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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?

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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.

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Well, How you derive this Wirtinger differential operator? I haven't found by searching from the internet. –  laovultai 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. –  laovultai 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? –  laovultai Nov 22 '13 at 7:58
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