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I have trouble understanding a whole array of things in complex analysis, which I have basically tracked to the statement "real and imaginary parts of a complex analytic function are not independent."

Because of that, I don't really understand the Cauchy-Riemann equations, the fact that for an analytic function, if its real part is constant, then the whole function is constant, and other fundamental things, such as Cauchy's Integral formula, Maximum modulus principle, etc. (the last two just make zero sense to me.)

The thing is, I pretty much understand the proofs, starting from the beginning, when we define differentiability of a complex function. I don't have any problems with the introduction of complex numbers as well, and different identities.

But I just don't have any intuition for why things are like that, and it's very frustrating, because I always feel like I don't understand complex numbers at all, and just do some standard exercises in class, relying on proven facts that I just assume to be true as a starting point.

But as soon as I go and try to understand the meaning of things we in the class work with, I just immediately stop understanding anything.

Can anyone help me understand why the real and imaginary parts of a complex function are not independent?

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    $\begingroup$ you mean when the complex function is differentiable? $\endgroup$ – J. W. Tanner Oct 29 '19 at 19:36
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    $\begingroup$ $\Bbb R^2$ is not a field, but $\Bbb C$ is. A '2D-field' is a really strong property, and, in fact, there are no more field extensions of $\Bbb R$ aside the trascendental ones like $\Bbb R(X)$. If Cauchy - Riemann equations astonish you, you should take a look at strong Picard's theorem... These properties of $\Bbb C$ make it really beautiful, if you like maths. I also felt shocked when I learnt these facts some years ago, but you will get used to them on time :) $\endgroup$ – ajotatxe Oct 29 '19 at 19:50
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    $\begingroup$ Let us suppose the "real and imaginary parts of a complex analytic function are independent". Let $f(z) = u(z) + \mathrm{i}v(z)$ be a complex analytic function with $u$ and $v$ both real valued. Could you append to your Question an example of computing the real and imaginary parts of $2f$ and $\mathrm{i}f$ that exhibits this independence? $\endgroup$ – Eric Towers Oct 30 '19 at 11:43
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    $\begingroup$ The Cauchy-Riemann equations are a special case that for a limit to be defined for a point in a multivariate function, it must be the same no matter along which trajectory we approach this point. $\endgroup$ – mathreadler Oct 30 '19 at 12:26
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    $\begingroup$ I felt many of the same frustrations, complex analysis is often taught from a formal analytical perspective and the geometric side is not mentioned. I recommend the book Visual Complex Analysis by Tristan Needham. A lot of the things you mention have great intuitive explanations. $\endgroup$ – Amos Joshua Oct 31 '19 at 15:16
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It's really just a question of the definition of the derivative. If $z=x+yi,$ $f(z)=u(x,y)+iv(x,y)$ can be any pair of functions $u,v.$

But if $f$ is differentiable, then:

$$f'(z)=\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}\tag{1}$$

then $h$ can approach $0$ in many different ways, since $h$ is complex.

For example, you can have $h\to 0$ on the real line. Then: $$f'(z)=\frac{\partial u}{\partial x} +i\frac{\partial v}{\partial x}$$

But if $h\to 0$ along the imaginary part, then:

$$\begin{align}f'(z)&=\frac{1}{i}\left(\frac{\partial u}{\partial y}+i\frac{\partial v}{\partial y}\right)\\ &=\frac{\partial v}{\partial y}-i\frac{\partial u}{\partial y} \end{align}$$

So for the limit to be independent of any path you take $h\to 0$ you must have at minimum that $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},\\\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}\tag{2}$$

So for (1) to be true, we need $u,v$ to satisfy the differential equations in (2).

It turns out that $(2)$ is enough to ensure that $(1)$ converges to a single value, but that is not 100% obvious.

The equations in (2) are called the Cauchy-Riemann equations.


