Application of the Argument Principle

[Tsunoyukami]

I've been studying for a test and have stumbled upon another type of problem I'm not comfortable with yet.

I need help applying the argument principle - the practice problems in my textbook all consider only polynomials in the first quadrant. In solving this type of problem I consider ##\gamma = \gamma_{1} + \gamma_{2} + \gamma_{3}## where ##\gamma_{1}## lies along the x-axis (positive real axis) from ##0## to ##R##, ##\gamma_{2}## is the quarter circle of radius ##R## with ##\theta \in [0, \frac{\pi}{2}]##, and ##\gamma_{3}## lies along the imaginary y-axis from ##R## to ##0##. I let ##R \to \infty## and this type of problem is 'easy' to solve (at least all the practice problems and examples).

However, I've been searching for additional problems online and a common one I've found is to determine the number of zeroes in a circle. I'm not sure how exactly I should do this. For example, consider Exercise 1 on page 3 of the following link: http://people.math.gatech.edu/~cain/winter99/ch11.pdf

I know the answer is 3 because the number of zeroes of this function inside the unit circle is one with multiplicity three, which means it is counted three times. I can check this using the argument principle and writing ##\gamma(t) = Re^{it}##, ##t \in [0, 2\pi]## and ##R=1##. Then I can write the following: ##f(z) = z^{3} \to f(Re^{it} = (Re^{it})^{3} = R^{3} \cdot e^{3it} = e^{3it}##. For ##t=0##, the argument of ##f(z)## is 0; for ##t=2\pi##, the argument of ##f(z)## is ##6\pi##. Then the winding number is three, and because there are no poles in this region there must be three zeroes as I predicted.

Now consider Exercise 6 on page 6 of the same link above.

"Show that the polynomial ##z^{6} +4z^{2} - 1## has exactly two zeros inside the circle ##|z| = 1##."

I used the same parametrization as above - ##e^{it}##, but because R is not large I can't factor our the term with the highest degree and approximate the other terms as 0. I'm not sure how to approach this problem...

$$p(z) = z^{6} +4z^{2} - 1 \to p(e^{it}) = (e^{it})^{6} +4(e^{it})^{2} - 1 = e^{6it} + 4e^{2it} - 1$$


Note: I can not use Rouche's Theorem as we have not yet covered it in class and it is not considered a valid approach on our test tomorrow. Is there a way to solve this problem only using the Argument Principle?

 

[Mandelbroth]

Now consider Exercise 6 on page 6 of the same link above.

"Show that the polynomial ##z^{6} +4z^{2} - 1## has exactly two zeros inside the circle ##|z| = 1##."

I used the same parametrization as above - ##e^{it}##, but because R is not large I can't factor our the term with the highest degree and approximate the other terms as 0. I'm not sure how to approach this problem...

$$p(z) = z^{6} +4z^{2} - 1 \to p(e^{it}) = (e^{it})^{6} +4(e^{it})^{2} - 1 = e^{6it} + 4e^{2it} - 1$$Note: I can not use Rouche's Theorem as we have not yet covered it in class and it is not considered a valid approach on our test tomorrow. Is there a way to solve this problem only using the Argument Principle?

Why not just evaluate ##\displaystyle \frac{1}{2\pi i}\int\limits_{[0,2\pi]}\frac{2ie^{2ix}(4+3e^{4ix})}{e^{6it}+4e^{2it}-1}~dt##?

 

[Tsunoyukami]

Can you explain how you got the numerator of your integral? I don't know how you got this expression at all...

 

[Mandelbroth]

Can you explain how you got the numerator of your integral? I don't know how you got this expression at all...

What is ##\frac{d}{dz}\left[z^6+4z^2-1\right]##?

I factored because I thought there was a cool way to solve it but the ##-1## in the denominator makes things interesting.

Think what the argument principle says.

 

[micromass]

Can you explain how you got the numerator of your integral? I don't know how you got this expression at all...

The numerator is just the derivative of the polynomial.
http://en.wikipedia.org/wiki/Argument_principle

 

[Tsunoyukami]

Oh right! Then it's in the form ##\frac{h'(z)}{h(z)}##. Obviously I'm getting tired. I'll work on finishing up this problem in a little while or tomorrow before the test. Thanks!

 

[Mandelbroth]

Oh right! Then it's in the form ##\frac{h'(z)}{h(z)}##. Obviously I'm getting tired. I'll work on finishing up this problem in a little while or tomorrow before the test. Thanks!

Again, you are most certainly welcome.

Good luck on your test.

 

[Tsunoyukami]

Okay, I have a couple hours left before my test and I would like to tackle this problem (even if I don't manage to get it before the test I'll be interested to know how to solve this type of problem in general).

