# In which sense does Cauchy-Riemann equations link complex- and real analysis?

On page 12 of Stein, Shakarchi textbook 'Complex analysis', the authors state that the Cauchy-Riemann equations link complex and real analysis. I have completed courses on real and complex analysis, but I feel that this is somewhat of an over-statement. But perhaps it is just me which doesnt have a good enough overview.

$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}$$

If anyone with a clear insight is able to concisely explain how one could justify writing something like this -- then that insight would be most valuable.

• They probably mean that first off, you have a real set of differential equations describing what a complex analytic function should be. Also, the CR equations are morally similar to Green's theorem in that analytic functions give zero integrals when integrated around a closed region: en.wikipedia.org/wiki/Green%27s_theorem – Alex R. Aug 3 '15 at 19:37
• In this book the connection is made. The author then says 'In classical complex analysis analyticity is not generally implied' a fact mentioned by Penrose in The Road to Reality, who seems unaware of the above proof - although he suspects it exists. – user301988 Dec 4 '16 at 19:30
• Would you like to elaborate on this? I will strongly consider accepting it as the answer if it can be understood by someone new to analysis. Some references to the book would also be good. – Mikkel Rev Dec 5 '16 at 10:40

This is slightly too long for a comment.

I think your problem is that you are looking for the statement to be deep. The CR equations simply provide that studying a differentiable complex valued function in one variables is the same as studying a pair of real valued functions in two variables satisfying the CR equations, and so theorems in real analysis have something to say about complex analysis, and everything in complex analysis can be converted into a real-analytic statement about a restricted class of functions. So there is a link. That doesn't mean that people are constantly jumping back and forth from one perspective to another, but without such a link, it would be odd to talk about contour integrals, as the paths of integration are 1-dimensional real-analytic objects, and cannot really be defined using only complex analytic objects.

It will be shown that a complex function $\,f = u + i.v\,$ is complex differentiable to $\,z = x + i.y\,$ if and only if it is real differentiable at $\,z\,$ and if the partial derivatives of $\,u\,$ and $\,v\,$ to $\,x\,$ and $\,y\,$ obey the so called Cauchy-Riemann Equations: $\;\partial u /\partial x = \partial v /\partial y\;$ and $\;\partial u /\partial y = - \partial v /\partial x$ .

### Two special cases

The complex derivative $f'(z)$ of a complex function $f(z)$ with $z$ complex is defined as follows (with $\Delta z$ complex as well). $$f'(z) = \lim_{\Delta z\rightarrow 0} \frac{f(z + \Delta z) - f(z)}{\Delta z}$$ A complex function can always be written in real valued components $(x,y,u,v)$ as: $$f(z) = u(x,y) + i.v(x,y) \qquad \mbox{where} \quad z = x + i.y$$ Complex differentiation is independent of the direction, the way $z$ approaches zero. Two special cases are distinguished: differentiation in the $x$-direction and differentiation in the $y$-direction. In the $x$-direction it is found that $\Delta z = \Delta x$ and the complex derivative is: $$f'(z) = \lim_{\Delta x\rightarrow 0} \frac{u(x + \Delta x,y) - u(x,y) + i\left[v(x + \Delta x,y) - v(x,y)\right]} {\Delta x} = \frac{\partial u}{\partial x} + i.\frac{\partial v}{\partial x}$$ In the $y$-direction it is found that $\Delta z = i.\Delta y$ and the complex derivative is: $$f'(z) = \lim_{\Delta y\rightarrow 0} \frac{u(x,y + \Delta y) - u(x,y) + i\left[v(x,y + \Delta y) - v(x,y)\right]} {i.\Delta y} = -i.\frac{\partial u}{\partial y} + \frac{\partial v}{\partial y}$$ Two complex numbers are equal if and only if the real and the imaginary parts are equal: $$f'(z) = \frac{\partial u}{\partial x} + i.\frac{\partial v}{\partial x} = -i.\frac{\partial u}{\partial y} + \frac{\partial v}{\partial y} \quad \Longrightarrow \quad \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \mbox{and} \quad \frac{\partial v}{\partial x} = - \frac{\partial u}{\partial y}$$ These are the well-known Cauchy-Riemann equations. Conclusion: if a function $\,f = u + i.v\;$ is complex differentiable, then the real and imaginary parts $\,u\,$ and $\,v\,$ of $\,f\,$ satisfy the Cauchy-Riemann equations at $\,z = x + i.y$ .

