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Let $E\subset \mathbb{C}$ be an open set and $f:E\to\mathbb{C}$ be a function. Let $S:=\{z\in\mathbb{C}:\left|z \right|=1\}$ be a circle in the complex plane. Then we can show that if $w\in \mathbb{C}$, then the directional derivative at $z$ in the direction $w$ satisfies the following equality:

$$ D_w f(z) = f'(z)w $$.

Now, I've been trying to deduce the Cauchy-Riemann equations from this, albeit unsuccessfully.

My attempt:

Let $f=u+iv=(u,v)$ for $u(x,y), v(x,y)$ some real functions in $\mathbb{R}^2$. Then the derivative of $f$ at $(x,y)$ is given by

\begin{bmatrix} u_x & u_y\\ v_x & v_y \end{bmatrix}

So, $$f'(z)w = \begin{bmatrix} u_x & u_y\\ v_x & v_y \end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix},$$

where $x^2+y^2=1$.

Unfortunately, I have no idea how to proceed from there. I believe there should be some property related directional derivatives, which would allow to proceed to the end.

I would definitely appreciate a hint.

2 Answers 2

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Write $f'(z) = a + ib$, so that for each $w=x +iy$ we have $$f'(z)w = (ax - by) + i(ay+bx) = (xu_x + yu_y) + i(xv_x + yv_y).$$

Equating components we get $$ax-by = xu_x + yu_y$$ and $$ay+bx = xv_x + yv_y.$$

Since this must be true for all $w$, we can equate components of $x,y$, yielding the equations $ a = u_x = v_y$ and $b = v_x = -u_y$ as desired.

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    How to reconcile that $f'(z)$ is a $2\times 2$ matrix, but can also be represented as the $2\times 1$ vector $a+ib$? – 2017-01-26
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    @sequence: My interpretation of your question is that you are *assuming* $f$ is complex differentiable; i.e. that there is some complex number $f'(z)$ such that $D_w f = f'(z)w$. A priori, when you write the matrix product in your question, this is $Df\ w = D_w f$, the product of a $2\times 2$ Jacobian matrix with a $2\times 1$ vector/complex number. It's only equal to $f'(z)w$ (the product of two complex numbers) due to the assumption of complex differentiability. The "reconciliation" that needs to be done is realizing that this is a stronger assumption than real differentiability. – 2017-01-26
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    i.e. that only some $2 \times 2$ matrices can have their multiplication action represented as multiplication by some complex number. – 2017-01-26
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    But can $f'(z)$ be represented by the Jacobian of the corresponding $f\in \mathbb{R}^2$? And if yes, then how does $a+ib$ equal this Jacobian? @AnthonyCarapetis – 2017-01-26
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    @sequence: $a+ib$ equals the complex number $f'(z)$. Any complex number $a+ib$ can be represented by the $2\times 2$ matrix $[a, -b; b, a]$, and multiplication of a $2 \times 1$ vector $[x, y]^T$ by this matrix gives the same answer as multiplication of the complex numbers $a+ib$ and $x+iy$. – 2017-01-26
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Let $f=u(x,y)+iv(x,y)$ be a differential function with directional derivative $$D_wf(z)=f'(z)w$$ and $w_1=(\cos\theta,\sin\theta)$, $w_2=(-\sin\theta,\cos\theta)$ be unit perpendicular vectors. $f$ is differentiable so $D_{w_1}f(z)=D_{w_2}f(z)$, this induce that $\dfrac{\partial u}{\partial {w_1}}=\dfrac{\partial v}{\partial {w_2}}$ and $\dfrac{\partial u}{\partial {w_2}}=\dfrac{\partial v}{\partial {w_1}}$ which by simplifying, from the first, we have $$u_x(-\sin\theta)+u_y(\cos\theta)=v_x(-\cos\theta)+v_y(-\sin\theta)\hspace{1cm};\hspace{1cm}\forall\theta$$ This leads us to Cauchy-Riemann equations.

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    How come $D_{w_1}f(z) = D_{w_2}f(z)$? Also, how do you get $v$ and $u$ from $w$? – 2017-01-26