Derivative of a complex function along different directions

[Swamp Thing]

TL;DR Summary

Plotted what should be an analytic function, as a function of real and imaginary part. The slopes are different. Why?

Below are plots of the function $e^{0.25(x-3)^{-2}} - 0.87 e^{(x-3.5)^{-2}}$
The first plot is for real values. It has a minimum at the red dot. The second plot has in its argument the same real part as the red dot, but has the imaginary part changing from -0.3 to 0.3. It shows the resulting change in the imaginary part of f(z).

For an analytic function (which I believe this one is) the derivatives around any point should be the same, approaching a point from the real and complex directions, but here obviously they are not. What have I got wrong?

 

[jasonRF]

You need to describe more carefully what you are actually plotting, because I cannot figure it out.

The first plot is for real values. It has a minimum at the red dot. The second plot has in its argument the same real part as the red dot, but has the imaginary part changing from -0.3 to 0.3.

If we do the usual thing and let $z=x+i\,y$ and $f(z)=u(x,y) + i\,v(x,y)$, where $x,y,u$ and $v$ are real, what exactly are you plotting in each of the two plots?

Then, once that is clear, you should write out explicitly what the derivative of $f$ is in each of the two directions, and see if you are actually plotting the relevant quantities.

jason

EDIT: also, you wrote your function as a function of $x$. Is $x$ a complex variable? Or is it the real part of a complex variable? Or neither? Also, whatever $x$ is, you need to think about what happens at $x=3$ and $x=3.5$. What do you think should happen at those locations?

 

[zinq]

"For an analytic function (which I believe this one is) the derivatives around any point should be the same, approaching a point from the real and complex directions, but here obviously they are not. What have I got wrong?"

I do understand at least this part of the original question. Yes — in fact that is the very definition of an analytic function: Let U be an open set in the complex plane. Then a function f : U → ℂ is analytic in U precisely when, for each z ∈ U, the limit as h → 0 of (f(z + h) - f(z)) / h exists.

Since h is complex, it can approach 0 in many different ways. But it's sufficient to prove that the above limit exists and is the same whether h approaches 0 through purely real values h = Δx, or through purely imaginary values h = iΔy. In fact, the equality of these two limits is equivalent to the truth of the famous Cauchy-Riemann equations.

And indeed any exponential function f(z) = exp(cz) and any complex linear combination of exponential functions is analytic.

 

[Swamp Thing]

Thanks for the responses. What I am trying to do is to eyeball the plots of Re( f(z) ) and I am ( f(z) ) as a function of z, and visually compare the slopes for real $\Delta z$ and imaginary $\Delta z$ around a point of interest. This comparison should be particularly easy at a min/max of the real part, where $Re(\Delta f)$ and $Im(\Delta f)$ are both zero.

My problem is with the plot of Im( f(z) ), where the $\Delta f$ is zero for real $\Delta z$ but non-zero for complex $\Delta z$ (or so it seems to me).

Here are 3D plots of the real and imaginary parts, with the changes from the point of interest shown along the real and imaginary directions (light blue and red respectively). The slope of the red curve in the second plot (imaginary part of f) is obviously different from the slope of the light blue one -- which is the cause of my confusion.

 

[jasonRF]

First, clearly your function is not the one as written since it has essential singularities at $x=3$ and $x=3.5$. Instead, I think it must be
$f(z) = e^{-0.25 \, (z-3)^2} - 0.87 \, e^{-(z-3.5)^2}$
Correct?

Given that, in your first post your first plot is clearly $u(x,0)$ and your second plot is $v(x0,y)$ for a value of $x0$ that is near the minimum of $u(x,0)$. I suspect you are picking an $x0$ that is not close enough to the true minimum of $u(x,0)$ to give you a zero slope for $v(x0,y)$ (note that your plot has a small slope, just not zero). Try using a more precise value for $x0$ and I suspect you will get the result you are expecting.

jason

 

[zinq]

For an analytic function is f(z) = u(x,y) + i v(x,y), the Cauchy-Riemann equations say that:

∂u/∂x = ∂v/∂y

and

∂u/∂y = -∂v/∂x

as functions of x and y. Are you sure that's not what you're seeing in your plots?

 

[Swamp Thing]

First, clearly your function is not the one as written

Yes, there should be negative signs in the exponents in the formula -- my bad.

