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I'm sorry if I sound too ignorant, I don't have a high level of knowledge in math.

The function $f(z)=z^2$ (where $z$ is a complex number) has a derivative equal to $2z$.

I'm really confused about this. If we define the derivative of $f(z)$ as the limit as $h$ approaches $0$ (being $h$ a complex number) of $(f(z+h)-f(z))/h$, then clearly the derivative is $2z$, but what does this derivative represent??

Also, shouldn't we be able to represent a complex function in 4-dimensional space, since our input and output have 2 variables each ($z=x+iy$) and then we could take directional derivatives...right?

But if we define the derivative as above, it would be the same if we approach it from all directions. That's what's bothering me so much.

I would really appreciate any explanation. Thanks!

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  • $\begingroup$ What's the question again? I am truly sorry but I don't quite understand what you were trying to ask. $\endgroup$ – BigbearZzz Aug 22 '16 at 21:29
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    $\begingroup$ The short answer is that complex analytic functions are really special compared to general differentiable functions $\mathbb{R}^2\to\mathbb{R}^2$. They are conformal (angle preserving), which most functions $\mathbb{R}^2\to\mathbb{R}^2$ are definitely not. $\endgroup$ – Harald Hanche-Olsen Aug 22 '16 at 21:30
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    $\begingroup$ @Harald well he didn't ask about why holomorphic functions are analytic though $\endgroup$ – Ant Aug 22 '16 at 21:31
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    $\begingroup$ @BigbearZzz. What I don´t get is how is it possible that complex function have just one derivative, in the sense that it does not matter where you approach them. $\endgroup$ – Leo Aug 22 '16 at 21:33
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    $\begingroup$ There is a very natural geometrical interpretation of the complex derivative, in terms of rotations and scaling; that is, the derivative just approximates your function by a linear function. If you want more detail, go read Needham's Visual Complex Analysis, it has very pretty pictures. $\endgroup$ – Javier Aug 23 '16 at 2:27
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If we define the derivative of $f(z)$ as the limit as $h$ approaches $0$ (being $h$ a complex number) of $(f(z+h)−f(z))/h$...

That's precisely what we do.

then clearly the derivative is $2z$, but what does this derivative represent??

Well, it represents what we have defined: the limit of the incremental ratio, same as in the real case. Probably you're wondering if we can interpret this complex derivative geometrically or visually, as we interpret the (real) derivative as the slope of the tangent line... There is no such a simple pictorial interpretation.

Also, shouldn't we be able to represent a complex function in 4-dimensional space, since our input and output have 2 variables each ... and then we could take directional derivatives...right?

Indeed, you could consider each of the two components (real and imaginary) separatedly, both for the variable and for the function, and then you'd get four (real) derivatives. And, yes, it's natural to ask how these four derivatives are related to the complex derivative, and if there are some (necessary and/or sufficient) restrictions on them so that the complex derivative gives the same value no matter the "direction" (as one would want)... Behold the Cauchy–Riemann equations and holomorphic functions.

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  • $\begingroup$ Ok! that was a pretty good explanation, I´ve seen the Cauchy-Riemman equations, but I didn´t understand them. Thank you! $\endgroup$ – Leo Aug 22 '16 at 22:20
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    $\begingroup$ Maybe we can't draws a picture of it, but we can locally linearly approximate the function in the same way we do for $\Bbb{R}$ functions: $f(z) \approx f(z_0) + f'(z_0)(z - z_0)$, or, with $f(z) = z^2$, $z^2 \approx z_0^2 + (2z_0)(z-z_0)$. This is implicit in OP's question and your answer, but is worth mentioning because it is an easy to see parallel between the two settings. $\endgroup$ – Eric Towers Aug 23 '16 at 15:06
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A complex-valued function $f$ is differentiable if near an arbitrary point $z_{0}$, the function $f$ acts like multiplication by a complex number $f'(z_{0})$ in the sense that $$ f(z_{0} + h) = f(z_{0}) + h\, f'(z_{0}) + o(|h|). $$ Since multiplication by a non-zero complex number $$ \alpha = |\alpha| \exp(i\theta) $$ rotates the plane (about the origin) through an angle $\theta$ and scales the plane (about the origin) by a factor $|\alpha|$, a complex-differentiable function is properly conformal (preserves both orientation and angle) at each point where the derivative is non-zero.

You're correct that drawing the graph of a complex-valued function of one complex variable entails four-dimensional plotting. Drawing the image of a Cartesian grid, however, is perfectly feasible in the plane:

A complex-differentiable map A smooth, non-holomorphic map

Each map above is defined by a pair of real cubic polynomials; the red curves are the images of vertical lines; the blue curves are images of horizontal lines. The maps differ in the sign of one component of the cubic term. One map is complex differentiable (or holomorphic), one is not.

