Green's Function Homework: Real & Imaginary Parts

In summary, the problem at hand is to calculate the real and imaginary parts of Green's function, which is given by the expression g00 = [1-(1-4t2(z-E0)-2)1/2]/2t2(z-E0)-1. To solve this, the expression is first simplified by multiplying the denominator and numerator by (z-E0). Then, using polar coordinates, the square root is expressed as (√r)cos(θ/2) and (√r)sin(θ/2) for its real and imaginary parts. This allows for the real and imaginary parts of g00 to be separated and calculated.
  • #1
Mancho
7
0

Homework Statement



I'm asked to calculate Green's function's real and imaginary parts.
The expression for the given Green's function is:

g00=[1-(1-4t2(z-E0)-2)1/2]/2t2(z-E0)-1 (1)
Where, z is the complex variable: z= E+iO+ (2)

Homework Equations



Complex number definition: Z = x + iy, where x is the real part and iy - imaginary.

The Attempt at a Solution


To separate real and imaginary parts I tried to express g00 in the form: g00= x+iy

First I simplified the (1) by multiplying the denominator and numerator by (z-E0).
The result is g00= 1/2t2[(z-E0)-(1-4t2)]1/2.

Then I'm stuck. I don't know how to remove the square root to divide real and imaginary parts. I'm not even sure if it is the pure math problem or if I have to take into consideration anything else.

I would appreciate any help.
 
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  • #2
Welcome to PF!

Mancho said:
I'm asked to calculate Green's function's real and imaginary parts.

I don't know how to remove the square root to divide real and imaginary parts.

Hi Mancho! Welcome to PF! :smile:

Hint: the square-root of re is (√r)eiθ/2, and so its real and imaginary parts are (√r)cos(θ/2) and (√r)sin(θ/2) :wink:
 
  • #3
Thanks a lot tiny-tim!

I guessed from your hint I had to use polar coordinates and it worked great! I got what I was supposed to. Maybe I will upload the solution later when I have time.

Thanks again! :)
 

1. What is a Green's function?

A Green's function is a mathematical tool used to solve differential equations in physics and engineering. It represents the response of a system to an impulse or point source input. It can also be thought of as a fundamental solution to a differential equation.

2. What are the real and imaginary parts of a Green's function?

The real and imaginary parts of a Green's function represent the solution to the real and imaginary parts of a differential equation, respectively. They can also be interpreted as the amplitude and phase of the system's response to an input.

3. How is a Green's function used in solving differential equations?

A Green's function is used as a fundamental solution to a differential equation. It is convolved with the input function to find the solution to the differential equation. This is known as the Green's function method or the method of convolution.

4. What is the physical significance of the real and imaginary parts of a Green's function?

The real and imaginary parts of a Green's function represent the physical behavior of a system in response to an input. The real part corresponds to the amplitude, or strength, of the response, while the imaginary part represents the phase, or timing, of the response.

5. How do the real and imaginary parts of a Green's function relate to each other?

The real and imaginary parts of a Green's function are related through the Hilbert transform. This mathematical operation converts the real part to the imaginary part and vice versa. In physical terms, this means that the amplitude and phase of a system's response are interdependent and cannot be separated from each other.

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