I suspect you are picking an x0x0 that is not close enough to the true minimum of u(x,0)u(x,0) to give you a zero slope for v(x0,y)v(x0,y)

I plotted the function and its derivative, and sure enough, the point I chose is not the true minimum.
The reason for my error was that I wrongly expected that the complex zeros would have the same real part as the real minimum -- which is a rather silly thing to assume, I now realize. So I solved numerically for the complex zero and assumed that the minimum would also lie there.

Thanks.

The function and its derivative:

 

[zinq]

Also be aware that an analytic function f(z) never achieves a minimum of |f(z)| at an interior point, only on the boundary (if any) of its domain . (In fact, f(z) is always the average of the values f(z + r e^iθ), 0 ≤ θ ≤ 2π that it takes on a small circle about z.)

 

[jasonRF]

The reason for my error was that I wrongly expected that the complex zeros would have the same real part as the real minimum -- which is a rather silly thing to assume, I now realize. So I solved numerically for the complex zero and assumed that the minimum would also lie there.

Now I don't understand what you are doing. I thought you were simply making plots to illustrate or understand what zinq posted:

∂u/∂x = ∂v/∂y

by finding a point where $\partial u(x_0,0)/\partial x = 0$. You almost did that in your first post, but I believe you didn't use an accurate enough value for $x_0$ for the derivative to be zero. Wolram Alpha says $x_0$ should be near 3.67973. If you plot $v(3.67973,y)$ for $y$ near zero, you should see that the slope is zero.

jason

 

[Swamp Thing]

Now I don't understand what you are doing.

Yes, it's a bit hard to explain the incorrect thought process that led me to choose the wrong point at which to examine the derivative. I had started out by asking Mathematica to find the complex pair of zeros near the minimum. Then I chose the real part of that pair of zeros as my "red dot" point where I was going to look at the derivatives. This is because I had a sort of subconscious "hidden assumption" in my mind that this pair of complex roots would actually coincide exactly with the minimum point.

I now see that the complex roots should be close to the minimum but not necessarily coincide with it. Perhaps the offset between the true minimum and the complex roots has something to do with the mirror symmetry of the function around the minimum. For example, they should coincide exactly for a parabola that doesn't cross the x-axis.

In any case, your hint helped me to see why my derivative in the complex direction was non-zero around my chosen point, so thanks again!

 

[FactChecker]

Also be aware that an analytic function f(z) never achieves a minimum of |f(z)| at an interior point, only on the boundary (if any) of its domain . (In fact, f(z) is always the average of the values f(z + r e^iθ), 0 ≤ θ ≤ 2π that it takes on a small circle about z.)

The modulus can have an interior minimum. Any interior zero of the function would be one.

 

[zinq]

Yikes, that is absolutely true. I spaced out and wrote "minimum" when I should have said "maximum". And I neglected to state that when the analytic function is constant, of course, it takes its minimum and maximum value everywhere.

(It is true that for an open set U on which an analytic function f(z) is non-zero everywhere, |f(z)| does take its minimum value at a boundary point. The condition that f(z) ≠ 0 on U is equivalent to saying there is some analytic function g(z) defined on U such that f(z) = exp(g(z)) for all z ∈ U.)

 

[mathwonk]

I think you need to be careful about the topology of the open set on which you try to define g(z) as log(f(z)). Maybe you were thinking of U as a disc, or another simply - connected open set?

 

[zinq]

Let's see, say we want the logarithm of f(z) = 1/z on the punctured plane U = ℂ - {0}.

Then log(r exp(iθ) would be ln(r) + iθ. So sure enough, this cannot be defined as a single-valued analytic function g(z) on U, contrary to what I wrote in the last paragraph of #12.

(I wish I could edit my mistakes so people don't read them and accidentally believe them.)

It is true that there exists a Riemann surface, the universal cover U^~ of U, on which log(z) is defined as a single-valued analytic function.

I think a correct statement is this: "If f is analytic on the open set U of ℂ and f(z) ≠ 0 for all z ∈ U, then on the universal cover U^~ of U (with covering projection π : U^~ → U) there exists a (single-valued) analytic function

g : U^~ → ℂ

such that for all w in U^~

f(π(w)) = exp(g(w))."

 

[mathwonk]

Yes, I agree, (because the universal cover is simply connected). My post has an "edit" button at the bottom left, so I assume yours does too.

 

[zinq]

I have edit buttons only for a short time after posting; after that, my edit buttons disappear.