Another way of looking at it is, given a function $f:\mathbb R^2\to\mathbb R^2$ mapping $\begin{pmatrix}x\\y\end{pmatrix}\mapsto \begin{pmatrix}u(x,y)\\v(x,y)\end{pmatrix}$ there is a matrix derivative standard from multi-variable calculus:

$$Df\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}\\\frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}\end{pmatrix}\tag{3}$$

For small vectors $$\mathbf h=\begin{pmatrix}h_1\\h_2\end{pmatrix}$$ you get $f\left(\begin{pmatrix}x\\y\end{pmatrix}+\mathbf h\right)\approx f\begin{pmatrix}x\\y\end{pmatrix}+Df\begin{pmatrix}x\\y\end{pmatrix}\mathbf h.$

In particular, $Df$ is in some sense the "best" matrix, $\mathbf A,$ for estimating $f(\mathbf v+\mathbf h)\approx f(\mathbf v)+\mathbf A\mathbf h.$

Now, these matrices are not complex numbers. But an interesting fact is that the set of matrices of the form:

$$\begin{pmatrix}a&-b\\b&a\end{pmatrix}\tag{4}$$

are a ring isomorphic to the ring of complex numbers. Specifically, the above matrix corresponds to $a+bi.$

We also have that:

$$\begin{pmatrix}a&-b\\b&a\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}= \begin{pmatrix}ax-by\\bx+ay\end{pmatrix}$$

compare that with:

$$(a+bi)(x+yi)=(ax-by)+(ay+bx)i.$$

So these matrices (4) act on $(x,y)^T$ the same way that $a+bi$ acts on $x+yi$ by multiplication.

The Cauchy-Riemann equations (2) just mean that $Df\begin{pmatrix}x\\y\end{pmatrix}$ is an example of (4) - that is, when the Cauchy-Riemann equations are true for $u,v$ then the multi-variate derivative (3) can be thought of as a complex number.

So we see that when we satisfy Cauchy-Riemann, $Df\begin{pmatrix}x\\y\end{pmatrix}\cdot\mathbf h$ can be seen as multiplication of complex numbers, $f'(z)$ and $h=h_1+h_2i.$ Then you have:

$$f(z+h)\approx f(z)+f'(z)h.$$

where $f'(z)$ is not just the best estimating complex number for this approximation, but also $f'(z)$ is the best linear operation on $h$ for this estimation.

So complex analysis is taking the vector function and asking, $f$ "when does it make sense to think of the derivative of the $\mathbb R^2\to\mathbb R^2$ as a complex number?" That is exactly when Cauchy-Riemann is true.

In the general case $f:\mathbb R^2\to\mathbb R^2,$ we can't really take the second derivative and get an estimate $f(z+h)\approx f(z)+Df(z)\cdot h +\frac{1}{2}D^2f(z)\cdot h^2+\cdots.$ We can't get easy equivalents to power series approximations of $f.$

But when $Df$ satisfies Cauchy-Riemann, we can think if $Df$ as a complex-valued function.

So complex analysis is a subset of the real analysis of functions $\mathbb R^2\to\mathbb R^2$ such that the derivative matrix $Df$ can be thought of as a complex number. This set of functions turns out to have a lot of seemingly magical properties.

This complex differentiability turns out to be fairly strong property on the functions we study. The niceness of the Cauchy-Riemann equations gives up some truly lovely results.

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    $\begingroup$ My feeling is that Nick the Dick already knows that. I think that he needs an intuitive explanation about the 'strange' properties of $\Bbb C$ because he thinks at it as $\Bbb R^2$. $\endgroup$ – ajotatxe Oct 29 '19 at 20:10
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    $\begingroup$ The difference comes in that you can actually take quotients in $\Bbb C$ where you cannot in $\Bbb R^2$ a priori, making derivatives in $\Bbb C$ very different from derivatives in $\Bbb R^2$---which involve dividing by the norm of a vector $\vec{h}$ instead of a plain complex number $h$ as with $\Bbb C$-derivatives. $\endgroup$ – Cameron Williams Oct 29 '19 at 20:12
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    $\begingroup$ @ajotatxe: My feeling is that Nick does not have the right starting point. If you start from the definition of "complex differentiable" that Thomas gives in (1), then you are naturally forced to conclude the CR equations must hold, and it is obvious that the real and imaginary parts are 'dependent' since they are just parts of the whole complex value. It should be clear that asking why they are not independent is just like asking why the $x$ and $y$ coordinates of a point moving along a smooth curve are not independent... $\endgroup$ – user21820 Oct 30 '19 at 6:29
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    $\begingroup$ @mbrig Sorry, I can't give a good reference. Any rigorous intro to complex analysis should include a proof, but the class I took on complex analysis was 36 years ago. $\endgroup$ – Thomas Andrews Nov 1 '19 at 20:03
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    $\begingroup$ @asmaier there are two ways in my answer. In the first way, you’d need a notion of division by $h$ where $h$ is a vector of three or more dimensions. There is a way to do that in four dimensions (using quaternions) but I don’t know whether you can do anything with that. The second way requires a notion of multiplying vectors of three dimensions or more, which you can only do with $4$ dimensions, again. I don’t know if it is useful. (It’s possible you can do 8 dimensions with octonians in both cases, but octonian multiplication is non-associative.) $\endgroup$ – Thomas Andrews Nov 1 '19 at 21:14
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The most sensible way, I believe, to understand this is that differentiation characterizes the micro-local behavior of a function at a point. In particular, one of several interpretations of the derivative of any function $f$ at some input point $x_0$ is that in a suitably tiny region around that point, $f$ "acts" like (up to some shifts to get things suitably centered) a multiplication by $f'(x_0)$.

