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

[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

 

[FactChecker]

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.

 

[fatpotato]

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.

 

[BvU]

(out of sheer curiosity) I wondered about

the complex valued function

because when looked at the expression I thought it's a real-valued function
$$\begin{align*} f(z) = Re\left(\frac{\cos z}{\exp{z}}\right) &= \frac{1}{2} Re\Biggl( \frac{\exp{iz} + \exp{-iz}}{\exp{z}} \Biggr) \\ &\ \\
&=\frac{1}{2} Re\Biggl( \exp(-z+iz) + \exp(-z-iz) \Biggr) \\ &\ \\
&=\frac{ \exp(-z+iz) + \bigl(\exp(-z+iz)\bigr)^* }{2}\end{align*}\\
$$

 

[fatpotato]

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 :

$$\begin{align*} f(z) = Re\left(\frac{\cos z}{\exp{z}}\right) &= \frac{1}{2} Re\Biggl( \frac{\exp{iz} + \exp{-iz}}{\exp{z}} \Biggr) \\ &\ \\
&=\frac{1}{2} Re\Biggl( \exp(-z+iz) + \exp(-z-iz) \Biggr) \\ &\ \\
&=\frac{ \exp(-z+iz) + \bigl(\exp(-z+iz)\bigr)^* }{2}\end{align*}\\
$$

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?

 

[BvU]

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 ...

 

[fatpotato]

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...

 

[BvU]

You are absolutely correct ! on me for not checking before posting the comment.