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The Wirtinger differential operators are introduced in complex analysis to simplify differentiation in complex variables. Most textbooks introduce them as if it were a natural thing to do. However, I fail to see the intuition behind this. Most of the time, I even think they tend to make calculations harder.

Is there a simple interpretation of these operators? What mental picture do you have when you use them?

Robin Wolffoot
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    You may read section 2.1 of http://www.researchgate.net/publication/282646505, "The Wirtinger partial derivatives: no need for the independence of a variable and its conjugate" – Arie ten Cate Jan 10 '16 at 10:40

2 Answers2

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It is natural when dealing with a complex-valued function $f(x,y)$ to think of a change of variables from $(x,y)$ to $(z,\bar z)$, where $z = x+i y$. We need to know how the derivatives transform and we find $$\begin{eqnarray*} \frac{\partial}{\partial x} &=& \frac{\partial z}{\partial x} \frac{\partial}{\partial z} + \frac{\partial \bar z}{\partial x} \frac{\partial}{\partial \bar z} \\ &=& \frac{\partial}{\partial z} + \frac{\partial}{\partial \bar z} \\ \frac{\partial}{\partial y} &=& \frac{\partial z}{\partial y} \frac{\partial}{\partial z} + \frac{\partial \bar z}{\partial y} \frac{\partial}{\partial \bar z} \\ &=& i\left(\frac{\partial}{\partial z} - \frac{\partial}{\partial \bar z}\right). \end{eqnarray*}$$ Solving for $(\partial/\partial z, \partial/\partial \bar z)$ we find the usual expressions for the Wirtinger derivatives. The derivatives act as they should, $$\frac{\partial z}{\partial z} = \frac{\partial \bar z}{\partial \bar z} = 1 \qquad \textrm{and}\qquad \frac{\partial \bar z}{\partial z} = \frac{\partial z}{\partial \bar z} = 0.$$

The derivative $\partial/\partial \bar z$ has the important property that if $$\frac{\partial}{\partial \bar z} f(z,\bar z) = 0$$ then $f$ satisfies the Cauchy-Riemann equations and so is holomorphic (and thus analytic). This is straightforward to verify, $$\begin{eqnarray*} \frac{1}{2}\left(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}\right) f(x,y) &=& \frac{1}{2}\left(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}\right) (u(x,y)+i v(x,y)) \\ &=& \frac{1}{2}((u_x - v_y) + i(u_y + v_x)), \end{eqnarray*}$$ where subscripts denote partial derivatives. Thus, if $\partial f(z,\bar z)/\partial \bar z = 0$ then the Cauchy-Riemann equations are satisfied, $u_x = v_y$ and $u_y = -v_x$. This gives us a very good intuition indeed. Roughly, if $f$ is not a function of $\bar z$, then $f$ is holomorphic.