Just to be sure - in your previous post you use x instead of t in the numerator; is this just a mistake or am I missing something important?

##\frac{1}{2\pi i} \int_{\gamma} \frac{h'(z)}{h(z)} dz## = (# zeroes inside \gamma) - (# poles inside \gamma)

Because ##h(z)## is a polynomial it has no poles, let alone poles inside ##\gamma##, so the above expression gives the number of zeroes inside ##\gamma##.


I will give a brief sketch of what my book does next. We let ##\gamma## be the circle ##\gamma(t) = e^{it}##, ##t \in [0,2\pi]##. Then, "ince ##h## is not zero on ##\gamma##, there is an annular region ##R-\delta ≤ |z| ≤ R+\delta## on which ##h## is not zero." (Complex Variables, 2nd edition by Stephen D. Fisher; pg. 174) We let D be this annular region with ##arg(z) \in (0, 2\pi)##. There is an analytic function ##f## in this region such that ##f=log h##. Consider a small positive value ##\epsilon##, then:

$$\int_{\gamma} \frac{h'(z)}{h(z)} dz = lim_{\epsilon \to 0} \int_{\epsilon}^{2\pi - \epsilon} \frac{h'(e^{it})}{h(e^{it})} ie^{it} dt = ... = lim_{\epsilon \to 0} log|h(e^{i(2\pi-\epsilon})| - log|h(e^{i\epsilon})| + iarg \left(h(e^{i(2\pi-\epsilon)})\right) - iarg \left(h(e^{i\epsilon})\right)$$

As ##\epsilon## approaches 0 the real part of this expression approaches zero and we are left only with the imaginary part. I will now apply this to my specific function.

$$\int_{\gamma} \frac{h'(z)}{h(z)} dz = iarg \left( e^{6i(2\pi-\epsilon)} +4e^{2i(2\pi-\epsilon)} -1\right) - iarg \left (e^{6i\epsilon} +4e^{2i\epsilon} -1 \right)$$

When ##\epsilon## approaches 0, we find:

$$\int_{\gamma} \frac{h'(z)}{h(z)} dz = iarg \left( e^{12\pi i} +4e^{4 \pi i} -1\right) - iarg \left (e^{0} +4e^{0} -1 \right)$$

But ##e^{2\pi i n}##, ##n \in Z##= 1. So it turns out both these expressions are equal and the change in argument is 0, which suggests that there are 0 zeros inside ##\gamma## = |z| = 1.

While this is entirely possible, I feel as if I've done something wrong.

Wolfram-Alpha tells me that there are two zeroes inside the unit circle: http://www.wolframalpha.com/input/?i=z^6+%2B+4z^2+-1


Where have I gone wrong?

 

[Mandelbroth]

Okay, I have a couple hours left before my test and I would like to tackle this problem (even if I don't manage to get it before the test I'll be interested to know how to solve this type of problem in general).

Just to be sure - in your previous post you use x instead of t in the numerator; is this just a mistake or am I missing something important?

Nope. That was my mistake. Sorry. The numerator should have been a function of t.

I will give a brief sketch of what my book does next. We let ##\gamma## be the circle ##\gamma(t) = e^{it}##, ##t \in [0,2\pi]##. Then, "ince ##h## is not zero on ##\gamma##, there is an annular region ##R-\delta ≤ |z| ≤ R+\delta## on which ##h## is not zero." (Complex Variables, 2nd edition by Stephen D. Fisher; pg. 174) We let D be this annular region with ##arg(z) \in (0, 2\pi)##. There is an analytic function ##f## in this region such that ##f=log h##. Consider a small positive value ##\epsilon##, then:

$$\int_{\gamma} \frac{h'(z)}{h(z)} dz = lim_{\epsilon \to 0} \int_{\epsilon}^{2\pi - \epsilon} \frac{h'(e^{it})}{h(e^{it})} ie^{it} dt = ... = lim_{\epsilon \to 0} log|h(e^{i(2\pi-\epsilon})| - log|h(e^{i\epsilon})| + iarg \left(h(e^{i(2\pi-\epsilon)})\right) - iarg \left(h(e^{i\epsilon})\right)$$

Something looks wrong with this. I'm off to dinner, but something tells me there is an error in here. I'll be back soon, so if you can't find the error, I'll look for it when I get back.

Edit: Let me see if I can do this.
We have that ##\displaystyle \int\limits_{\gamma}\frac{h'(z)}{h(z)}~dz=\int\limits_{\gamma}d\left( \ln(h(z))\right)=i\int\limits_{(0,2\pi)}e^{it}\frac{d(\ln(h(e^{it})))}{dt}~ dt##, where the parentheses under the last integral indicate we are integrating over an open interval. Does that look like it will be equal to your last expression in the quote above? (Hint: ...no.)

