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I understand the process my book took to get there.

They used the limit process from the "real" direction, and the "complex direction."

To me, it seems obvious that if $z(x,y)=x-iy$, $\displaystyle \frac{\partial z}{\partial x}=1$, and $\displaystyle \frac{\partial z}{\partial (iy)}=-1$.

However, the book uses this to justify why the function is not differentiable, because it approaches different values from different directions.

Clearly then, I am missing something. I am confident that this intuition is correct for multivariable functions, but it doesn't work for $f(z)=\bar{z}$?

Why can a function not have different derivatives in different directions?

I was thinking about this some more and I'm wondering if multivariable functions DO have the "same" derivative in any direction. For example, consider the function $f(x,y)=3x^2+6y^2$; it's derivative is $6x\,dx+12y\,dy$ in every direction.

Does that mean, that $\displaystyle \frac{d\bar{z}}{z}=\,dx-\,d(iy)$?

Can someone comment on my reasoning or explain the difference between multivariable derivatives and complex derivatives?

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    $\begingroup$ For a function to be complex differentiable, its various partial derivatives have to be related by the Cauchy-Riemann Equations. I've always understood the reason for this is because we want to consider limits like $f(z) / z$ as $z \to 0$, rather than $f(z) / |z|$ as $z \to 0$ as in normal calculus for functions $\mathbb{R}^2 \to \mathbb{R}^2$. $\endgroup$
    – Joppy
    Jun 18, 2019 at 2:28
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    $\begingroup$ It's differentiable as a real function, but not as a complex function: $\lim_{z\to0}\frac{f(z)-f(0)}{z-0}$ doesn't exist. $\endgroup$ Jun 18, 2019 at 2:28
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    $\begingroup$ A way of viewing it is that $z \mapsto \overline{z}$ is differentiable as a map from $\mathbb{R}^2$ to $\mathbb{R}^2$, in fact it is a linear map so its derivative is the matrix $\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$. But this operator is not represented by multiplication by any complex number, which is what is required for complex differentiability. The operators that are represented by a complex number are those of the form $\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$ for real numbers $a,b$ (this is the operator represented by multiplication by $a+bi$). $\endgroup$
    – Ian
    Jun 18, 2019 at 2:29
  • $\begingroup$ @LordSharktheUnknown so what's the difference b4etween differentiability and complex diferentiabilyt $\endgroup$ Jun 18, 2019 at 2:38
  • $\begingroup$ Your book does not give a definition of the derivative of a function $f(z)$ with respect to the complex variable $z$? What book is that??? $\endgroup$
    – bof
    Jun 18, 2019 at 4:29

2 Answers 2

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The basic reason is that $z \mapsto \bar{z}$ doesn't "locally look like the action of a complex number" - at any point.

To understand what that means, we need to actually better understand what is often meant by a "derivative" and, to do that, we must revisit first the simple real-number case, in much better depth. In the sadly all-too-common introductions to this topic, the derivative is introduced as being something like the "slope of a tangent line", and this often justified by imagining points on the graph of the function moving together and showing how when you do that, the line between them seems to approach another line, which is taken to be the tangent line, either as a definition, or as a realization of something talked about only in rather vague terms like a "line that 'touches' at a point" - and I've always found this approach deeply unsatisfying.

For one, why the heck are we doing this procedure? Why are we interested in this kind of line? And why do we call it "tangent" - after all, this word comes from a Latin root meaning "touch", and then we're back to this "touching" thing again, but what on Earth does that mean? Moreover, we then run into things like wondering why that a function like $x \mapsto |x|$ cannot be "differentiated" at $x = 0$, and why perhaps we could not consider it to have infinitely many "tangent lines" there. Or why that $x \mapsto x^3$ has a tangent line at $x = 0$ when that line cuts through the graph there, and doesn't simply "touch" it.

Yet, after much frustration with this, I eventually happened on what I think is a much neater, more intuitive way to imagine the derivative, and which also allows us to rigorously define the notion of a tangent line and derive or prove the derivative from it. That is, the tangent line comes first, the derivative, next.

And the way you do this is to consider the idea of magnification. (I'd like to put some pictures here, but unfortunately don't have access to any drawing facilities nice enough to really do this the justice it deserves.) "Magnification" is just that: you've probably also heard it called "zooming" or "zooming in" - think about, say, the zoom feature on a camera or, perhaps more germanely, the zoom feature in a computer image manipulation program like Photoshop. But, whereas zooming into a picture typically inevitably results in its becoming pixelated, what we're going to be considering here now is zooming into the ideal graph of a mathematical function, something which has infinite detail and infinite resolution - hence, it never, ever becomes pixelated.