When $f$ is a complex function, then complex differentiability means that it must act like multiplication by a complex number, namely the complex derivative $f'(z_0)$ for a now complex test point $z_0$. And multiplication by a general complex number scrambles together the real and imaginary parts of the number so multiplied.

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Since the independence was nicely addressed in the accepted answer, I address the relationship between the Cauchy integral formula and the maximum principle first. (Below the line there's a very similar explanation for why real and imaginary parts are not independent)

The Cauchy integral formula intuitively states that the value at a point is the average of the values in a circle around a point, weighted somehow according to their distance (and angle). This is not easy to see directly from the Cauchy-Riemann equations (i.e., linearity of the differential). It is quite a deep theorem that $f$ is holomorphic if and only if $f$ satisfies the Cauchy integral formula (the proof usually involves this cycle: holomorphic $\Rightarrow$ integral formula $\Rightarrow$ power series $\Rightarrow$ holomorphic). The nicest proof of the Cauchy integral formula I have seen so far just uses homotopy invariance and the fact that integrals along contractible curves are zero, but this is a story for another question I guess. Let me just remark that the Cauchy integral formula is true more generally, e.g.,

  1. The Mean Value Equality for harmonic functions
  2. The Stokes Equation for differential forms on smooth manifolds with boundary

Both of these results may offer geometric insight when the time comes. Perhaps it is best for now to think about "functions satisfying the Cauchy integral formula" just as "some class of functions that has this mean value property", knowing in the back of your head that you will eventually understand that these functions are actually the same as "functions that have $\mathbb C$-linear differential".

Believing in the Cauchy integral formula has the advantage of making the maximum principle seem intuitive: If the function is the weighted average of itself at all the circles around it, it cannot get bigger inside these circles than it is on the circles. If you look closely, this is probably basically the argument in your proof of the maximum principle.


Recall that the differential of a (real-)differentiable function $f: \mathbb R^2 \rightarrow \mathbb R^2$ at a point $x \in \mathbb R^2$ is the $\mathbb R$-linear map $Df_x: \mathbb R^2 \rightarrow \mathbb R^2$ that approximates $f$ best at the point $x$. You might have seen this as $$\lim_{h \rightarrow 0}\frac{f(x + h) - f(x) - Df_x(h)}{\Vert h \Vert} = 0.$$ This could be visualised by saying that every (real-)differentiable function $f: \mathbb R^2 \rightarrow \mathbb R^2$, after zooming in enough, looks like a linear transformation. Maybe check out some visualizations by 3blue1brown if you do not have a concrete picture in mind.

Now consider a holomorphic function $f: \mathbb C \rightarrow \mathbb C$. In this case, the differential of $f$ at $z$ is the $\mathbb C$-linear map $Df_z: \mathbb C \rightarrow \mathbb C$ that approximates $f$ best at $z$. Emphasis lies on the fact that $Df_z$ is $\mathbb C$-linear this time, i.e., $Df_z$ is just multiplication by some complex number denoted $f'(z)$.

Now comes the important observation: Let $a \in \mathbb C$ be a complex number. Then the map $z \mapsto az$ is given by stretching and rotating, but not shearing. As a map $\mathbb R^2 \rightarrow \mathbb R^2$, it has the form $$\begin{pmatrix} x & -y \\ y & x \end{pmatrix},$$ where $a = x + iy$ if that helps you to visualize it (otherwise, please ignore the matrix).

Why is this important? This shows, that when zooming in, $f$ looks like stretching and rotating. But, as you may gather from the above matrix, from intuition or just by blind faith, if you know what stretching and rotating does to one vector, then you also know what it does to all other vectors. This is why real and imaginary parts are not independent.

