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/
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