Analytic Functions - Palka, Ch. III, Section 1.3 ....

[Math Amateur]

I am reading Bruce P. Palka's book: An Introduction to Complex Function Theory ...

I am focused on Chapter III: Analytic Functions ...

I need help with fully understanding some remarks by Palka regarding an analytic function in Chapter III, Section 1.3 ...

The remarks I refer to from Palka read as follows:View attachment 7392In the above text from Palka Chapter III, Section 1.3 we read the following:

" ... ... For instance $$h(z) = \sqrt{ 1 + z^2 }$$ is analytic in $$U = \mathbb{C} \sim \{ z : \text{ Re } z = 0 \text{ and } \lvert \text{ I am } z \rvert \ge 1 \}$$ ... ... "Can someone please show the explicit calculations that show that we need to exclude the points in the set $$\{ z : \text{ Re } z = 0 \text{ and } \lvert \text{ I am } z \rvert \ge 1 \}$$ from $$\mathbb{C}$$ ...
Help will be much appreciated ... ...

Peter====================================================================================

Readers of the above post may be helped by access to Palka's introduction to and definition of analytic functions ... so I am providing the same ... as follows:
View attachment 7393
Readers of the above post may be helped by access to Palka's Example 1.5, Ch. III, Section 1.2 ... so I am providing the same ... as follows:
View attachment 7394
https://www.physicsforums.com/attachments/7395

 

[Euge]

The function $h(z) = \sqrt{1 + z^2}$ is analytic on the set of all $z$ such that $1 + z^2$ lies outside the negative real axis. So to find the "bad" points, we set $\operatorname{Re}(1 + z^2) \le 0$ and $\operatorname{Im}(1 + z^2) = 0$. If $z = x + yi$, then by algebra $1 + z^2 = (1 + x^2 - y^2) + 2xyi$. So $\operatorname{Re}(1 + z^2) \le 0$ becomes $1 + x^2 - y^2\le 0$, and $\operatorname{Im}(1 + x^2) = 0$ becomes $2xy = 0$. Since $2xy = 0$, either $x = 0$ or $y = 0$. If $y = 0$, the inequality $1 + x^2 - y^2 \le 0$ becomes $1 + x^2 \le 0$, which is absurd. If $x = 0$, we obtain $1 - y^2 \le 0$, or $y^2 \ge 1$, i.e., $\lvert y \rvert \ge 1$. This means that the bad points consists of all $z = x + yi$ such that $x = 0$ and $\lvert y \rvert \ge 1$, as desired.

 

[Math Amateur]

The function $h(z) = \sqrt{1 + z^2}$ is analytic on the set of all $z$ such that $1 + z^2$ lies outside the negative real axis. So to find the "bad" points, we set $\operatorname{Re}(1 + z^2) \le 0$ and $\operatorname{Im}(1 + z^2) = 0$. If $z = x + yi$, then by algebra $1 + z^2 = (1 + x^2 - y^2) + 2xyi$. So $\operatorname{Re}(1 + z^2) \le 0$ becomes $1 + x^2 - y^2\le 0$, and $\operatorname{Im}(1 + x^2) = 0$ becomes $2xy = 0$. Since $2xy = 0$, either $x = 0$ or $y = 0$. If $y = 0$, the inequality $1 + x^2 - y^2 \le 0$ becomes $1 + x^2 \le 0$, which is absurd. If $x = 0$, we obtain $1 - y^2 \le 0$, or $y^2 \ge 1$, i.e., $\lvert y \rvert \ge 1$. This means that the bad points consists of all $z = x + yi$ such that $x = 0$ and $\lvert y \rvert \ge 1$, as desired.

Thanks for the help, Euge ...

Just working through your post now ...

Peter