Finding the real and imaginary parts of a function

In summary, when given a holomorphic function f:C-->C, with real and imaginary parts uf and vf, the real and imaginary parts of g(x,y) can be found by setting u_g=\text{Re}(g) and v_g=\text{Im}(g) using the formula g(x,y) = u(x,y)+iv(x,y) and starting with u_f = \frac{f(z)+\overline{f(z)}}{2}. Furthermore, for clarification, \overline{f(\overline{z})} is the complex conjugate of f(\overline{z}).
  • #1
shen07
54
0
If f:C-->C is holomorphic and View attachment 1263 , find the real and imaginary parts ug and vg of g in terms of the real and imaginary parts uf and vf of f.
 

Attachments

  • CodeCogsEqn.gif
    CodeCogsEqn.gif
    490 bytes · Views: 91
Physics news on Phys.org
  • #2
Re: Please can you give me some hint to do this exercise

For clarification you mean by \(\displaystyle u_g=\text{Re}(g) \) and \(\displaystyle v_g=\text{Im}(g)\) using that \(\displaystyle g(x,y) = u(x,y)+iv(x,y)\) , right?
 
  • #3
Re: Please can you give me some hint to do this exercise

ZaidAlyafey said:
For clarification you mean by \(\displaystyle u_g=\text{Re}(g) \) and \(\displaystyle v_g=\text{Im}(g)\) using that \(\displaystyle g(x,y) = u(x,y)+iv(x,y)\) , right?

yeas right
 
  • #4
Re: Please can you give me some hint to do this exercise

I would suggest starting by

\(\displaystyle u_f = \frac{f(z)+\overline{f(z)}}{2}\)
 
  • #5
Re: Please can you give me some hint to do this exercise

ZaidAlyafey said:
I would suggest starting by

\(\displaystyle u_f = \frac{f(z)+\overline{f(z)}}{2}\)
One more question what is \(\displaystyle \overline{f(\overline{z})}\) actually?? i don't quite understand this!
 
  • #6
Re: Please can you give me some hint to do this exercise

shen07 said:
One more question what is \(\displaystyle \overline{f(\overline{z})}\) actually?? i don't quite understand this!
Consider a simple example:
[tex]f(z) = u(z) + i v(z)[/tex] with z = x + iy.

Then
[tex]f(z) = u(x + iy) + i v(x + iy)[/tex]

[tex]f( \overline{z} ) = u(x - iy) + i v(x - iy)[/tex]

[tex]\overline{f( \overline{z} ) } = u(x - iy) - i v(x - iy)[/tex]

Is this what you are looking for? Or something more conceptual?

-Dan
 
Last edited by a moderator:
  • #7
Another example:
$$f(z)=z^2+i \Rightarrow \overline{f\left({\bar{z}}\right)}=\overline{(\bar{z})^2+i}=\overline{ \overline{z^2}+i}=\overline{\overline{z^2}}+\bar{i}=z^2-i$$
 

FAQ: Finding the real and imaginary parts of a function

What does it mean to find the real and imaginary parts of a function?

Finding the real and imaginary parts of a function involves breaking down a complex function into two separate parts: the real part, which represents the horizontal or vertical component of the function, and the imaginary part, which represents the vertical or horizontal component.

Why is it important to find the real and imaginary parts of a function?

Knowing the real and imaginary parts of a function is important in solving complex mathematical problems, particularly in fields such as engineering, physics, and signal processing. It also helps in understanding the behavior and properties of a function.

How do you find the real and imaginary parts of a function?

To find the real and imaginary parts of a function, you can use the algebraic method, which involves separating the function into real and imaginary terms and then solving for each part separately. Another method is using the polar form, which converts the complex function into polar coordinates and then identifies the real and imaginary parts.

Are there any common mistakes when finding the real and imaginary parts of a function?

Yes, one common mistake is forgetting to distribute the imaginary unit (i) to the imaginary term when using the algebraic method. Another mistake is mixing up the order of the terms when converting from rectangular form to polar form.

Can you give an example of finding the real and imaginary parts of a function?

For example, in the complex function f(z) = 3 + 4i, the real part is 3 and the imaginary part is 4i. Using the polar form, the function can be written as f(z) = 5(cosθ + isinθ), where the real part is 5cosθ and the imaginary part is 5sinθ.

Back
Top