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    $\begingroup$ "[...] when zooming in, $f$ looks like stretching and rotating. But [...] if you know what stretching and rotating does to one vector, then you also know what it does to all other vectors." Very nice! $\endgroup$ – user21820 Oct 30 '19 at 12:53
  • $\begingroup$ @EricWofsey I agree with you in the case that the center of the disk bounded by the circle in question agrees with the point you are applying the formula to. But if you choose a point that is not the center, then this cannot be true, since otherwise the function would be constant inside the disk. $\endgroup$ – Levi Nov 1 '19 at 15:42
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    $\begingroup$ Yup, the root of all this is your "important observation": A complex number's action on the plane is to rotate and scale the plane. The complex numbers "are" rotations and scalings, and differentiable mappings between them have to respect this structure. $\endgroup$ – nomen Nov 1 '19 at 23:10
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Consider the linear function,

$$ f(z) = m z + b.$$

The transformations of the complex plane which can be written this way include only the following:

  1. Rotations.
  2. Translations.
  3. Dilations.

The Cauchy-Riemann equations are the conditions that a function $f(z)$ must meet so that locally its transformations are a combination of the above transformation. If a function doesn't obey the Cauchy-Riemann equations then it may introduce some shear or change angles between curves at points of intersection.

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As others have said, one way to look at why the real and imaginary parts of a complex differentiable function are not independent is because the derivative of such a function is required to be a complex number.

Geometrically complex numbers represent combinations of rotations and scaling, and if you start modifying the real part independently of the imaginary part, the derivative (which still exists as a vector function) may no longer represent a rotation-scaling combination and therefore cannot be written as a complex number, and therefore the function is not complex differentiable at that point.

Complex analysis can be very geometrical and intuitive. As an aside I highly recommend the book Visual Complex Analysis by Tristan Needham.

So let’s try an example. Consider the function $f(z) = z^2$. This is a complex differentiable function which sends the complex plane $\mathbb{C}$ onto itself with a twist. If we consider the point $p=i$, it’s clear that $f$ sends $p$ to $f(p) = -1$:

enter image description here

Now let’s consider a small piece of the complex plane around $i$ (here shown in blue). I’m going to call this a neighbourhood. Since $f$ is continuous it sends this neighbourhood of $i$ to a (possibly deformed) neighbourhood of $f(i)$:

enter image description here

Questions about whether $f$ is differentiable, or complex differentiable amount to how $f$ deforms this blue blob. If we want $f$ to be just regular differentiable (not complex differentiable) then $f$ has to transform this blob using a linear transformation - any old linear transformation will do. This is a stronger condition than simple continuity but it’s not that strong (relatively speaking). But if we want it to be complex differentiable (a requirement for being complex analytical), then it has to transform this blob using only a rotation and a scaling transformation - this is a very strong condition.

So let’s take a closer look at this. The derivative of $f$ is $f’(z) = 2z$, which at $p = i$ gives us $f’(p) = 2i$.

What does it mean to say that the derivative of $f$ at $p$ is $2i$ ?

The derivative at $p$ describes how the blue blob around $p$ is sent to a blue blob around $f(p)$. the complex number $i$ represents a 90 degree counter-clockwise rotation, and so in this case we’re saying that $f$ deforms the blob around $p$ by rotating it by 90 degrees and scaling it by 2 (zooming in so everything is twice as big). It looks something like this (where we zoomed in really really close to $p$ on the left and $f(p) = -1$ on the right):

enter image description here

So far so good. But why can’t we change the real and imaginary parts independently?

To get an intuitive understanding of this, and to make the real and imaginary parts more visible, let’s look at $f$ not as a complex function, but as a function of the euclidean plane $\mathbb{R}^2$ into itself. It’s important to understand that complex numbers can be viewed as an “extra” feature that exists on top of the regular 2D euclidean plane. So any time you have a complex function, you also have a regular 2D function and it’s useful to be able to jump between the two representations. In our case If we write:

$$z = x + i y$$

Then we can write $f$ (still as a complex function):

$f(x + iy) = (x + iy)^2 = x^2 + 2i xy - y^2 = (x^2 - y^2) + 2 xy i$

So as a function of vectors from $\mathbb{R}^2$ to $\mathbb{R}^2$ $f$ can be written as:

$f(\begin{bmatrix}x\\y\end{bmatrix}) = \begin{bmatrix}x^2 - y^2\\2xy\end{bmatrix}$

Viewed as $\mathbb{R}^2$ $i$ is $\begin{bmatrix}0\\1\end{bmatrix}$ and we can confirm that $f(\begin{bmatrix}0\\1\end{bmatrix}) = \begin{bmatrix}-1\\0\end{bmatrix}$ which corresponds to $-1$ in the complex plane.

