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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?

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  • $\begingroup$ Consider the examples $f=x+iy$, $f=ix-y$. $\endgroup$ Commented May 15, 2020 at 12:47
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    $\begingroup$ 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 $u_y=-v_x$. $\endgroup$
    – Koro
    Commented May 15, 2020 at 21:29
  • $\begingroup$ You have had three answers, asamsa, and you have not engaged with any of them. What's up? $\endgroup$ Commented May 16, 2020 at 22:56
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    $\begingroup$ It is not polite, asamsa, to post a question and then disengage from those who have tried to help you by posting answers. $\endgroup$ Commented May 18, 2020 at 13:20
  • $\begingroup$ you can easily remember the condition in this way: math.stackexchange.com/a/4439868/532409 $\endgroup$
    – Quillo
    Commented Apr 30, 2022 at 14:30

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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*}

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

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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 $v_x = -u_y$.

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