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How to show the transition between $(4)$ to $(5)$ in here?

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up vote 2 down vote accepted

Equation $4$ is just an application of the multivariable chain rule, while equation $5$ uses the definition of $\dfrac{d}{dz}$.

Don't think of equation $5$ as having been derived from equation $4$. Instead, think of the right hand sides of equations $4$ and $5$ as two different expressions for the same quantity, namely $\dfrac{df}{dz}$.

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Where do you get this multivariable chain rule? Any references please? – alvoutila Nov 5 '13 at 16:23
@laovultai: It is pretty standard, it should appear in any multivariable calculus text. There are also lots of sources on the internet; this for example. – Michael Albanese Dec 22 '13 at 4:54
The multivariable chain rule would give $\frac {df} {dz} = \frac {\partial f} {\partial x} \frac {dx} {dz} + \frac {\partial f} {\partial y} \frac {dy} {dz}$, not $\frac {df} {dz} = \frac {\partial f} {\partial x} \frac {\partial x} {\partial z} + \frac {\partial f} {\partial y} \frac {\partial y} {\partial z}$. – Sidd Jun 30 '15 at 18:17
Also, the $\frac \partial {\partial z}$ is a specially defined derivative, the Wirtinger derivative. – Sidd Jun 30 '15 at 18:19

$\frac{\partial}{\partial z} = \frac{1}{2}\left(\frac{\partial }{\partial x} - i \frac{\partial }{\partial y} \right)$ is the Wirtinger derivative and therefore $$\frac{\partial x}{\partial z} = \frac{1}{2}, \quad \frac{\partial y}{\partial z} = -\frac{i}{2}.$$

Edit: Here are the intermediate steps (requested in a comment): $$ \frac{\partial x}{\partial z} = \frac{1}{2}\left(\frac{\partial x }{\partial x} - i \frac{\partial x}{\partial y} \right) = \frac{1}{2} \left(1 -i\cdot0 \right) = +\frac{1}{2} $$ $$ \frac{\partial y}{\partial z} = \frac{1}{2}\left(\frac{\partial y }{\partial x} - i \frac{\partial y}{\partial y} \right) = \frac{1}{2} \left(0 -i\cdot 1 \right) = -\frac{i}{2} $$

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so if $f(z)=z^2$, then $\frac{\partial f}{\partial z} = 2z$ and on the other hand $z^2=x^2-y^2+i2xy$, then $1/2 (\frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y}) =1/2 (2x+i2y-i(-2y+i2x))=1/2(2x+i2y+i2y+2x)=2x+i2y=2z=\frac{\partial f}{\partial z}$ Ok. – alvoutila Sep 4 '13 at 15:52
@gammatester.. Can you explain me those two equations, why you get $1/2$ when you derivate $x$ with respect to $z$ and why you get $-i/2$ when you derivate $y$ with respecto to $z$? – alvoutila Nov 7 '13 at 14:21
@laovultai: I have just edited-in the intermediate steps – gammatester Nov 7 '13 at 14:43
@gammatester.. As I have asked above about multivariable chain rule, I ask from you now that do they interpret in equation (4) complex function $f$ to be function of two variables $x$ and $y$ and also vector function $\mathbb{R}^2 \mapsto \mathbb{R}^2$ and if yes, why it is valid to interpret complex function as vector function $\mathbb{R}^2 \mapsto \mathbb{R}^2$ and why do multivariable chain rule applies to such complex function as $f$? As I know it multivariable chain rule applies only to multivariable functions like $f(x,y)=xe^y$, where $x(t)=sin(t)$ and $y(t)=e^{2t}$. – alvoutila Nov 8 '13 at 11:09
There is a higher dimensional chain rule, see I don't know how they interpret the equation, I only gave you the hint. The Cauchy-Riemann equations are about partial differential equations and to quote Wikipedia once more "The Cauchy-Riemann equations [...] consist of a system of two partial differential equations which, together with certain continuity and differentiability criteria, form a necessary and sufficient condition for a complex function to be complex differentiable" – gammatester Nov 8 '13 at 11:45

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