- #1
Tsunoyukami
- 215
- 11
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?
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?