Toronto Math Forum
MAT3342018F => MAT334Tests => Quiz7 => Topic started by: Victor Ivrii on November 30, 2018, 03:49:48 PM

Using argument principle along line on the picture, calculate the number of zeroes of the following function in the given annulus:
$$
z^4  2z  2 \qquad \text{in }\ \bigl\{\frac{1}{2}< z < \frac{3}{2}\bigr\}.
$$

Using Rouche's Theorem, which is a corollary of the Argument Principle.
$\displaystyle f( z) =\ z^{4} \ \ 2z\ \ 2$ in $\displaystyle \frac{1}{2} \ < \ z\ < \ \frac{3}{2}$
$\displaystyle Let\ g( z) \ =\ 2$
For the smaller curve we have $\displaystyle z=\frac{1}{2} e^{i\theta }$,
$\displaystyle f( z) ( 2) \ \leq \ z^{4} \ +\ 2z\ \ =\ \frac{17}{16} \ < \ g( z) \ $
So, f(z) and g(z) have the same number of zeros in the smaller circle, which is 0.
$\displaystyle Now,\ let\ g( z) \ =\ z^{4}$
For the larger curve we have $\displaystyle z=\frac{3}{2} e^{i\theta }$
$\displaystyle f( z) \ \ z^{4} \ \leq 2z\ +\ 2\ \ \ =\ 5\ < \ g( z) \ =\ \frac{81}{16} \ \ $
So, f(z) and g(z) have the same number of zeros in the larger circle, which is 4.
Thus, all four zeros of f(z) are in $\displaystyle \frac{1}{2} \ < \ z\ < \ \frac{3}{2}$