Question

Let f(z) = [ Sin ^2 z . (1+cosz) ] / [ {(z-1)^2}-1 ]
1- find the zeros of f
2- classify the zero z=0 , z= pi, z=2pi

Answer 1

Answer on Question#50048 - <math> - <complex analysis="">

Let $f(z) = \frac{1 + \cos z}{(z - 1)^2 - 1} \sin^2 z$

1) find zeroes of $f$

Solution. Let's solve equation:

$$f(z) = 0;$$ $$\frac{1 + \cos z}{(z - 1)^2 - 1} \sin^2 z = 0 \Rightarrow \left[ \begin{array}{c} 1 + \cos z = 0 \\ \sin^2 z = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{c} z = \pi + 2\pi k, k \in \mathbb{Z} \\ z = \pi k, k \in \mathbb{Z} \end{array} \right., \text{ where } (z - 1)^2 - 1 \neq 0 \Leftrightarrow z_1 \neq 0, z_2 \neq 2$$

Answer: $z = \pi k, k \in \mathbb{Z} \setminus \{0\}$ zeroes of $f$

2) classify the zero $z = 0$, $z = \pi i$, $z = 2\pi i$

Solution. First of all, we must note that

$f(\pi i) = \frac{1 + \cos\pi i}{(\pi i - 1)^2 - 1} \sin^2 \pi i \neq 0$ and $f(2\pi i) = \frac{1 + \cos 2\pi i}{(2\pi i - 1)^2 - 1} \sin^2 2\pi i \neq 0$. So, this means that in the task was a mistake. We assume that there was $z = 0$, $z = \pi$, $z = 2\pi$

Let's consider $z = 0$:

Since, $f$ didn't determined at $z = 0$, but holomorphic in neighborhood around $z = 0$, excluding $z = 0$, then $z = 0$ is the singular point.

Since, $\lim_{z\to 0}f(z) = \lim_{z\to 0}\frac{1 + \cos z}{(z - 1)^2 - 1}\sin^2 z = \lim_{z\to 0}\frac{1 + \cos z}{(z - 2)}\frac{\sin z}{z}\sin z = 0$, then $z = 0$ is a removable singularity.

Let's consider $z = \pi$:

$$f(z) = \frac{1 + \cos z}{(z - 1)^2 - 1} \sin^2 z = (z - \pi)^4 \frac{1 + \cos z}{z(z - 2)} \frac{\sin^2 z}{(z - \pi)^4} = (z - \pi)^4 h(z), \text{ where } h(z) = \frac{1 + \cos z}{z(z - 2)} \frac{\sin^2 z}{(z - \pi)^4}$$

Since, $\lim_{z\to \pi}(z - \pi)h(z) = \lim_{z\to \pi}\frac{1 + \cos z}{z(z - 2)}\frac{\sin^2 z}{(z - \pi)^3} = \lim_{z\to \pi}\frac{1}{z(z - 2)}\frac{\sin^2 z}{z - \pi}\frac{1 + \cos z}{(z - \pi)^2} = \frac{0}{2\pi(\pi - 2)} = 0$,

but $\lim_{z\to \pi}h(z) = \lim_{z\to \pi}\frac{1 + \cos z}{z(z - 2)}\frac{\sin^2 z}{(z - \pi)^4} = \lim_{z\to \pi}\frac{1}{z(z - 2)}\frac{\sin^2 z}{(z - \pi)^2}\frac{1 + \cos z}{(z - \pi)^2} = \frac{1}{2\pi(\pi - 2)} \neq 0$, then $z = \pi$ is 4 order zero of $f$.

Let's consider $z = 2\pi$:

$$f(z) = \frac{1 + \cos z}{(z - 1)^2 - 1} \sin^2 z = (z - 2\pi)^2 \frac{1 + \cos z}{z(z - 2)} \frac{\sin^2 z}{(z - 2\pi)^2} = (z - 2\pi)^2 h(z), \text{ where}$$ $$h(z) = \frac{1 + \cos z}{z(z - 2)} \frac{\sin^2 z}{(z - 2\pi)^2}$$

Since, $\lim_{z\to 2\pi}(z - 2\pi)h(z) = \lim_{z\to 2\pi}\frac{1 + \cos z}{z(z - 2)}\frac{\sin^2 z}{z - 2\pi} = \frac{2}{2\pi(2\pi - 2)} 0 = 0$,

but $\lim_{z\to 2\pi}h(z) = \lim_{z\to 2\pi}\frac{1 + \cos z}{z(z - 2)}\frac{\sin^2 z}{(z - \pi)^2} = \lim_{z\to \pi}\frac{1 + \cos z}{z(z - 2)}\frac{\sin^2 z}{(z - 2\pi)^2} = \frac{2}{2\pi(2\pi - 2)} \neq 0$, then $z = 2\pi$ is 2 order zero of $f$.

Answer: $z = 0$ is a removable singularity, $z = \pi$ is 4 order zero of $f$, $z = 2\pi$ is 2 order zero of $f$.

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