How do you convert $(-13,12)$ to polar form?

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Answer 1 · 2017-02-25T14:22:53

$sqrt 313 [cos (pi - arctan frac{12}{13}) + i sin (pi - arctan frac{12}{13})]$

$x = r cos t, y = r sin t$

$-13 = r cos t, 12 = r sin t$

$13^2 + 12^2 = r^2 = 313$

$tan t = -12/13$

$t = - arctan frac{12}{13} + pi$

$-13 + 12 i = r (cost + i sin t)$

How do you convert (-13,12)(-13,12) to polar form?