How do you convert # -6i # to polar form?

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Answer 1 · 2016-05-22T12:26:42

Polar form of #-6i# is #(6,-pi/2)#

A complex number #a+ib# in polar form is written as

#rcostheta+irsintheta#

Hence #r=sqrt(a^2+b^2)#

As #-6i# can be written as #0-6i#

#r=sqrt()^2+(-6)^2)=sqrt36=6# and hence

#costheta=0/6=0# and #sintheta=-6/6=-1#

Hence, #theta=-pi/2# and

Polar form of #-6i# is #(6,-pi/2)#