The Cauchy Riemann Equations for this Function

In summary, when you are composing two functions and one of them is not entire, you must multiply the functions and not compose them.
  • #1
RJLiberator
Gold Member
1,095
63

Homework Statement



Verify that each of the following functions is entire:
f(z)=(z^2-2)e^(-x)e^(-iy)

Homework Equations



The Cauchy Riemann equations u(x,y) = ______ and v(x,y) = ______
u_y=-v_x
u_x=v_y

The Attempt at a Solution



So, I've done a few of these problems and understand that to prove it's entire we need to show that the C-R equations exist for all z and are satisfied. Also, noting that Ux Uy Vx Vy are continuous.

My problem is, I'm not sure how to get this into correct form. What is u(x,y) =?
Do I distribute first? How to change this function into something that works?

Would it be
(z^2-2)/(e^x*e^iy)
Is that a good start? :/
 
Physics news on Phys.org
  • #2
RJLiberator said:

Homework Statement



Verify that each of the following functions is entire:
f(z)=(z^2-2)e^(-x)e^(-iy)

Homework Equations



The Cauchy Riemann equations u(x,y) = ______ and v(x,y) = ______
u_y=-v_x
u_x=v_y

The Attempt at a Solution



So, I've done a few of these problems and understand that to prove it's entire we need to show that the C-R equations exist for all z and are satisfied. Also, noting that Ux Uy Vx Vy are continuous.

My problem is, I'm not sure how to get this into correct form. What is u(x,y) =?
Do I distribute first? How to change this function into something that works?

Would it be
(z^2-2)/(e^x*e^iy)
Is that a good start? :/
I don't believe so.

For starters, write z = x + iy, and split ##(z^2 - 2)e^{-x}e^{-iy}## into its real and imaginary parts. u(x, y) is the real part of f(z) and v(x, y) is the imaginary part.

A useful formula that you omitted, and that might come in handy, is ##e^{it} = \cos(t) + i\sin(t)##.
 
  • Like
Likes RJLiberator
  • #3
Thanks Mark.

So we do have to expand the z then.

So this would mean:

(x^2+2yxi-y^2-2)/(e^xcos(y)+e^x(isin(y)))

The thing that is throwing me off here is the negative exponents, I'm not sure what to do with those, so I put the term in the denominator.

That doesn't seem right however.
 
  • #4
Moving them to the denominator is no help. Leave the e-x factor as is, and rewrite the e-iy factor as below.
##e^{-x}e^{-iy} = e^{-x}(\cos(y) - i\sin(y))##
 
  • #5
Rather than put "[itex]e^xcos(y)+e^x(isin(y))[/itex]" in the denominator, I think I would write [itex]e^{-x- iy}= e^{-x}e^{-iy}= e^{-x}(cos(-y)+ i sin(-y))= e^{-x}(cos(x)- i sin(x))[/itex]
 
  • #6
That was indeed the element of this question that I was missing out/overlooked.

So with that in play, we see that
u(x,y) = (x^2-y^2-2)e^(-x)cos(x)+2ye^(-x)sin(y)
v(x,y) = 2ye^(-x)cos(x)-(x^2-y^2-2)e^(-x)sin(y)
 
  • #7
Re: this thread, I found out that the composition of two functions that are entire is entire.

So One can look at (z^2-2) and e^(-x)e^(-iy) as two different functions, evaluate their cauchy-riemann equations and see that each is entire to conclude the composition is entire.

=)
 
  • #8
But notice you are not composing the functions, you are multiplying them ( though the result is true that the product of entires is entire )
 
  • Like
Likes RJLiberator

Related to The Cauchy Riemann Equations for this Function

1. What are the Cauchy Riemann Equations for this function?

The Cauchy Riemann Equations are a set of two partial differential equations that must be satisfied for a complex-valued function to be analytic. These equations are:
∂u/∂x = ∂v/∂y
∂u/∂y = -∂v/∂x
where u(x,y) and v(x,y) are the real and imaginary parts of the complex function f(z), respectively.

2. How are the Cauchy Riemann Equations derived?

The Cauchy Riemann Equations are derived from the Cauchy-Riemann conditions, which state that a function is analytic at a point if and only if it is differentiable at that point. The equations are derived by expressing the complex function f(z) in terms of its real and imaginary parts and then using the limit definition of the derivative to determine the conditions for differentiability.

3. What is the significance of the Cauchy Riemann Equations?

The Cauchy Riemann Equations are significant because they provide a necessary and sufficient condition for a complex function to be analytic. This means that if a function satisfies these equations, it is guaranteed to be analytic and therefore has many useful properties, such as being infinitely differentiable and having a Taylor series representation. Additionally, these equations are used extensively in the study of complex analysis and have applications in many areas of mathematics and physics.

4. Can the Cauchy Riemann Equations be used to determine whether a function is analytic?

Yes, the Cauchy Riemann Equations can be used to determine whether a function is analytic. If a function satisfies these equations, it is guaranteed to be analytic. However, if a function does not satisfy these equations, it does not necessarily mean that it is not analytic. Further analysis is required to determine the analyticity of a function in this case.

5. How are the Cauchy Riemann Equations used in practice?

The Cauchy Riemann Equations are used in practice to determine whether a function is analytic and to find the analyticity of a given function. They are also used to find the properties of analytic functions, such as the location of singularities and the behavior of the function near these points. Additionally, these equations are used in the development of various numerical methods for solving complex-valued differential equations and in the study of conformal mappings, which have many applications in physics and engineering.

Similar threads

  • Calculus and Beyond Homework Help
Replies
27
Views
825
  • Calculus and Beyond Homework Help
Replies
19
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
  • Calculus and Beyond Homework Help
Replies
17
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
14
Views
2K
Back
Top