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I came across this the other day: $$ \frac{\partial f}{\partial \bar{z}} = \frac12\left(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\right) $$ I decided to attempt to work it out myself to better understand it. I know $2x = z + \bar{z}$, and $2iy = z - \bar{z}$, and using the total derivative we have $$ \frac{\partial f}{\partial \bar{z}} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial \bar{z}} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial \bar{z}} $$ and this is about where I got stuck. How exactly am I supposed to calculate $\frac{\partial x}{\partial \bar{z}}$? My confusion doesn't lie in the notation, but in the mechanics of the very thing I'm being asked to differentiate. Look at $x$: $$ x(\bar{z}) = \frac{\bar{z} + z}{2} = \frac{\bar{z}+\bar{\bar{z}}}{2} $$ if we label $Z = \bar{z}$, then $\frac{\partial x}{\partial \bar{z}} = \frac{\partial x}{\partial Z}$, and $x(Z) = \frac{Z+\bar{Z}}2$. However, as far as I can tell, $\frac{Z+\bar{Z}}{2}$ isn't even complex differentiable, because $\bar{Z}$ isn't complex differentiable with respect to $Z$. $x(Z)$ doesn't satisfy the CR equations: $$ x(X+iY) = X + i0 = u(X, Y) + iv(X, Y) \\ u_X = 1 \neq 0 = v_Y \\ v_X = 0 \neq -1 = -u_Y $$ so how could I possibly take the complex derivative of it? That doesn't make any sense.

What exactly am I missing here? Is the derivative $\frac{\partial x}{\partial \bar{z}}$ a different kind of derivative? Are we not supposed to do the complex derivative but instead something else?

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  • $\begingroup$ This is a similar question to yours. $\endgroup$ – Mark Feb 2 '17 at 23:48
  • $\begingroup$ @Mark Ahh, I hadn't heard the name Wirtinger derivative before. Thanks for the reference. $\endgroup$ – user3002473 Feb 2 '17 at 23:52
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It is a different way of thinking.

Think of a function of two variables. The variables may be $x,y$. But you can write $$ z = x+iy,\qquad \overline{z} = x - i y $$ and get two new variables $z, \overline{z}$. You can write $x$ and $y$ in terms of $z$ and $\overline{z}$. You can write $z$ and $\overline{z}$ in terms of $x$ and $y$. Thus, you can do a change of variables. When $f$ is considered a function of $z$ and $\overline{z}$ in this way, then of course the two partial derivatives make sense.

Some day perhaps you will study differential geometry, where you will learn a more complete way of doing this.

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  • $\begingroup$ Hmm, ok, so we're treating $\bar{z}$ more as a symbol, and not so much as a function of $z$. That clears up another confusion I had about these derivatives, is that why can't we simply write $\frac{\partial F}{\partial \bar{z}} = \frac{\partial F}{\partial z}\frac{\partial z}{\partial \bar{z}}$. $\endgroup$ – user3002473 Feb 2 '17 at 23:56

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