The holomorphic map sends a square grid to an "infinitesimally square grid". (There is one critical point, whose image is visually apparent.) The derivative $f'(z_{0})$ at a point determines the size of the image (quasi-)square and the angle of rotation compared to the original Cartesian square.

The non-holomorphic map sends a square grid to a grid of curvilinear parallelograms that are not generally squares. This is the geometric manifestation of the directional derivatives at $z_{0}$ depending on the direction of approach to $z_{0}$: The Cartesian grid in the domain "gets stretched and rotated by differing amounts in different directions at a single point."

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    $\begingroup$ I like this answer, but you should clarify that the holomorphic map is on the left, and non-holomorphic on the right. $\endgroup$ – Daniel R. Collins Aug 23 '16 at 16:25
  • $\begingroup$ How did you plot the deformed Cartesian grid? $\endgroup$ – ablmf Jan 17 '18 at 13:49
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Let $w=f(z) = z^2$. Then $f(2+i)= 3+4i$ and $\dfrac{dw}{dz} = f'(z) = 2z$ and $f'(2+i) = 4+2i$.

The means an infinitely small change in $z$, from $2+i$ to $2+i+dz$, has a corresponding infinitely small change in $w$, from $3+4i$ to $3+4i + dw = 3+4i + (4+2i)\,dz$.

The change $dz$ gets multiplied by $4+2i$. Geometrically, what is a multiplication by $4+2i$? We have $$ 4+ 2i = \sqrt{4^2 + 2^2}\ \left( \cos\theta+i\sin\theta \right) \text{ where } \theta = \arctan\frac 2 4. $$ Thus, multiplying $dz$ by $4+2i$ amounts to rotating $dz$ through that angle $\theta$ and multiplying its absolute value by $\sqrt{4^2+2^2} = 2\sqrt{5} \approx 4.472.$

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The complex derivative shouldn't be thought of as representing a "gradient" as such, but rather as a set of instructions for approximating a function near a point where it is differentiable.


Suppose $f\colon\mathbb{C}\to\mathbb{C}$ is complex differentiable at a point $z_{0}\in\mathbb{C}.$ This means that the limit $$\lim_{h\to0}\frac{f(z_{0}+h)-f(z_{0})}{h}$$ exists and is finite. In turn, this means there must be some "error" function $\epsilon,$ with $\epsilon(h)\to0$ as $h\to0,$ such that $$f(z_{0}+h)-f(z_{0}) = h[f^{\prime}(z_{0}) + \epsilon(h)].$$ Rearranging this, we find that, for small $\lvert h\rvert,$ $$f(z_{0}+h) = f(z_{0}) + hf^{\prime}(z_{0}) + h\epsilon(h).$$ Since $\epsilon\to0$ as $h\to0,$ we can make an approximation of sorts: when $\lvert h \rvert$ is small, so too is $\lvert \epsilon(h) \rvert.$ Then $\lvert h\epsilon(h) \rvert = \lvert h \rvert \lvert \epsilon(h) \rvert$ must be very small indeed. Thinking of complex numbers as vectors in the plane (as in this image), we have \begin{equation} f(z_{0}+h) \approx f(z_{0}) + hf^{\prime}(z_{0}), \end{equation} where by "$\approx$" I mean the pointy ends of the vectors are close together. So what does this all mean?

Recall that if you multiply a complex number $a$ by another complex number $b,$ the product $ab$ has magnitude $\lvert a \rvert \lvert b \rvert$ and has argument $\arg(a)+\arg(b).$ The approximation above tells us that if $z_{1}$ is close to $z_{0},$ so that $z_{1}=z_{0}+h$ for some "small" number $h,$ then $$f(z_{1}) \approx f(z_{0}) + (z_{1}-z_{0})f^{\prime}(z_{0}).$$ That is, to compute $f(z_{1})$ as a vector, you take $f^{\prime}(z_{0})$ as a vector, stretch it (that is, multiply its magnitude by $\lvert z_{1}-z_{0}\rvert$), rotate it (that is, add the arguments), and finally you add on the vector $f(z_{0}).$

So, what is the complex derivative? It's a complex scaling factor, just like the real derivative is a real scaling factor. The only difference is that complex scaling factors introduce rotations. It shouldn't be thought of as representing a "gradient" as such, but rather as a set of instructions for approximating a function near a point where it is differentiable.


