Hi : I am trying to learn about complex variables so I've been working my way through a book called An Introduction to Complex Analysis and Geometry by John D'angelo. It's a very nice basic book but I'm having trouble with a lot of the exercises. I emailed the author for a solutions manual but it doesn't exist. So what I planned to do was ask questions here when I can't solve the exercise on my own.
I am not a regular member of this list so if I don't follow the proper etiquette please let me know.
My first question is:
For $t - \alpha^2 > 0$, show that there is an $n \in \boldsymbol{N}$, such that $\frac{2 \alpha}{n} < \alpha^2 - 2$.
I think this question is related to the Archimedean property which is stated as Proposition 3.2 in the book and says that, given positive real numbers $x$ and $\epsilon$, there exists a positive integer $n$ such that $n \epsilon > x$. Equivalently, given $y > 0$, there is an $n \in \boldsymbol{N}$ such that $\frac{1}{n} < y$.
Thank you very much to anyone who can help. This is not homework. Just some random person trying to learn the basics of complex numbers.
I'm not sure where the $t$ relates here, but you are right that this relies on the Archimedean Property. It can be stated more simply that if $x$ is a real number, there exists an $n \in \mathbb{N}$ such that $n > x$. Here, let $x = \frac{2\alpha}{\alpha^2 - 2}$. Provided $\alpha^2 \neq 2$ (which is what I think you might have meant instead of $t$), then that is a real number, and so there is an $n \in \mathbb{N}$ such that
$$\frac{2\alpha}{\alpha^2 - 2} < n$$
Dividing both sides by $n$ and multiplying both sides by the denominator on the left gives you the inequality. Keep in mind that you need the denominator to be $> 0$ for the sign of the inequality to remain the same, which is what I think the first statement with $t$ was supposed to say.