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 · 2017-01-13T16:24:38

Given: $z = a + b i$ $z = a + bi$
$r = \sqrt{a^{2} + b^{2}}$ $r = sqrt(a^2 + b^2)$
$θ = 2 π + {tan}^{- 1} (\frac{b}{a})$ $theta = 2pi + tan^-1(b/a)$ (because it is quadrant 4)
$z = r (cos (θ) + i sin (θ)$ $z = r(cos(theta) + isin(theta)$

Explanation:

For the number $z = 3 - 3 i$ $z = 3 - 3i$

$r = \sqrt{3^{2} + 3^{2}}$ $r = sqrt(3^2 + 3^2)$

$r = 3 \sqrt{2}$ $r = 3sqrt2$

$θ = 2 π + {tan}^{- 1} (- \frac{3}{3})$ $theta = 2pi + tan^-1(-3/3)$

$θ = 2 π + {tan}^{- 1} (- 1)$ $theta = 2pi + tan^-1(-1)$

$θ = 2 π - \frac{π}{4}$ $theta = 2pi - pi/4$

$θ = \frac{7 π}{4}$ $theta = (7pi)/4$

$z = 3 \sqrt{2} (cos (\frac{7 π}{4}) + i sin (\frac{7 π}{4}))$ $z = 3sqrt2(cos((7pi)/4) + isin((7pi)/4))$

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

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