Q7 TUT 0101

[Victor Ivrii]

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

[Jeffery Mcbride]

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 $\displaystyle \frac{1}{2} \ < \ |z|\ < \ \frac{3}{2}$