To that end, suppose that you take a real function $f: \mathbb{R} \mapsto \mathbb{R}$. For an illustration, let's consider $f := \sin$. Technically, this function cannot "really" be defined without calculus, but meh, it's familiar even before then due to our logically broken way of teaching and, nonetheless, given the level of this question, should be familiar to its audience anyways even did we have a properly-structured way - just pointing it out because the derivative is a fundamental of calculus and given before the integral. We plot the graph in a square area - say with upper-left corner $(-5, 5)$ and lower-right corner $(5, -5)$ - which we call a window. The function looks like this. We will be considering the point $(0, 0)$ on the graph as our zoom-in point. The point is highlighted below. The graphing facilities I have are limited, but I hope this graph looks okay.

enter image description here

This is, of course, the typical sine wave pattern we all know and love. Let us, now, change the size of the window, so as to constrict it around that point. Take instead from UL $(-1, 1)$ to LR $(1, -1)$, so it is five times smaller, or a zoom of 5x. The point $(0, 0)$ is still centered, but now the graph has changed.

enter image description here

Note how it looks more linear. In fact, we can zoom in again - now the window has corners with coordinates $\pm 0.1$. The zoom factor is now 10 times more, or 50x, magnification.

enter image description here

Indeed, at this point, it is almost indistinguishable from a straight line. One more zoom to 500x looks like

enter image description here

effectively, the graph looks essentially exactly like that of a line with the equation $y = x$.

Moreover, were we to do this at any other point on the sine graph, we would find the same thing. The line would be through a different point, and perhaps with a different slope, but it would still end up looking like a line if we zoomed in far enough.

However, not all functions have this property. Consider the fresh calc bogeyman $x \mapsto |x|$. This puppdog is pictured below.

enter image description here

As you can see, it has a "V" shape, with the angle at $(0, 0)$. If we zoom into that point as before, we see this:

enter image description here

That is 500x zoom, but it wouldn't matter if we made it five googolplex. The graph, in fact, would look exactly the same: a "V" shape with a corner. Yet if we zoomed anywhere else, it would, again, look like a line.

In other words, only some functions at some points have the property they look like lines when zoomed in. Others, don't.

And it is this property that we call "differentiability". The line that is formed upon zooming in "by an infinitely large amount" is what we call the tangent line. That is, the tangent line is, effectively, an "infinitely small piece" of the graph of the function - blown up infinitely large, so as to be a perfect straight line. This idea is absolutely crucial to making sense of derivatives in all advanced contexts, including even the notion of "tangent spaces" in differential geometry: the tangent space is, effectively, in the same way, an "infinitely small piece" now of a differentiable manifold - a "surface" described wholly on its own terms, without reference to any exterior space in which it may or may not reside.

In fact, we can easily work with this to come up with a rigorous formal definition of the tangent line and then prove that its slope must be the derivative, as an honest theorem, not some out-of-the-hat postulate. Moreover, with such definition in hand, we can then prove that a function like $x \mapsto \sin(x)$ is differentiable but $x \mapsto |x|$ is not differentiable at $x = 0$: the tangent line doesn't exist there, not because there are infinitely many "touching lines", but because the graph never resolves into a line no matter how far we zoom in (heck, screw googolplex, try Rayo's number, the largest "officially" named number, as the zoom factor! It will still be a "V"-shape, even if we then have to "take maths' word for it" as we can't "calculate" that on any buildable computer). It is a very nice exercise to go through this; I won't bloat the post with more details here.

The same can be done now with a complex function, only here, it gets a bit more subtle. Because a complex function's graph "lives" in four dimensions (okay, if you want, two complex dimensions, but that is both also four real dimensions and moreover a space with topological dimension 4, hence requires 4 real dimensions to draw no matter how you slice it), we need to resort to a different method of picturing. In this case, the way we should picture such a function (and now I REALLY don't have the graphing facilities I'd need) as a transformation of the plane - stretching, twisting, bending, folding, etc. it. And in this view, we can "zoom in" to a small piece of the transformed plane in like manner, and ask how that it was transformed from the original.

And what complex differentiability means, now, in this case, is that the small piece of transformed plane "looks" like one that has been transformed only by rotation and uniform scaling - that is, effectively, the same thing as multiplication by a complex number.

This can be formalized as follows. Whereas a real-differentiable function in one dimension "magnified" to a line, a differentiable plane transformation, we expect, should "magnify" to a linear transformation. That is, near the point, we should be able to "capture" the behavior by some matrix

$$\begin{bmatrix} a_{00} & a_{01} \\ a_{10} & a_{11} \end{bmatrix}$$

acting on a plane column vector by left-multiplication of said vector, as usual.

