Resposta :

Celio

Olá, Sevalho.

 

[tex]z=-7+7i \\\\ \text{Forma trigonom\'etrica}: z=|z|(\cos \theta + i \sin \theta)\\\\ |z| = \sqrt{(-7)^2+7^2} = \sqrt{2 \cdot 49}=7\sqrt2\\\\ -7 + 7i = 7\sqrt2 (\cos \theta + i \sin \theta) \Rightarrow \\\\ 7(-1+i)=7(\sqrt2 \cos \theta + i \sqrt2 \sin \theta) \Rightarrow \\\\ \begin{cases}\sqrt2\cos \theta =-1 \\ \sqrt2\sin \theta=1 \end{cases} \Rightarrow \begin{cases}2\cos \theta =-\sqrt2 \\ 2\sin \theta=\sqrt2 \end{cases} \Rightarrow [/tex]

 

[tex]\begin{cases} \cos \theta =-\frac{\sqrt 2}{2} \\ \sin \theta=\frac{\sqrt 2}{2} \end{cases} \Rightarrow \theta=45\º+90\º=135\º[/tex]

 

Portanto, a forma trigonométrica (ou polar) de  [tex]z=-7+7i[/tex]  é:

 

[tex]\boxed{z=7\sqrt2(\cos135\º + i\sin 135\º)}[/tex]

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