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Roots of Complex Numbers in Polar Form

Date: 09/15/2004 at 20:35:59
From: Steve
Subject: Proof Polar Form of Complex number

Dear Dr. Math:

I am having trouble proving that the polar form of every complex 
number with z not equal to 0 has two square roots.  I am not exactly
sure where to start.  I have tried several different techniques, but
do not know where to go from there.  Can you help me?



Date: 09/15/2004 at 23:02:27
From: Doctor Luis
Subject: Re: Proof Polar Form of Complex number

Hi Steve,

The essential problem is that e^z is a periodic function.  The period
happens to be 2pi*i, which is complex.  You can see that from the
following:

  e^(z + 2pi*i) = e^(x + i*y + 2pi*i)
                = (e^x) * e^[i*(y + 2pi)]
                = (e^x) * [cos(y + 2pi) + i*sin(y + 2pi)]
                = (e^x) * [cos(y) + i*sin(y)]
                = (e^x) * (e^(i*y))
                = e^(x + i*y)
                = e^z

Do you see how it doesn't make a difference when I add 2pi*i to z?
Essentially, that number comes from the periodicity of sin and cos 
(both of which have period of 2pi).

So, how is this related to square roots?

As you noted, you can also express any complex number in polar form:

  z = r*e^(i*theta)

for suitable radius and angle theta.  Of course, this number just has
the square root

  z^(1/2) = [r*e^(i*theta)]^(1/2)
          = r^(1/2) * e^(i*theta/2)


Now, remember the bit about the e^z function?  It's periodic with 
period 2pi*i.  So, the same number z can be represented as

  z = r*e^[i*theta + 2pi*i]

and it's still the same number!  Look at what happens when we take a
square root:

  z^(1/2) = [r*e^(i*theta + 2pi*i)]^(1/2)
          = r^(1/2) * e^[(i*theta/2) + pi*i]
          = r^(1/2) * e^(i*theta/2) * e^(pi*i)
          = r^(1/2) * e^(i*theta/2) * [cos(pi) + i*sin(pi)]
          = r^(1/2) * e^(i*theta/2) * (-1 + i*0)
          = (-1) * [r^(1/2) * e^(i*theta/2)]

And we get an extra factor of (-1) times the other sqare root of z.
That's the (complex) origin of the "plus or minus" that we use with
square roots. 

You could probably convince yourself that every complex number has 
three cube roots, four fourth roots, five fifth roots, and so on. 
Just keep adding enough 2pi*i periods, and you can get as many as n
distinct n-th roots. 

Now, why do we get several answers?  Well, that just goes back to my
original point.  e^z is a periodic function, and so its inverse is not
very well defined.  When you take logarithms of complex functions, you
actually get an infinite number of answers (all of which just differ
by a multiple of 2pi*i).  And for roots?  What happens there is that
some of the roots overlap and you end up with a finite number of
solutions.

Does this help any?  Let us know if you have any more questions.

- Doctor Luis, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Imaginary/Complex Numbers

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