# How to remember the Cauchy-Riemann equations?

A complex function $$f$$ with real and imaginary parts $$u$$ and $$v$$ respectively is holomorphic in some domain $$\Omega$$ iff $$u$$ and $$v$$ satisfy the Cauchy-Riemann equations in $$\Omega$$: $$\frac {\partial u}{\partial x} = \frac {\partial v}{\partial y} \\ \frac {\partial v}{\partial x} = -\frac {\partial u}{\partial y}.$$ I am looking for different ways of recalling or producing this result. One way is to begin with the requirement $$\bar \partial f= 0,$$ with $$\bar \partial = \partial_x + i \partial_y.$$ The real and imaginary parts of $$\partial_x(u+iv) + i\partial_y(u+iv) = 0$$ are then the Cauchy-Riemann equations.

Another well-known heuristic is to compare the Jacobian $$\begin{pmatrix} u_x & u_y\\ v_x & v_y\end{pmatrix}$$ with the matrix representation of a complex number. How else do you produce these equations when needed?

• Consider the examples $f=x+iy$, $f=ix-y$. Commented May 15, 2020 at 12:47
• One mnemonic could be this: Write the determinant of the Jacobian matrix : $u_xv_y-u_yv_x= u_x v_y+(-u_y) (v_x)$ See minus sign is coming in the second pair i.e. $(u_y, v_x)$ hence $u_y=-v_x$.
– Koro
Commented May 15, 2020 at 21:29
• You have had three answers, asamsa, and you have not engaged with any of them. What's up? Commented May 16, 2020 at 22:56
• It is not polite, asamsa, to post a question and then disengage from those who have tried to help you by posting answers. Commented May 18, 2020 at 13:20
• you can easily remember the condition in this way: math.stackexchange.com/a/4439868/532409 Commented Apr 30, 2022 at 14:30

If $$\frac{\partial f}{\partial z}$$ is well-defined for complex $$z$$, then for real $$x$$ and $$y,$$ $$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial(iy)} = -i\frac{\partial f}{\partial y}$$ That is, $$\frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x} =-i\left(\frac{\partial u}{\partial y} + i\frac{\partial v}{\partial y}\right) = \frac{\partial v}{\partial y} - i\frac{\partial u}{\partial y}.$$ That's easy to remember, I think.
It can be made into a rigorous proof by applying the Chain Rule to the composites of $$f$$ with paths: $$\begin{gather*} \xi \colon [-\delta, \delta] \to \mathbb{C}, \ t \mapsto (x + t) + iy, \\ \eta \colon [-\delta, \delta] \to \mathbb{C}, \ t \mapsto x + i(y + t), \end{gather*}$$ for small $$\delta > 0,$$ thus: $$\begin{multline*} \frac{\partial u}{\partial y} + i\frac{\partial v}{\partial y} = (f \circ \eta)'(0) \\ = f'(\eta(0))\eta'(0) = if'(x + iy) = if'(\xi(0))\xi'(0) \\ = i(f \circ \xi)'(0) = i\left(\frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x} \right). \end{multline*}$$
Consider $$f(z)=z^2=(x+iy)^2=u(x,y)+iv(x,y)$$ where $$u(x,y)=x^2-y^2$$ and $$v(x,y)=2xy$$.
Calculate $${\partial u\over\partial x}=2x,\ {\partial u\over\partial y}=-2y,\ {\partial v\over\partial x}=2y,\ {\partial v\over\partial y}=2x$$ and it's clear $${\partial u\over\partial x}={\partial v\over\partial y},\ {\partial u\over\partial y}=-{\partial v\over\partial x}$$
Here's a way of seeing the Cauchy-Riemann equations I find memorable. For a function to be complex differentiable, the limit $$\lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0}$$ must exist, and hence be independent of the direction by which $$z$$ approaches $$z_0$$.
Mentally fix $$z_0 = (x,y)$$ in $$\mathbb{C}$$. Writing $$f = u + iv$$ with $$u, v$$ real, approaching $$z_0$$ from above gives the limit $$\lim_{\epsilon \to 0} \frac{f(x, y + \epsilon) - f(x,y)}{i\epsilon} = -i(u_y + iv_y) = v_y -iu_y,$$ and approaching from the right gives us $$\lim_{\epsilon \to 0} \frac{f(x + \epsilon, y) - f(x,y)}{\epsilon} = u_x + iv_x.$$ For $$f$$ to be complex differentiable, these must exist and be equal, so their real and imaginary parts must be equal and we have $$u_x = v_y$$ and $$v_x = -u_y$$.