Rouche’s Theorem applied to the complex valued function $f(z) = z^6 + \cos z$

Let $f(z) = z^6 + \cos z$. Apply Rouche’s theorem to find the change in argument of $f(z)$ as $z$ travels once around the circle of radius 2, center zero, in the positive direction.


By the argument principle, the change in argument of $f(z)$ as $z$ travels around the circle is equal to 2π  times the number of zeros minus the number of poles of $f(z)$ inside the circle. The function $f(z)$ does not have any poles. To find the number of zeros of $f(z)$ inside the circle we compare $f(z)$ with the function $z^6$ and apply Rouche’s Theorem. So let $g(z) = z^6$ . On the circle we have
\[|g(z)| = |z| ^6 = 26 = 64\] and
\[|f(z) − g(z)| = | \cos(z)|= \frac{|e ^{iz} + e^{ −iz}|}{ 2}\leq \frac{|e ^{iz}| + |e^{ −iz}|}{ 2} =e^{ |z|} = e ^2 < 64.\]
Thus, Rouche’s Theorem applies and $f(z)$ has the same number of zeros inside the circle as $z^6$ , which is $6$. Thus, the change in argument is $2π × 6 = 12π$.

Erdos Erdos
The answer is accepted.
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