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

2 Answers
Douglas K.
Mar 23, 2017

Please see the explanation.

Explanation:

Because the real part (a) of the complex number is zero, you cannot use #theta = tan^-1(b/a)#; you must know that the angle is either #pi/2# or #3pi/2#. Because the sign of the complex part is negative, you must know that this makes the angle the latter, #3pi/2#.

#theta = 3pi/2#

You can see that the magnitude is 2.

The polar form is:

#2(cos(3pi/2)+isin(3pi/2))#

kamil9234
Mar 23, 2017

I hope a get it right: #2*(cos(270)+isin(270))#

Explanation:

#r=sqrt(x^2+y^2)=sqrt(4)=2#
#alpha=ctg^(-1)(0/-2)=90#
#alpha=360-90=270#

In polar:
#2*(cos(270)+isin(270))#

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