Show that the real part of a certain complex function is harmonic

I will edit the original post so as to not misguide the readers. Thank you for pointing this out.In summary, the complex valued function $$f(z) = Re\big(\frac{\cos z}{\exp{z}}\big)$$ is proven to be harmonic on the whole complex plane. This can be shown by using the definition of harmonic functions and the fact that the function $$\frac{\cos z}{\exp z}$$ is entire and its real and imaginary parts are harmonic. Although there may seem to be a shortcut, it ultimately depends on what theorems and lemmas have been established.
  • #1
fatpotato
Homework Statement
Prove that ##f(z) = Re\big(\frac{\cos z}{\exp{z}}\big)## is harmonic on the whole complex plane.
Relevant Equations
Extension of usual functions to the complex plane
Definition of ##\cos(z)## and ##\exp(z)## where ##z \in \mathbb{C}##
Cauchy-Riemann equations
Laplace equation
Hello,

I have to prove that the complex valued function $$f(z) = Re\big(\frac{\cos z}{\exp{z}}\big) $$ is harmonic on the whole complex plane.

This exercice immediately follows a chapter on the extension of the usual functions (trigonometric and the exponential) to the complex plane, so I tend to believe I have to use the definitions to extract the real part of ##f(z)## and prove that it solves the Laplace equation on ##\mathbb{C}##.

However, I have the feeling there is a more elegant solution than going through the tedious computation needed, involving the expansion of the cosine in exponential form : $$ f(z) = Re\big(\frac{\cos z}{\exp{z}}\big) = \frac{1}{2} Re\big( \frac{\exp{iz} + \exp{-iz}}{\exp{z}} \big)$$

Is there a subtlety I am not seeing? For example, I can see that the derivative of ##f(z)## exists, is always continuous and that the denominator is never zero, and I would like to think ##f## is analytic, therefore proving without further computation that its real part indeed solves the Laplace equation, but I fear using this shortcut without showing that the partial derivatives of ##f## exist, are continuous and satisfy the Cauchy-Riemann equations.

Is it proof enough?

Thank you
 
Physics news on Phys.org
  • #2
The validity of a proof depends on what statements you make are theorems and lemmas that have already been established. For example, if you have proven that the quotient of analytic functions with a non-zero denominator is analytic then the proof is easier. Otherwise, you should use the definitions directly and calculate all the partial derivatives needed.
 
  • Like
Likes fatpotato
  • #3
In this case, the proof was indeed given back a few chapters.

If we take ##g(z)## being the function ##\frac{\cos z}{\exp z}##, it has been proven that its numerator and denominator are analytic everwhere, and that the denominator is never zero on the whole complex plane, so ##g## is entire.

This means that since ##g## is entire, its real and imaginary part are harmonic (and harmonic conjugates), so ##f(z) = Re(g(z))## is harmonic everywhere.

Thank you for your message.
 
  • Like
Likes FactChecker
  • #4
(out of sheer curiosity) I wondered about
fatpotato said:
the complex valued function

because when looked at the expression I thought it's a real-valued function
$$\begin{align*} f(z) = Re\left(\frac{\cos z}{\exp{z}}\right) &= \frac{1}{2} Re\Biggl( \frac{\exp{iz} + \exp{-iz}}{\exp{z}} \Biggr) \\ &\ \\
&=\frac{1}{2} Re\Biggl( \exp(-z+iz) + \exp(-z-iz) \Biggr) \\ &\ \\
&=\frac{ \exp(-z+iz) + \bigl(\exp(-z+iz)\bigr)^* }{2}\end{align*}\\
$$
 
  • Like
Likes fatpotato
  • #5
You are certainly right. Furthermore, taking the real part of ##f## will always yield a real number, so my sentence is incorrect. What I should rather say, is simply that ##f(z)## is a fonction of a complex number.

Now, I don't understand how you went from the second line to the third :
BvU said:
$$\begin{align*} f(z) = Re\left(\frac{\cos z}{\exp{z}}\right) &= \frac{1}{2} Re\Biggl( \frac{\exp{iz} + \exp{-iz}}{\exp{z}} \Biggr) \\ &\ \\
&=\frac{1}{2} Re\Biggl( \exp(-z+iz) + \exp(-z-iz) \Biggr) \\ &\ \\
&=\frac{ \exp(-z+iz) + \bigl(\exp(-z+iz)\bigr)^* }{2}\end{align*}\\
$$
I did the computation by expanding ##z## into ##z = x + iy## which isn't very fun to do and takes time, do you have an insight here?
 
  • #6
fatpotato said:
from the second line to the third
I figured ##\ \exp(-z+iz) \ ## is the complex conjugate of ##\ \exp(-z-iz) \ ## so the sum is real ...
 
  • #7
But is it? This would be the case if ## z \in \mathbb{R}##, but given that ##z \in \mathbb{C}##, I believe your statement isn't true :

$$\exp (-z +iz)* = \exp[( -x -iy )+(ix-y)]* = \exp(-x-y)\exp i(x-y)* = \exp(-x-y) [ \cos(x-y) - i\sin(x-y)]$$

And expanding ##\exp(-z-iz)## should result in an equality, however :

$$\exp (-z -iz) = \exp[( -x -iy ) - (ix-y)] = \exp[( -x+y -i(x+y)] = \exp(-x+y) [ \cos(x+y) + i\sin(x+y)]$$

As far as I look, these two results are not the same...
 
  • Like
Likes BvU
  • #8
You are absolutely correct ! o:) o:) o:) on me for not checking before posting the comment.
 
  • Like
Likes fatpotato

Related to Show that the real part of a certain complex function is harmonic

1. What is a complex function?

A complex function is a mathematical function that takes a complex number as its input and produces a complex number as its output. It can be represented in the form f(z) = u(x,y) + iv(x,y), where z = x + iy, u(x,y) is the real part, and v(x,y) is the imaginary part.

2. What does it mean for a function to be harmonic?

A function is said to be harmonic if it satisfies Laplace's equation, which states that the sum of the second-order partial derivatives of the function with respect to its independent variables is equal to zero. In simpler terms, this means that the function is smooth and has no sudden changes or discontinuities.

3. How can you show that the real part of a complex function is harmonic?

To show that the real part of a complex function is harmonic, we can use the Cauchy-Riemann equations, which state that if a complex function is differentiable at a point, then its partial derivatives must satisfy certain conditions. If these conditions are met, then the function is said to be analytic, and its real part will be harmonic.

4. What is the significance of the real part of a complex function being harmonic?

The real part of a complex function being harmonic is significant because it means that the function is differentiable and can be represented as a power series. This allows us to use techniques from calculus and complex analysis to solve problems involving the function.

5. Can you provide an example of a complex function whose real part is harmonic?

One example of a complex function whose real part is harmonic is f(z) = z^2 + 2z + 5. Its real part is u(x,y) = x^2 + 2x + 5, which satisfies Laplace's equation and is therefore harmonic.

Similar threads

  • Calculus and Beyond Homework Help
Replies
17
Views
1K
  • Calculus and Beyond Homework Help
Replies
27
Views
838
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
486
  • Calculus and Beyond Homework Help
Replies
19
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
970
Back
Top