Author Topic: 2020F-Test4-MAIN-A-Q1  (Read 2404 times)

Xuefen luo

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« on: December 09, 2020, 01:16:20 PM »
Problem 1. (a) Find all zeroes of the function $f(z) = \frac{sin(2\pi \sqrt{z})}{sin(\pi z)}$ in domain $D=\mathbb{C}\ (-\infty,0]$.
(b)Also find all singular points of this function and determine their types(removable, pole (in which case what is it's order), essential singularity, not
isolated singularity).
(c) In particular, if $\infty$ is in the domain: check whether it is a zero.
(d) Draw these points on the complex plane.

Answer: Notice that $sin(2\pi \sqrt{z})$ has zeroes only at $z=\frac{n^2}{4}, n \in \mathbb{Z^+}$, and $sin(\pi z)$ has zeroes only at $z=m, m \in \mathbb{Z^+}$ in domain $D=\mathbb{C}\ (-\infty,0]$. Since all these zeroes are simple, we can conclude that:
-$z=\frac{n^2}{4}$ with $n \in \mathbb{Z^+}$ and $ n^2$ is a multiple of 4 such that $z=\frac{n^2}{4}=m$ for some $m\in \mathbb{Z^+}$, then $z$ are removable singularities;
-$z=\frac{n^2}{4}$ with $n \in \mathbb{Z^+}$  and  $n^2$ is not a multiple of 4 such that $z \notin \mathbb{Z^+}$, then $z$ are simple zeroes;
-$z=m$ for $m \in \mathbb{Z^+}$ and $m \neq \frac{n^2}{4}$ for any $n \in \mathbb{Z^+}$, then $z$ are simple poles.
-there are no essential singularities;
-$z=\infty$ is a non-isolated singularity.