How do you convert $- 3 - 3 i$ $-3-3i$ to polar form?

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How do you convert $- 3 - 3 i$ $-3-3i$ to polar form?

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Answer 1 · 2016-05-26T10:19:31

In polar form the equation is $Z = 3 \sqrt{2} (cos 225 + i sin 225)$ $Z=3sqrt2(cos 225 +isin225)$

Explanation:

Let $Z = - 3 - 3 i;$ $Z= -3 -3i ;$ Modulas $Z : r = \sqrt{{(- 3)}^{2} + {(- 3)}^{2}} = 3 \sqrt{2}$ $Z: r =sqrt((-3)^2+(-3)^2)=3sqrt2$
Argument $Z; θ = 180^{0} + {tan}^{- 1} (\frac{3}{3}) = 180 + 45 = 225^{0} (180^{0}$ $Z; theta = 180^0 + tan^-1(3/3) = 180+45=225^0 (180^0$ added as the point lies in 3rd quadrant). In polar form the equation is $Z = 3 \sqrt{2} (cos 225 + i sin 225)$ $Z=3sqrt2(cos 225 +isin225)$ [Ans]

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How do you convert $- 3 - 3 i$ $-3-3i$ to polar form?

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How do you convert $- 3 - 3 i$ $-3-3i$ to polar form?