user26872
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  • Well, How you derive this Wirtinger differential operator? I haven't found by searching from the internet. –  Nov 15 '13 at 15:25
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    @laovultai: As described above, if you solve $\partial_x=\partial_z+\partial_{\bar z}$ and $\partial_y = i(\partial_z-\partial_{\bar z})$ for $\partial_z$ and $\partial_{\bar{z}}$ you will find the usual expressions for the Wirtinger derivatives, $\partial_z = \frac{1}{2}(\partial_x-i\partial_y)$ and $\partial_{\bar z} = \frac{1}{2}(\partial_x+i\partial_y)$. – user26872 Nov 16 '13 at 19:04
  • @oen.. Do you agree that if $f$ is analytic, then $\frac{df}{dz}=\frac{1}{2}(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y})$, but if eq. $f(x,y)=x^2-y^2+i(y^2-x^2) $ then $ 0=\frac{df}{dz} \neq \frac{1}{2}(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}) $? So you would need assumption that $f$ is analytic in order that Wirtinger derivatives would hold. –  Nov 21 '13 at 11:57
  • @laovultai: If you calculate carefully you will find that $\partial f/\partial \bar z = (1-i)\bar z = (1-i)(x-i y)$. You will find this result whether you use the Wirtinger derivative or instead making the transformation $(x,y)\to(z,\bar z)$ in $f$. – user26872 Nov 21 '13 at 15:33
  • @oen.. So $f$ can be expressed as $f(z)$ in this case? What about in case if $f(x,y)=x^2+i(y^2-x^2)$ then $0=\frac{df}{dz} \neq \frac{1}{2}(\partial/\partial x + i \partial / \partial y)$ and you need analyticity in order WDO to hold. Right? –  Nov 22 '13 at 07:58
  • @laovultai: If $f(x,y)=x^2-y^2+i(y^2-x^2)$, then $f(z,\bar z) = \frac{1}{2}(1-i)(z^2+\bar z^2)$. Thus, $f$ cannot be expressed in terms of $z$ only. In fact, $\partial f/\partial \bar z = (1-i)\bar z = (1-i)(x-i y)$. This is the result you should find using the Wirtinger derivative: $$\begin{eqnarray} \partial_{\bar z}f &=& \frac{1}{2}(\partial_x + i \partial y)(x^2-y^2+i(y^2-x^2)) \ &=& \frac{1}{2}(\partial_x + i \partial y)((1-i)x^2-(1-i)y^2) \ &=& \frac{1}{2}((1-i)2x-i(1-i)2y) \ &=& (1-i)(x-iy) \ &=& (1-i)\bar z. \end{eqnarray}$$ – user26872 Nov 23 '13 at 20:45
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    For the last statement "Roughly, if f is not a function of z¯, then f is holomorphic." @user26872, whys is it roughly? Is there a case where that is not true? – ions me Sep 03 '21 at 16:31
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    @IonSme: Let $f(z,\bar z) = 1/\bar z-1/\bar z$. This function has a dependence on $\bar z$: it is undefined if $\bar z=0$, otherwise it is zero. However, $f$ is holomorphic on $C-{0}$ (and can be extended to the holomorphic function $0$ on all of $C$ since the singularity is removeable). – user26872 Sep 06 '21 at 18:14
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In the first place, as noted above: the Wirtinger partial derivatives "act as they should''. The proof is not trivial, but it is easily seen with simple examples. Not only the two examples above, below the cited phrase, but also somewhat more complicated: $$ \frac{\partial z\overline{z}}{\partial z} = \overline{z} $$ and $$ \frac{\partial z\overline{z}}{\partial \overline{z}} = z.$$ For instance: $$ \frac{\partial z\overline{z}}{\partial z} = \frac{1}{2}\left( \frac{\partial \, x^2+y^2}{\partial x} - i \frac{\partial \, x^2+y^2}{\partial y}\right) = \frac{1}{2}\left(2x- i 2y\right) = x- i y = \overline{z}$$ Of course, the whole point of the Wirtinger partial derivatives is their behavior as ordinary partial derivatives, with a jump immediately from $\partial z\overline{z}/\partial z$ to $\overline{z}$. This is my "mental picture".

In the second place, this mental picture of ordinary partial derivatives can be applied to prove the rule that a function $f$ is complex-differentiable with respect to $z$ if and only if ${\partial f}/{\partial \overline{z}} = 0$, as follows. Start with

\begin{align*} \frac{\mathrm{d} f}{\mathrm{d} z} &= \frac{\partial f}{\partial z} + \frac{\partial f}{\partial \overline{z}} \, \frac{\mathrm{d} \overline{z}}{\mathrm{d} z}\\ &= \frac{\partial f}{\partial z} + \frac{\partial f}{\partial \overline{z}} \, \lim_{\Delta z \rightarrow 0} \frac{\Delta \overline{z} }{\,\Delta z \,}\\ &= \frac{\partial f}{\partial z} + \frac{\partial f}{\partial \overline{z}} \, \lim_{\Delta z \rightarrow 0} \frac{\overline{\Delta z} }{\,\Delta z \,}\\ &= \frac{\partial f}{\partial z} + \frac{\partial f}{\partial \overline{z}} \, \lim_{\Delta z \rightarrow 0} e^{- i 2\arg(\Delta z )}\\ & = \frac{\partial f}{\partial z} + \lim_{\Delta z \rightarrow 0} \left( \frac{\partial f}{\partial \overline{z}} \, e^{- i 2\arg(\Delta z )} \right). \end{align*}

Then, if ${\partial f}/{\partial \overline{z}} = 0$ holds then we have indeed: $$ \frac{\mathrm{d} f}{\mathrm{d} z} = \frac{\partial f}{\partial z} + \lim_{\Delta z \rightarrow 0} \left( 0 \times e^{- i 2\arg(\Delta z )} \right) = \frac{\partial f}{\partial z} + \lim_{\Delta z \rightarrow 0} 0 = \frac{\partial f}{\partial z} $$ Otherwise, indeed, ${\mathrm{d} f}/{\mathrm{d} z}$ does not exist, because then the $\lim_{\Delta z \rightarrow 0}$ does not exist.

Michael Albanese
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Arie ten Cate
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