Question about partial derivative relations for complex numbers

[binbagsss]

Apologies this is probably a very bad question but it's been a while since I have seen this.

I have $z=x+iy$. I need to convert $\frac{\partial \psi(z)}{\partial z}$ , with $\psi$ some function of $z$, in terms of $x$ and $y$
I have $dz=dx+idy$. so
$\frac{\partial \psi }{\partial z}=\frac{\partial \psi}{\partial x+i \partial y}$
But I am not sure how to simplify futher as I don't think you can't write
$\frac{\partial \psi}{\partial x+i \partial y} = \frac{\partial \psi}{\partial x} +\frac{1}{i}\frac{\partial \psi}{ \partial y}$?

Thanks

 

[fresh_42]

Some function is little information. Say it is complex differentiable and $\psi\, : \,\mathbb{C}\longrightarrow \mathbb{C}.$ Then we have $\psi(z)=\psi(x+iy)=u(x+iy)+i v(x+iy).$ Written as real functions, we get $u,v\, : \,\mathbb{R}^2\longrightarrow \mathbb{R}$ with $u(x,y)=u(x+iy)$ and $v(x,y)=v(x+iy).$ It makes no sense to use different letters, but you can use a prime if you like since these equalities are not proper equalities, just correspondences. Now, you can consider the partial derivatives
$$
\dfrac{\partial u}{\partial x}\, , \,\dfrac{\partial u}{\partial y}\, , \,\dfrac{\partial v}{\partial x}\, , \,\dfrac{\partial v}{\partial y}
$$
In order to be complex differentiable, the complex derivative has to be complex linear, not only real linear. Thus we have the additional requirements
$$
\dfrac{\partial u}{\partial x}=\dfrac{\partial v}{\partial y}\;\text{ and }\;\dfrac{\partial u}{\partial y}=-\dfrac{\partial v}{\partial x}\quad (2)
$$

I would have approached it without the deviation along the real numbers and would have used the Weierstraß formula and complex linearity, but the above is how it is done with real components and the Cauchy-Riemann equations (2).

https://www.physicsforums.com/insights/an-overview-of-complex-differentiation-and-integration/
could also be helpful.

 

[martinbn]

Apologies this is probably a very bad question but it's been a while since I have seen this.

I have $z=x+iy$. I need to convert $\frac{\partial \psi(z)}{\partial z}$ , with $\psi$ some function of $z$, in terms of $x$ and $y$
I have $dz=dx+idy$. so
$\frac{\partial \psi }{\partial z}=\frac{\partial \psi}{\partial x+i \partial y}$
But I am not sure how to simplify futher as I don't think you can't write
$\frac{\partial \psi}{\partial x+i \partial y} = \frac{\partial \psi}{\partial x} +\frac{1}{i}\frac{\partial \psi}{ \partial y}$?

Thanks

$\frac{\partial f}{\partial z}=\frac12\left( \frac{\partial f}{\partial x} -i\frac{\partial f}{\partial y}\right)$

 

[weirdoguy]

I have dz=dx+idy. so

Your "so" does not follow. It does not work that way for multiple reasons, but the very first thing you should be concerned with is that you use $d$ before "so" and $\partial$ after.

 

[binbagsss]

Some function is little information. Say it is complex differentiable and $\psi\, : \,\mathbb{C}\longrightarrow \mathbb{C}.$ Then we have $\psi(z)=\psi(x+iy)=u(x+iy)+i v(x+iy).$ Written as real functions, we get $u,v\, : \,\mathbb{R}^2\longrightarrow \mathbb{R}$ with $u(x,y)=u(x+iy)$ and $v(x,y)=v(x+iy).$ It makes no sense to use different letters, but you can use a prime if you like since these equalities are not proper equalities, just correspondences. Now, you can consider the partial derivatives
$$
\dfrac{\partial u}{\partial x}\, , \,\dfrac{\partial u}{\partial y}\, , \,\dfrac{\partial v}{\partial x}\, , \,\dfrac{\partial v}{\partial y}
$$
In order to be complex differentiable, the complex derivative has to be complex linear, not only real linear. Thus we have the additional requirements
$$
\dfrac{\partial u}{\partial x}=\dfrac{\partial v}{\partial y}\;\text{ and }\;\dfrac{\partial u}{\partial y}=-\dfrac{\partial v}{\partial x}\quad (2)
$$

I would have approached it without the deviation along the real numbers and would have used the Weierstraß formula and complex linearity, but the above is how it is done with real components and the Cauchy-Riemann equations (2).

https://www.physicsforums.com/insights/an-overview-of-complex-differentiation-and-integration/
could also be helpful.

in the context of the problem I have, I am pretty certain one does not want to introduce $\Re(\psi)$ and $\Im(\psi)$

 

[fresh_42]

in the context of the problem I have, I am pretty certain one does not want to introduce $\Re(\psi)$ and $\Im(\psi)$

Then why do you want to split $z$?

 

[martinbn]

in the context of the problem I have, I am pretty certain one does not want to introduce $\Re(\psi)$ and $\Im(\psi)$

The steps you wrote are meaningless. The answer you wrote is of by a factor of one half. I wrote the usual form. Doesn't that answer your question?

 

[binbagsss]

The steps you wrote are meaningless. The answer you wrote is of by a factor of one half. I wrote the usual form. Doesn't that answer your question?

yes it does, yes apologies i did miss the factor of 1/2- had a bad signal your reply did not load. ty. can you show me how you get to that properly, if it is meaningless to write:

$\frac{\partial \psi}{\partial z}=\frac{\partial \psi}{\frac{1}{2}\left(\partial x + i \partial y\right)}$ and try to simplify this..

 

[fresh_42]

yes it does, yes apologies i did miss the factor of 1/2- had a bad signal your reply did not load. ty. can you show me how you get to that properly, if it is meaningless to write:

$\frac{\partial \psi}{\partial z}=\frac{\partial \psi}{\frac{1}{2}\left(\partial x + i \partial y\right)}$ and try to simplify this..

https://en.wikipedia.org/wiki/Wirtinger_derivatives

It comes from $x=\dfrac{1}{2}(z+\overline{z})$ and $y=\dfrac{i}{2}(\overline{z}-z)$ and $d\psi =\dfrac{\partial \psi}{\partial x}dx +\dfrac{\partial \psi}{\partial y}dy.$

 

[binbagsss]

https://en.wikipedia.org/wiki/Wirtinger_derivatives

It comes from $x=\dfrac{1}{2}(z+\overline{z})$ and $y=\dfrac{i}{2}(\overline{z}-z)$ and $d\psi =\dfrac{\partial \psi}{\partial x}dx +\dfrac{\partial \psi}{\partial y}dy.$

ty