Partial derivative with respect to complex conjugate

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Main Question or Discussion Point

So, my complex analysis professor defined [tex] \partial f / \partial z^*[/tex] as
[tex]\frac {\partial f}{\partial z^*} = \frac {1}{2} \left( \left(\frac {\partial u}{\partial x}-\frac {\partial v}{\partial y}\right) + i\left(\frac {\partial u}{\partial y} + \frac {\partial v}{\partial x}\right)\right)[/tex]
where [tex]z = x + iy[/tex] and [tex]f(z) = u(x,y) + iv(x,y)[/tex]. My prof then showed that [tex] \partial f / \partial z^* = 0[/tex] when f is differentiable.

So my question is, where does this definition come from? It behaves exactly as you would expect it to by the notation (ie [tex] (\partial / \partial z^*) zz^* = z[/tex]), but the definition doesn't make this obvious.
 
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  • #2
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I know this thread is over half a year old, but I happened to come across it while searching some stuff about Complex Analysis on Google. It's possible that you may have already gotten an answer somewhere else, but for others who would like to know why this is and may end up here after a search on Google, I'll post my reasoning here.

Let's look at what makes a function [itex]f(z)[/itex]. Well, [itex]z = x + iy[/itex], so we can define it in terms of functions of [itex]x[/itex] and [itex]y[/itex]:

[tex]f(z) = f(x + iy) = u(x, y) + iv(x, y)[/tex]

We need to remember what [itex]z^*[/itex] is. It is simply [itex]z[/itex] whose imaginary part's sign is changed, [itex]z^* = x-iy[/itex].

Also, remember that we can define [itex]x[/itex] and [itex]y[/itex] in terms of [itex]z[/itex] and [itex]z^*[/itex]:

[tex]x = Re(z) = \frac{z + z^*}{2}[/tex]
[tex]y = Im(z) = \frac{z - z^*}{2i}[/tex]

Now, we can differentiate with respect to [itex]z[/itex] and [itex]z^*[/itex]. We will use the http://www.math.hmc.edu/calculus/tutorials/multichainrule/multichainrule.pdf" [Broken] (go to page 2 for the chain rule used here) for this.

[tex]\frac{\partial f}{\partial z} = \frac{\partial u}{\partial z} + i\frac{\partial v}{\partial z} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial z} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial z} + i \left( \frac{\partial v}{\partial x} \frac{\partial x}{\partial z} + \frac{\partial v}{\partial y} \frac{\partial y}{\partial z} \right)[/tex]

Since [itex]\frac{\partial x}{\partial z}=\frac{1}{2}[/itex] and [itex]\frac{\partial y}{\partial z}=\frac{1}{2i} = -i \frac{1}{2}[/itex], we have that

[tex]\frac{\partial f}{\partial z} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial z} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial z} + i \left( \frac{\partial v}{\partial x} \frac{\partial x}{\partial z} + \frac{\partial v}{\partial y} \frac{\partial y}{\partial z} \right) = \frac{1}{2} \frac{\partial u}{\partial x} - i \frac{1}{2} \frac{\partial u}{\partial y} + i\left( \frac{1}{2} \frac{\partial v}{\partial x} - i \frac{1}{2} \frac{\partial v}{\partial y} \right) = \frac{1}{2}\left[ \left(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}\right) + i \left( \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right) \right][/tex]


Now, we can apply the same to [itex]\frac{\partial f}{\partial z^*}[/itex]. Having [itex]\frac{\partial x}{\partial z^*}=\frac{1}{2}[/itex] and [itex]\frac{\partial y}{\partial z^*}=-\frac{1}{2i} = i \frac{1}{2}[/itex], we get

[tex]\frac{\partial f}{\partial z^*} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial z^*} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial z^*} + i \left( \frac{\partial v}{\partial x} \frac{\partial x}{\partial z^*} + \frac{\partial v}{\partial y} \frac{\partial y}{\partial z^*} \right) = \frac{1}{2} \frac{\partial u}{\partial x} + i \frac{1}{2} \frac{\partial u}{\partial y} + i\left( \frac{1}{2} \frac{\partial v}{\partial x} + i \frac{1}{2} \frac{\partial v}{\partial y} \right) = \frac{1}{2}\left[ \left(\frac{\partial u}{\partial x} - \frac{\partial v}{\partial y}\right) + i \left( \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} \right) \right][/tex]


The reason for why [itex]f(z)[/itex] is differentiable when [itex]\frac{\partial f}{\partial z^*} = 0[/itex] comes from the Cauchy-Riemann Equations.

[tex]\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}[/tex]
and
[tex]\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}[/tex]

You see, if you plug these into [itex]\frac{\partial f}{\partial z^*}[/itex], you will get 0. If you do not, then [itex]f(z)[/itex] does not satisfy the Cauchy-Riemann Equations, and that means it is not differentiable.
So there you have it. :smile:

Edit: Oh I see that over 600 people have viewed this thread, so I guess it was a good idea to bump it.
 
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  • #3
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I ended up getting it myself, but thanks for your reply anyways!
 

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