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Convert -6- 8i into polar form, r(cos(0) + i sin(0)):

Convert -6- 8i into polar form, r(cos(0) + i sin(0)):-example-1

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ANSWER:


10(\cos(233.13\operatorname{\degree})+\imaginaryI\sin(233.13\operatorname{\degree}))

Explanation:

We have the following expression of a complex number:


-6-8i

We must convert it to its polar form, for this we must calculate its magnitude and angle, like this:


\begin{gathered} \text{ Magnitude: }||-6-8i||=√((-6)^2+(-8)^2)=√(36+64)=√(100)=10 \\ \\ \text{ Angle: }\theta=\arctan\left((-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}

User Jay Doshi
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