Author Topic: Q7 TUT 0101  (Read 2339 times)

Victor Ivrii

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Q7 TUT 0101
« 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:
$$z^4 - 2z - 2 \qquad \text{in }\ \bigl\{\frac{1}{2}< |z| < \frac{3}{2}\bigr\}.$$

Jeffery Mcbride

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Re: Q7 TUT 0101
« Reply #1 on: November 30, 2018, 04:01:32 PM »
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 $\displaystyle \frac{1}{2} \ < \ |z|\ < \ \frac{3}{2}$