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

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Answer 1 · 2017-01-13T16:24:38

Given: $z = a + bi$
$r = sqrt(a^2 + b^2)$
$theta = 2pi + tan^-1(b/a)$ (because it is quadrant 4)
$z = r(cos(theta) + isin(theta)$

For the number $z = 3 - 3i$

$r = sqrt(3^2 + 3^2)$

$r = 3sqrt2$

$theta = 2pi + tan^-1(-3/3)$

$theta = 2pi + tan^-1(-1)$

$theta = 2pi - pi/4$

$theta = (7pi)/4$

$z = 3sqrt2(cos((7pi)/4) + isin((7pi)/4))$

How do you convert 3-3i3-3i to polar form?