To take a specific example, the function $f\colon z\mapsto z^{2}$ is differentiable at the point $1+i$, and it's derivative there is $2(1+i)$ (it's also differentiable everywhere else, I just want a concrete example). By the above, this means that if I have a complex number $z_{1}$ which is close to $1+i,$ then $z_{1}^{2}\approx (1+i)^{2} + (z_{1}-1-i)(2(1+i)),$ i.e., $$z_{1}^{2} \approx -2i + 2(1+i)z_{1}.$$ We can check this numerically: let's agree that $1.1+i$ is close to $1+i.$ Then, setting $z_{1} = 1.1+ i$ gives $$z_{1}^{2} = 0.21 + 2.2 i,$$ and meanwhile $$-2i + 2(1+i)z_{1} = 0.2 + 2.2 i.$$ I think if we agree that $1.1+i$ is close to $1+i,$ then we should also agree that $0.2+2.2i$ is close to $0.21+2.2i.$

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  • $\begingroup$ Note that in this case the approximation is not subtle; if $z_{1}\approx1+i$ then $(z_{1}-1-i)^{2}\approx0$, and rearranging this gives the above approximation. But, of course, differentiation remains useful for approximating other, more complicated functions, and really comes into its own when we extend the approximations to get Taylor polynomials. $\endgroup$ – Will R Aug 23 '16 at 16:12
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Treating the complexes as two real dimensions is an important idea. If we relate the standard coordinates as $z=x+\mathbf{i}y$, then you'd have $\mathrm{d}z = \mathrm{d}x + \mathbf{i} \mathrm{d}y$. It's convenient to also use $\mathrm{d}\bar{z} = \mathrm{d}x - \mathbf{i} \mathrm{d}y $.

Every differential is a linear combination of $\mathrm{d}z$ and $\mathrm{d}\bar{z}$. The complex differentiable functions are the ones for which the coefficient on $\mathrm{d}\bar{z}$ is zero, so that we have $\mathrm{d}f(z) = g(z) \mathrm{d} z $, and we'd call $g$ the complex derivative of $f$.

That is, the complex differentiable functions $f$ are for which the "rate of change" in $f(z)$ is proportional to the "rate of change" in $z$.


If you prefer partial derivatives to differentials, we define $$\frac{\partial }{\partial z} = \frac{1}{2} \left( \frac{\partial}{\partial x} - \mathrm{i} \frac{\partial}{\partial y} \right) \qquad \qquad \frac{\partial }{\partial \bar{z}} = \frac{1}{2} \left( \frac{\partial}{\partial x} + \mathrm{i} \frac{\partial}{\partial y} \right)$$ As for why these particular formulas are the way they are, they are the ones that satisfy $$ \frac{\partial z}{\partial z} = 1 \qquad \qquad \frac{\partial \bar{z}}{\partial z} = 0 $$ $$ \frac{\partial z}{\partial \bar{z}} = 0 \qquad \qquad \frac{\partial \bar{z}}{\partial \bar{z}} = 1 $$

The complex differentiable functions are the real differentiable functions for which $$\frac{\partial f(z)}{\partial \bar{z}} = 0$$ so that the derivative is completely determined by the derivative in the "$z$ direction".


When working with more general functions, it is actually somewhat useful to treat $z$ and $\bar{z}$ as 'independent' variables, so that you can do calculus more conveniently. e.g. rather than working with $|z|^2$, you'd use $z \bar{z}$, whose differential $\mathrm{d}(|z|^2)$ is easy to compute as $z \mathrm{d}\bar{z} + \bar{z} \mathrm{d}z$. This can be more convenient to work with that $\mathrm{d}x$ and $\mathrm{d}y$, because you get to exploit complex derivatives and arithmetic.

In these terms, the complex differentiable functions are precisely those functions that 'depend' only on $z$ and not $\bar{z}$.

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From the point of view of the Taylor's theorem (which, I think, is geometrically the “correct” one), the derivative of a complex function (along with its value) represents exactly the same thing as the derivative of a real function: a linear approximation of the function at the point. If $f$ is differentiable at $x_0$, then $$ f(x_0+\Delta x)=f(x_0)+f'(x_0)\cdot \Delta x+o(\lvert \Delta x\rvert). $$ The theorem holds over reals and over the complex numbers, along with its generalizations, to higher dimensions and higher derivatives, which motivates the respective definitions.

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Perhaps I'm repeating some things already said, but here's my two cents.

In a sense, your question of "why shouldn't the derivative change when you approach from different directions" in the reason why complex analysis works the way it does. In the real case, $f'(x) =a$ means that for small changes $\Delta x$ of $x$, we get small changes $a\Delta x$ of $y$. It's the same in the complex case, but now $a$ is a complex number, so $\Delta x \mapsto a\Delta x$ behaves, as a map $R^2 \to R^2$, as a rotation and scaling of the complex number $\Delta x$, which you can think of as a vector in $R^2$. Thus multiplication by $a$ can be thought of a linear transformation transformation $R^2\to R^2$, just as the usual derivative of a function $R^2 \to R^2$.

This ultimately means that complex differentiable functions send little circles to little circles (a rotation and scaling surely does just this), which for one thing explains why complex differentiable functions are conformal, which in turn helps explain why they are in a sense "so rigid".

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