Now what is the derivative of f in this picture? It’s the jacobian matrix:

$Df_{(x,y)} = \begin{bmatrix}\frac{\partial f_x}{\partial x}&&\frac{\partial f_x}{\partial y}\\ \frac{\partial f_y}{\partial x}&&\frac{\partial f_y}{\partial y}\end{bmatrix} = \begin{bmatrix}2x&&-2y\\2y&&2x\end{bmatrix}$

This is like the derivative in 1D in that it tells us how $f$ changes near a point, but instead of a number it’s a matrix - or well, it’s a formula for a matrix at any given point $(x, y)$. We agreed that $i$ corresponds to $x = 0, y= 1$ so at this point we get this specific matrix

$Df_{(0,1)} = \begin{bmatrix}0&&-2\\2&&0\end{bmatrix} = 2 \begin{bmatrix}0&&-1\\1&&0\end{bmatrix}$

This matrix tells us how the blue blob around $\begin{bmatrix}0\\1\end{bmatrix}$ is transformed into the blue blob around $\begin{bmatrix}-1\\0\end{bmatrix}$.

Note that the matrix

$\begin{bmatrix}0&&-1\\1&&0\end{bmatrix}$

Represents a 90 degree rotation, so again we see that $f$ transforms the blob around $p$ by rotating by 90 deg and scaling by a factor of 2 (this matrix can be seen as the matrix version of the imaginary number i). And in fact it had better be the case, because $f$ is the same function! We’re just writing it down in two equivalent ways, and if they showed us two different behaviours we’d be in trouble.

So now let’s try modifying the real part and see where we lose complex differentiability. Suppose we modify the real part from $2xy$ to $xy$ and we call this new function $g$:

$g(x + iy) = (x^2 - y^2) + i (xy)$

Note that this is a perfectly valid function from the complex plane to itself. It’s not complex differentiable (as we’ll see in a moment), but it does map $\mathbb{C}$ to $\mathbb{C}$. Now let’s forget the “extra” complex “structure” of $g$ and write it as a function of $\mathbb{R}^2->\mathbb{R}^2$:

$g(\begin{bmatrix}x\\y\end{bmatrix}) = \begin{bmatrix}x^2 - y^2\\xy\end{bmatrix}$

The derivative is:

$Dg_{(x,y)} = \begin{bmatrix}2x&&y\\-2y&&x\end{bmatrix}$

And at $\begin{bmatrix}0\\1\end{bmatrix}$:

$Dg_{(0,1)} = \begin{bmatrix}0&&1\\-2&&0\end{bmatrix}$

So far so good. Again, note that there is nothing wrong with this derivative - g is differentiable as function from $\mathbb{R}^2$ to $\mathbb{R}^2$. Now let's try to jump back into the complex picture and write this derivative as a complex number. To help us let's visualize what this matrix does around $p$:

enter image description here

Yikes, we're stuck! See how one axis is elongated but the other isn’t? This cannot be represented as a complex number - it cannot be achieved by a rotation followed by scaling because rotating and scaling always affect both axes equally (imagine rotating and scaling a photo in photoshop, but where you're not allowed to change the height without changing the width or vice-versa). This means the derivative of $g$ does not correspond to a complex number. So $g$ is not complex-differentiable at this point - we started with a complex differentiable function, modified just the imaginary part, and we "broke" the complex differentiability, which shows some form of dependence between the real and imaginary parts.

This is a very specific example, but the same intuition applies generally. Since locally (when we zoom in very close to a point) a complex differentiable function is only allowed to rotate and scale it places strong constraints on the real and imaginary parts. Of course you could have seen that with the Cuachy-Riemman equations more quickly, but that skips a lot of the intuitive geometry. And actually a lot of crazy-sounding phenomena in complex analysis can be traced back to geometric explanations such as this one.

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While others have already given excellent detailed answers, I want to contribute a more "basic" idea, which hopefully can help you.

The special thing about complex analysis as compared to real analysis is that you have multiple "directions" from which you can approach a certain point (remember that a derivative says something about the vicinity of a point in a certain direction). The fact that you cannot only approach the point from the purely imaginary axis or the purely real axis, but from all directions in between imposes certain conditions on the function and thus interdependence between real and imaginary parts. All these derivatives have to exist and they have to "smoothly" blend into each other - somewhat like your right-side derivative and left-side derivative have to correspond in real analysis for the function to be considered differentiable at this point.

