# Why the real and imaginary parts of a complex analytic function are not independent?

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?

It’s really just a question of the definition of the derivative. If $$z=x+yi,z=x+yi,$$ $$f(z)=u(x,y)+iv(x,y)f(z)=u(x,y)+iv(x,y)$$ can be any pair of functions $$u,v.u,v.$$

But if $$ff$$ is differentiable, then:

$$f′(z)=limh→0f(z+h)−f(z)hf'(z)=\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}\tag{1}$$

then $$hh$$ can approach $$00$$ in many different ways, since $$hh$$ is complex.

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

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

f′(z)=1i(∂u∂y+i∂v∂y)=∂v∂y−i∂u∂y\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 $$h→0h\to 0$$ you must have at minimum that $$∂u∂x=∂v∂y,∂v∂x=−∂u∂y\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},\\\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}\tag{2}$$

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

It turns out that $$(2)(2)$$ is enough to ensure that $$(1)(1)$$ 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 $$f:R2→R2f:\mathbb R^2\to\mathbb R^2$$ mapping $$(xy)↦(u(x,y)v(x,y))\begin{pmatrix}x\\y\end{pmatrix}\mapsto \begin{pmatrix}u(x,y)\\v(x,y)\end{pmatrix}$$ there is a matrix derivative standard from multi-variable calculus:

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

For small vectors $$h=(h1h2)\mathbf h=\begin{pmatrix}h_1\\h_2\end{pmatrix}$$ you get $$f((xy)+h)≈f(xy)+Df(xy)h.f\left(\begin{pmatrix}x\\y\end{pmatrix}+\mathbf h\right)\approx f\begin{pmatrix}x\\y\end{pmatrix}+Df\begin{pmatrix}x\\y\end{pmatrix}\mathbf h.$$

In particular, $$DfDf$$ is in some sense the “best” matrix, $$A,\mathbf A,$$ for estimating $$f(v+h)≈f(v)+Ah.f(\mathbf v+\mathbf h)\approx f(\mathbf v)+\mathbf A\mathbf h.$$

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

$$(a−bba)\begin{pmatrix}a&-b\\b&a\end{pmatrix}\tag{4}$$

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

We also have that:

$$(a−bba)(xy)=(ax−bybx+ay)\begin{pmatrix}a&-b\\b&a\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}= \begin{pmatrix}ax-by\\bx+ay\end{pmatrix}$$

compare that with:

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

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

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

So we see that when we satisfy Cauchy-Riemann, $$Df(xy)⋅hDf\begin{pmatrix}x\\y\end{pmatrix}\cdot\mathbf h$$ can be seen as multiplication of complex numbers, $$f′(z)f'(z)$$ and $$h=h1+h2i.h=h_1+h_2i.$$ Then you have:

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

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

So complex analysis is taking the vector function and asking, $$ff$$ “when does it make sense to think of the derivative of the $$R2→R2\mathbb R^2\to\mathbb R^2$$ as a complex number?” That is exactly when Cauchy-Riemann is true.

In the general case $$f:R2→R2,f:\mathbb R^2\to\mathbb R^2,$$ we can’t really take the second derivative and get an estimate $$f(z+h)≈f(z)+Df(z)⋅h+12D2f(z)⋅h2+⋯.f(z+h)\approx f(z)+Df(z)\cdot h +\frac{1}{2}D^2f(z)\cdot h^2+\cdots.$$ We can’t get easy equivalents to power series approximations of $$f.f.$$

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

So complex analysis is a subset of the real analysis of functions $$R2→R2\mathbb R^2\to\mathbb R^2$$ such that the derivative matrix $$DfDf$$ 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.