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))#