50}}[/tex]
[tex]\mathsf{\rho = \sqrt{2.5^2}}[/tex]
[tex]\mathsf{\rho = 5\sqrt{2}}[/tex]
[tex]\mathsf{cos\:\Theta = \dfrac{a}{\rho} = \dfrac{-5}{5\sqrt{2}} = -\dfrac{\sqrt{2}}{2}}[/tex]
[tex]\mathsf{sen\:\Theta = \dfrac{b}{\rho} = \dfrac{5}{5\sqrt{2}} = \dfrac{\sqrt{2}}{2}}[/tex]
[tex]\mathsf{\Theta = 135\textdegree = \dfrac{3\pi }{4}}[/tex]
[tex]{\mathsf{z = \rho\:(cos\:\Theta + i\:sen\:\Theta)}}}[/tex]
[tex]\boxed{\boxed{\mathsf{z = 5\sqrt{2}\:(cos\:\dfrac{3\pi }{4} + i\:sen\:\dfrac{3\pi}{4})}}}[/tex]