### Real Differentiable

The complex derivative $f'(z)$ of a complex function $f(z)$ with $z$ complex is defined as follows (with $\Delta z$ complex as well). $$f'(z) = \lim_{\Delta z\rightarrow 0} \frac{f(z + \Delta z) - f(z)}{\Delta z}$$ Consequently: $$f(z + \Delta z) = f(z) + f'(z) \Delta z \qquad \mbox{for} \quad z \rightarrow 0$$ A complex function can always be written in real valued components $(x,y,u,v)$ as: $$\begin{cases} f(z) = u(x,y) + i.v(x,y) \\ f'(z) = u'(x,y) + i.v'(x,y) \end{cases} \qquad \mbox{where} \quad z = x + i.y$$ Consequently, for $\Delta x \rightarrow 0$ and $\Delta y \rightarrow 0$ : $$u(x + \Delta x,y + \Delta y) + i . v(x + \Delta x,y + \Delta y) = u(x,y) + i . v(x,y)$$ $$+ \left[ u'(x,y) + i . v'(x,y) \right] \left[ \Delta x + i \Delta y \right] \quad \Longrightarrow \quad$$ $$u(x + \Delta x,y + \Delta y) = u(x,y) + u'(x,y)\Delta x - v'(x,y)\Delta y$$ $$v(x + \Delta x,y + \Delta y) = v(x,y) + v'(x,y)\Delta x + u'(x,y)\Delta y$$ Meaning that, if a function $f = u + i.v$ is complex differentiable at $z = x + i.y$ , then its real and imaginary parts $(u,v)$ are real differentiable at ($x,y)$ . On the other hand it is known from real analysis that: $$u(x + \Delta x,y + \Delta y) = u(x,y) + \frac{\partial u}{\partial x}\Delta x + \frac{\partial u}{\partial y}\Delta y \qquad \mbox{for} \quad (\Delta x,\Delta y) \rightarrow (0,0)$$ $$v(x + \Delta x,y + \Delta y) = v(x,y) + \frac{\partial v}{\partial x}\Delta x + \frac{\partial v}{\partial y}\Delta y \qquad \mbox{for} \quad (\Delta x,\Delta y) \rightarrow (0,0)$$ This can only be consistent if the Cauchy-Riemann equations are indeed valid: $$\frac{\partial u}{\partial x} = u'(x,y) = \frac{\partial v}{\partial y} \qquad ; \qquad \frac{\partial v}{\partial x} = v'(x,y) = - \frac{\partial u}{\partial y}$$ It is known from real analysis that a function $\,f(x,y)\,$ is total differentiable at $(a,b)$ , if and only if the partial derivatives $\,\partial f /\partial x\,$ and $\,\partial f /\partial y\,$ exist in a neighborhood of $(a,b)$ and both are continuous there. Thus making the picture complete.

### Independent of angle

The reverse question is: if the Cauchy-Riemann equations hold for the real and imaginary parts of a complex function $\,f$ , is $\,f\,$ complex differentiable then ? Let's write the complex derivative with $\,\Delta z = r.e^{i\theta}$ . The reason is that the complex derivative must be independent of any (real) angle $\theta$ while the (real) distance $r$ from $z + \Delta z$ to $z$ approaches zero. So this is what we do: $$f'(z) = \lim_{r\rightarrow 0} \frac{f(z + r.e^{i\theta}) - f(z)}{r.e^{i\theta}}$$ Remember that $f(z) = u(x,y) + i.v(x,y)$ where $z = x + i.y$ . Also remember the jewel formula by Euler $\;e^{i\theta} = \cos(\theta) + i.\sin(\theta)$ . Giving: $$f'(z) = \lim_{r\rightarrow 0} \frac{u(x + r\cos(\theta),y + r\sin(\theta)) - u(x,y)} {r\cos(\theta) + i.r\sin(\theta)}$$ $$+ \; i \; \lim_{r\rightarrow 0} \frac{v(x + r\cos(\theta),y + r\sin(\theta)) - v(x,y)} {r\cos(\theta) + i.r\sin(\theta)}$$ Given the real differentiable functions $[u,v](x,y)$ , for $[\Delta x,\Delta y] \rightarrow 0$ : $$[u,v](x+\Delta x,y+\Delta y) - [u,v](x,y) = \frac{\partial [u,v]}{\partial x}(x,y)\, \Delta x + \frac{\partial [u,v]}{\partial y}(x,y)\, \Delta y$$ With $\left[\Delta x,\Delta y\right] = \left[r\cos(\theta), r\sin(\theta)\right]$ : $$f'(z) = \lim_{r\rightarrow 0} \frac{\partial u /\partial x.r\cos(\theta) + \partial u /\partial y.r\sin(\theta)} {r\cos(\theta) + i.r\sin(\theta)}$$ $$+ \, i \, \lim_{r\rightarrow 0} \frac{\partial v /\partial x.r\cos(\theta) + \partial v /\partial y.r\sin(\theta)} {r\cos(\theta) + i.r\sin(\theta)}$$ Taking the limit for $r\rightarrow 0$ is easy: $$f'(z) = \frac{\partial u /\partial x.\cos(\theta) + \partial u /\partial y.\sin(\theta)} {\cos(\theta) + i\sin(\theta)} \; + \; i \; \frac{\partial v /\partial x.\cos(\theta) + \partial v /\partial y.\sin(\theta)} {\cos(\theta) + i\sin(\theta)}$$ Substitute herein the Cauchy-Riemann equations (by a copy and paste from the preceding subsection): $$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \mbox{and} \quad \frac{\partial v}{\partial x} = - \frac{\partial u}{\partial y}$$ Just doit: $$f'(z) = \frac{\partial u /\partial x.\cos(\theta) - \partial v /\partial x.\sin(\theta)} {\cos(\theta) + i\sin(\theta)} \; + \; i \; \frac{\partial v /\partial x.\cos(\theta) + \partial u /\partial x.\sin(\theta)} {\cos(\theta) + i\sin(\theta)}$$ $$f'(z) = \frac{\partial u}{\partial x} \; \frac{\cos(\theta) + i\sin(\theta)} {\cos(\theta) + i\sin(\theta)} \; + i \; \frac{\partial v}{\partial x} \; \frac{\cos(\theta) + i\sin(\theta)} {\cos(\theta) + i\sin(\theta)} = \frac{\partial u}{\partial x} + i \, \frac{\partial v}{\partial x}$$ Which indeed is independent of any angle $\theta$ . Consequently, if the function $\,f = u + i.v\,$ is real differentiable at $\,z = x + i.y\,$ and if the partial derivatives of $\,u\,$ and $\,v\,$ to $\,x\,$ and $\,y\,$ obey the Cauchy-Riemann Equations, then the complex derivative of $\,f\,$ is independent of the direction in which we differentiate.