Complex Variables. Problem about complex sine.

In summary, the function sin(z) is inyective on a vertical band of the complex plane, but it is possible to show that it maps the strip \{x + iy : -\pi/2 \le x \le \pi/2 and y > 0\} to the upper half-plane bijectively and the strip \{x + iy : -\pi/2 \le x \le \pi/2 and y < 0\} to the lower half-plane bijectively.
  • #1
ELESSAR TELKONT
44
0
[SOLVED] Complex Variables. Problem about complex sine.

Homework Statement



Proof that the function
[tex]\begin{displaymath}
\begin{array}{cccc}
f: &A=\left\{z\in\mathbb{C}\mid-\frac{\pi}{2}<\Re z<\frac{\pi}{2}\right\} &\longrightarrow &B=\mathbb{C}-\left\{z\in\mathbb{C}\mid \Im{z}=0,\,\left\vert\Re z\right\vert\geq 1\right\}\\
&z &\mapsto &\sin(z)
\end{array}
\end{displaymath}[/tex]
is a biyection.

Homework Equations





The Attempt at a Solution



I have already proven (on the previous exercise of my book) that the complex sine is inyective on a vertical band of the complex plane of width [tex]\pi[/tex]. Then I have only to proof that the function is surjective or that there exists left inverse (more difficult since its multivaluated in general).

My problem is that I have no idea how to proof that since I have saw that the complex sine maps vertical lines in the complex plane to hyperbolas and horizontal ones to ellipses. Then I think it's imposible that the sine could map the set [tex]A[/tex] to the set [tex]B[/tex] one-one because the hyperbolas and the ellipses I have mentioned get out from [tex]B[/tex].
 
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  • #2
I did this very same problem recently. Fortunately, the book had already proven that sin(z) maps the strip [itex]\{x + iy : 0 \le x \le \pi/2[/itex] and [itex]y \ge 0\}[/itex] in a one-to-one fashion onto the first quadrant.

Then, using the fact that [itex]\sin(-\bar{z}) = -\overline{\sin(z)}[/itex], one may show that sin(z) maps the strip [itex]\{x + iy : -\pi/2 \le x \le \pi/2[/itex] and [itex]y > 0\}[/itex] to the upper half-plane bijectively.

Similarly, one may show that sin(z) maps the strip [itex]\{x + iy : -\pi/2 \le x \le \pi/2[/itex] and [itex]y < 0\}[/itex] to the lower half-plane bijectively.

Now all that's left is to consider the strip [itex]\{x + iy : -\pi/2 \le x \le \pi/2[/itex] and [itex]y = 0\}[/itex] which I leave to you.
 
  • #3
Thanks for your help
 

Related to Complex Variables. Problem about complex sine.

1. What are complex variables?

Complex variables refer to mathematical expressions that involve both real and imaginary numbers. They are often used in advanced calculus, physics, and engineering.

2. What is the difference between a complex number and a complex variable?

A complex number is a single number that contains both a real and an imaginary component, while a complex variable is a function that takes in a complex number as its input and outputs another complex number.

3. How do complex variables relate to the concept of analytic functions?

Analytic functions are functions that can be expressed as a power series, and they are essential in the study of complex variables. In fact, every complex variable is an analytic function, but not all analytic functions are complex variables.

4. What is the complex sine function?

The complex sine function, denoted as sin(z), is an analytic function that is defined for all complex numbers. It is the complex extension of the familiar sine function from real analysis.

5. What are some applications of complex variables in real life?

Complex variables have numerous applications in physics, engineering, and other fields. They are used in signal processing, fluid dynamics, electromagnetism, and quantum mechanics, among others. They are also essential in solving complex integration problems and in understanding the behavior of systems with oscillatory components.

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