 

[jackmell]

Okay, I have a couple hours left before my test and I would like to tackle this problem (even if I don't manage to get it before the test I'll be interested to know how to solve this type of problem in general).

Just to be sure - in your previous post you use x instead of t in the numerator; is this just a mistake or am I missing something important?

##\frac{1}{2\pi i} \int_{\gamma} \frac{h'(z)}{h(z)} dz## = (# zeroes inside \gamma) - (# poles inside \gamma)

Because ##h(z)## is a polynomial it has no poles, let alone poles inside ##\gamma##, so the above expression gives the number of zeroes inside ##\gamma##.


I will give a brief sketch of what my book does next. We let ##\gamma## be the circle ##\gamma(t) = e^{it}##, ##t \in [0,2\pi]##. Then, "ince ##h## is not zero on ##\gamma##, there is an annular region ##R-\delta ≤ |z| ≤ R+\delta## on which ##h## is not zero." (Complex Variables, 2nd edition by Stephen D. Fisher; pg. 174) We let D be this annular region with ##arg(z) \in (0, 2\pi)##. There is an analytic function ##f## in this region such that ##f=log h##. Consider a small positive value ##\epsilon##, then:

$$\int_{\gamma} \frac{h'(z)}{h(z)} dz = lim_{\epsilon \to 0} \int_{\epsilon}^{2\pi - \epsilon} \frac{h'(e^{it})}{h(e^{it})} ie^{it} dt = ... = lim_{\epsilon \to 0} log|h(e^{i(2\pi-\epsilon})| - log|h(e^{i\epsilon})| + iarg \left(h(e^{i(2\pi-\epsilon)})\right) - iarg \left(h(e^{i\epsilon})\right)$$

As ##\epsilon## approaches 0 the real part of this expression approaches zero and we are left only with the imaginary part. I will now apply this to my specific function.

$$\int_{\gamma} \frac{h'(z)}{h(z)} dz = iarg \left( e^{6i(2\pi-\epsilon)} +4e^{2i(2\pi-\epsilon)} -1\right) - iarg \left (e^{6i\epsilon} +4e^{2i\epsilon} -1 \right)$$

When ##\epsilon## approaches 0, we find:

$$\int_{\gamma} \frac{h'(z)}{h(z)} dz = iarg \left( e^{12\pi i} +4e^{4 \pi i} -1\right) - iarg \left (e^{0} +4e^{0} -1 \right)$$

But ##e^{2\pi i n}##, ##n \in Z##= 1. So it turns out both these expressions are equal and the change in argument is 0, which suggests that there are 0 zeros inside ##\gamma## = |z| = 1.

While this is entirely possible, I feel as if I've done something wrong.

Wolfram-Alpha tells me that there are two zeroes inside the unit circle: http://www.wolframalpha.com/input/?i=z^6+%2B+4z^2+-1


Where have I gone wrong?

Ok, you're completely ignoring Complex Analysis dealing with multivalued functions, in particular, the multi-valued complex logarithm function, and are not applying correctly, the Fundamental Theorem of Calculus as it applies to multivalued functions.

Alice looks up ahead and sees the signpost:


[tex]\mathop\oint\limits_{|z|=1} \frac{1}{z}dz=\log(z)\biggr|_1^1[/tex]

and thinks to herself, "well ain't that zero?" but then remembers how the Fundamental Theorem of Calculus applies to line-integrals over multi-valued functions: The integral of the function over the path is equal to the difference of the function's antiderivative at the end points of the contour along an analytically-continuous path over the antiderivative.

Alice then remembers what a plot of the real and imaginary parts of the log function look like: The real part is a single-valued function that looks like a funnel. The imaginary part however is multivalued and looks like a twisted helical sheet. Therefore, when a [itex]2\pi[/itex] path over the unit circle is mapped onto these two surfaces in an analytically-continuous fashion, the path returns to the starting point over the real sheet but does not return to the starting part over the imaginary sheet: the imaginary sheet has twisted by [itex]2\pi[/itex] (and onto another single-valued determination). Alice then took out her marks-a-lot and finished the problem:

[tex]\mathop\oint\limits_{|z|=1} \frac{1}{z}dz=\log(z)\biggr|_1^1=2\pi i=i\Delta_{\gamma} \arg z[/tex]

Here's what Im(log) looks like and as you can see, if you go around once, as in a contour around the unit circle, the end point is not the same as the starting point, in fact, since Im(log(z))=arg(z), the sheet just winds around the origin, with it's value equal to the total angular displacement or [itex]\Delta \arg z[/itex]