How do you convert $-3 – 6i$ to polar form?

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Answer 1 · 2018-04-09T04:36:59

To convert to polar form we need to find modulus and argument of $-3 - 6i$

Modulus:
$|-3-6i| = sqrt((-3)^2+(-6)^2)$
$|-3-6i| = sqrt(45)$

Argument:
$theta = tan^-1(6/3)$
$theta = tan^-1(2)$

Then $-3-6i$ can be rewritten as
$sqrt(45)(cos(tan^-1(2))+isin(tan^-1(2)))$

or

$sqrt(45)cis(tan^-1(2))$

How do you convert -3 – 6i -3 – 6i to polar form?