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How do you write the complex number #-2-3i# in polar form?

1 Answer

sankarankalyanam

Jul 2, 2018

Polar form of #-2 -3i = (3.6056, 236.31^@)#

Explanation:

#z =a + i b = r (cos theta + i sin theta)#

#r = |sqrt(a^2 + b^2)| = sqrt (-2^2 + -3^2) = 3.6056#

# theta = tan^-1 (b/a) = tan ^-1 (-3/-2) = 236.31^@ " Point in III Quadrant"#

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