In a way, this concept of "no matter from which side I approach the point" HOLDING FOR EVERY POINT is a very strong requirement for a function, which means that if these requirements hold (=function is complex differentiable) this allows us to deduce surprisingly strong statements about the properties of such a function.

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    $\begingroup$ I might be confused, but isn't this also true of differentiable functions $\mathbb R^2 \to \mathbb R^2$? $\endgroup$ – Mees de Vries Nov 1 '19 at 9:33
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    $\begingroup$ @MeesdeVries in the definition of Derivative of a function from $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ we divide by the norm of the vector $h$, since there is no meaning of division by $h$. But the difference in $\mathbb{C}$ is that you can divide by $h$ itself (when $h$ is not zero.). This is possible because $\mathbb{C}$ has a field structure. This is the reason why the Complex differentiability is a very strong condition. I hope you understand the point. $\endgroup$ – S.Sundara Narasimhan Nov 6 '19 at 9:22
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    $\begingroup$ I don't think that is usefully expressed as "derivatives approaching from every direction", though: differentiable functions $\mathbb R^2 \to \mathbb R^2$ also have "derivatives in every direction", not just along the two standard axes. $\endgroup$ – Mees de Vries Nov 6 '19 at 9:43
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Other people have answered this question very nicely. Here is another way of saying the same things again, in a more geometric way.

When you differentiate a function $f:\mathbb{R}\to\mathbb{R}$, you're trying to approximate its graph by a straight line.

When you differentiate a function $f:\mathbb{R}^2\to\mathbb{R}^2$, you're trying to approximate its graph (which now lives in $\mathbb{R}^2\times\mathbb{R}^2=\mathbb{R}^4$) by a plane. Indeed, the two components of the function are completely independent, i.e. the projections of this plane to the two copies of $\mathbb{R}^3$ (that you get by forgetting one of the two components of the target) are completely independent of one another.

When you differentiate a function $f:\mathbb{C}\to\mathbb{C}$, you're trying to approximate its graph by a complex line. Complex lines in $\mathbb{C}^2$ are, in particular, planes in $\mathbb{R}^4$, but they are a restricted class of planes (not every 2-plane is a complex line). This restriction is precisely expressed by the Cauchy-Riemann equations.

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I think you first need to see why the real and imaginary parts of a complex number are not independent -- indeed could not be dependent upon each other. Then why the same goes for functions follows immediately.

So why is it that the real and imaginary parts of a complex number have to depend on each other? The simple reason is that both coordinates are necessary to define the complex number, since this is perforce two-dimensional.

Geometrically speaking, we do need two numbers to define a point in the plane. Similarly, we need two displacements to define a vector in the plane.

Since these coordinates define a single object, it is clear that they have to be related. That is indeed why they are called co-ordinates -- they together order the point to which they refer. Suppose that there is no relationship between these numbers, then they obviously couldn't be referring to the same object -- I hope you see this now.

So the parts of a complex number are perforce related since they refer to the self-same object.

Now to functions. The parts of a complex function in rectangular form must also be related since they they refer to a single object. This should now be clear.

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    $\begingroup$ Could you elaborate on how the situation in $\mathbb C$ is different than in $\mathbb R^2$? $\endgroup$ – Levi Oct 30 '19 at 7:46
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    $\begingroup$ @Levi I didn't refer to $\mathrm R^2$ at all; as OP desired, I was talking about the complex plane. Of course everything I've said also applies to points in $\mathrm R^2,$ but not only that -- for points in any space, and even more broadly for any object defined by more than one object or property. OP wanted to know why the parts of a complex expression in rectangular coordinates are not independent. I have given the most basic reason why this is so -- I have not explicitly stated what these relationships may be, but this is not necessary to see why the coordinates have to be related. $\endgroup$ – Allawonder Oct 30 '19 at 8:00
  • $\begingroup$ You are wrong about this. Read a book about analytic (holomorphic) functions! $\endgroup$ – Lehs Nov 15 '19 at 21:09
  • $\begingroup$ @Lehs What exactly is this this that I am wrong about (sic), especially since this is a long dead post whose original context you're probably ignorant of? $\endgroup$ – Allawonder Nov 15 '19 at 21:59
  • $\begingroup$ @Allawonder - for example, holomorphic functions are all about the derivative and you didn't even mention this essential part. $\endgroup$ – Lehs Nov 16 '19 at 10:38

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