Now, it turns out also that complex numbers can be represented as matrices: in particular, real 2x2 matrices of the form

$$\begin{bmatrix} a && -b \\ b && a \end{bmatrix}$$

when combined using matrix addition and multiplication, behave exactly like complex numbers $a + bi$. Moreover, when we apply them as linear transformations, they act as the complex multiplication does: scaling and rotating the plane.

Thus, the notion of complex differentiability is that the local linear transformation at each point of the differentiable function, looks like the action of a complex number. In particular, if a real plane transformation $f(\mathbf{r}) := (u(r_x, r_y), v(r_x, r_y))$ is differentiable, its "derivative" is given by the Jacobian matrix

$$J[f] := \begin{bmatrix} \frac{\partial u}{\partial x} && \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} && \frac{\partial v}{\partial y} \end{bmatrix}$$

and if we combine that with the matrix representation of a complex number, we see that the criterion for complex differentiability should be as follows:

  • From that the entries in the main diagonal ("$a$" in the illustration) must be the same: $$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$$
  • From that the entries in the anti-diagonal must be opposite: $$-\frac{\partial u}{\partial y} = \frac{\partial v}{\partial x}$$

Yet these are exactly the famous Cauchy-Riemann equations! And this is what their "true" meaning is. And what is the complex number that the Jacobian matrix represents? Well, it is the derivative - $f'(z)$!

And from this, right away we see why $z \mapsto \bar{z}$ is not complex-differentiable: geometrically, it is a reflection, and since the action of any complex number can only ever be a rotation plus a scaling at best, then there cannot be any complex number which represents the action of this function!

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    $\begingroup$ +1 for a very intuitive way to view the derivative and Cauchy-Riemann equations. I like this idea more than the “tangent line” idea; it generalizes much better. $\endgroup$
    – csch2
    Jun 18, 2019 at 4:41
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    $\begingroup$ this is definitely a lot to swallow, but i think i now undestand the difference between a function of two variables and ONE function of z $\endgroup$ Jun 18, 2019 at 6:26
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    $\begingroup$ @SakethMalyala : Well, technically, any function $f: \mathbb{C} \rightarrow \mathbb{C}$ "is a function of $z$". What, however, a holomorphic function has that generic functions of this type lack, is that in a certain sense they incorporate the *multiplicative* structure of the complex numbers - which is the quintessential feature that sets $\mathbb{C}$ apart from $\mathbb{R}^2$. $\endgroup$ Jun 18, 2019 at 6:36
  • $\begingroup$ @The_Sympathizer i think that just about did it. because f(x,y) cannot replicate the multiplication that involves in computing f(z) so they are not the same, and have different rules $\endgroup$ Jun 18, 2019 at 6:44
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    $\begingroup$ @Saketh Malyala : Yes, the complex limit is effectively "coming in from all directions". However, actually, I'd say, that's not the best way to think of it. The best way is to say that instead of dealing with a one-dimensional "confidence interval" around a point, we are dealing with a two-dimensional "confidence region", and as we shrink that region in the plane of function inputs around the point of approach, the range of outputs therein also shrinks to the limit value. $\endgroup$ Jun 18, 2019 at 6:50
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The derivative at point $z_2$ is defined as $$\lim_{z_1 \to z_2} \dfrac {f(z_1)-f(z_2)}{z_1-z_2}$$.

Notice that this is perfectly legitimate since for $z_1,z_2 \in \mathbb C$ and $z_1 \neq z_2$ we have that $\dfrac {f(z_1)-f(z_2)}{z_1-z_2} \in \mathbb C$.

So, by forming these quotients we do not leave the field of complex numbers.

In the definition of $$\lim_{z_1 \to z_2} \dfrac {f(z_1)-f(z_2)}{z_1-z_2}$$ it should be understood that a function is differentiable at $z_2$ if this limit is independent of the path that point $z_1$ travels in order to reach a point $z_2$, and this requirement is very strong.

So, if some function at some point has two different values of this limit on two different paths then the function is not differentiable at that point.

As an exercise, for a function you mentioned, function $f(x+yi)=x-yi$, calculate the limit at zero when $x+yi$ approaches zero on the path where $y=x$ and the limit at zero when $x+yi$ approaches zero on the path where $y=-x$, and you will see a two different values, so, $f(x+yi)=x-yi$ is not differentiable at zero.

In the real case, the derivative at a point is a measure of the rate of change of a function at that point, but in the complex case I do not know what the derivative really "measures" (since complex numbers have polar representation it could be that the derivative "measures" stretching and rotating at a point), but as a concept it is useful and generalization of the derivative in the real case.

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