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How do you convert #-5-5i# to polar form?

1 Answer

May 19, 2018

#(5sqrt(2), pi/4)#

Explanation:

Polar form: #(r, theta)#

#a + bi#

#a = -5#
#b=-5#

#r = sqrt(a^2 + b^2) = sqrt((-5)^2 + (-5)^2) = sqrt(50)= 5sqrt(2)#

#theta = tan(b/a) = tan((-5)/-5) = tan(1) = pi/4#

#(5sqrt(2), pi/4)#

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