Derivazione parziale complessa

[tommy1996q]

Vi avviso in anticipo che questa sarà una domanda molto banale, ma non sono sicuro di aver capito un argomento basilare di analisi complessa. Procedo con ordine. Abbiamo un complesso $z=x+iy$, definiamo il differenziale di $z$ e $\overline{z}$ come $dz=dx +idy$ e $d\overline{z}=dx-dy$. Da ciò ricaviamo che $dx=\frac{dz+d\overline{z}}{2}$ e $dy= \frac{dz-d\overline{z}}{2i}$. Ora, vorremmo definire cosa significano $\partial _z$ e $\partial _{\overline{z}}$. Visto che devo poter scrivere $df$ equivalentemente con $x$ e $y$ o con $z$ e $\overline{z}$, facendo due conti vedo che deve essere $\partial _z=\frac{\partial _x - i \partial _y}{2}$ e $\partial_{ \overline{z}}=\frac{\partial _x + \partial_ y}{2}$. Veniamo all'esercizio. Mi viene richiesto di calcolare la derivata parziale rispetto a $y$, $x$ e $\overline{z}$ di $z^2$. Se ho ben capito, per avere $\partial_y$ rispetto a $z$ e $\overline{z}$ dovrei invertire le formule sopra, e a questo punto otterrei $\partial_y=\frac{\partial_{\overline{z}}- \partial_z}{i}$, da cui otterrei che $\partial_y z^2=2iz$. Analogamente troverei che la derivata parziale rispetto a x torna $2z$, e questo tornerebbe col fatto che $\partial_{\overline{z}} z^2=0$, se la scrivo secondo quanto ho detto prima.

Il ragionamento è corretto? Nel caso, c'è un modo più rapido di ragionare o devo sempre fare queste sostituzioni?

[javicemarpe]

The partial derivative with respect to $y$ is correct (and also the partial derivative with respect to $x$). Anyways, I think that these identities are only useful when you want to compute $\patial_{\overline{z}}$ or $\partial_z$. I did my computations like this:

First, I write $f(z)=z^2$ as $f(x,y)=(x+iy)^2=x^2-y^2+2ixy$. Then, I compute the partial derivatives:
$$\partial_xf(x,y)=2x+2iy,\qquad \partial_y(x,y)=-2y+2ix.$$

Now, using the identities (which, actually, are definitions), we get:
$$\partial_\overline{z}f(x,y)=\frac{1}{2}(2x+2iy-2iy-2x)=0;\qquad \partial_zf(x,y)=\frac{1}{2}(2x+2iy+2iy+2x)=2x+2iy=2z.$$

That's the way it I do these things. I think that it's easier to think about a complex function $f$ (which we don't know whether it's holomorphic or not) as a vector function of the variables $x,y$. The operators $\partial_z$ and $\partial_\overline{z}$ are useful in order to translate this information in terms of the complex variable $z$.

By the way, it's easy to prove (and it's a good exercise to get used to these operators) that, if $f$ is holomorphic in $z_0$, then $\partial_{\overline{z}}(z_0)=0$ and $\partial_z(z_0)=f'(z_0)$.

[tommy1996q]

Thanks, at the end of your reply you meant $\partial_z (f(z_0))=f'(z_0)$ right? Also, do you consider $f'$ as the derivative of $f$ seen in $\mathbb{R}^2$? So something like $f(x,y)=u(x,y)+iv(x,y)$ and then $f'=u'+iv'$ with $u,v$ seen as functions from $\mathbb{R}^2$ to $\mathbb{R}$?

[javicemarpe]

Yes, I wanted to say that. On the other hand, I thought you knew what the derivative of a complex function is. The definition of derivative is the following:
$$f'(z_0)=\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}.$$

It is the same as the definition for real functions, but the fact that you're working with complex numbers gives you too many good properties that differentiable real functions don't have. It has something to do with differentiation in the sense of functions from $\matbb{R}^2$ to $\matbb{R}^2$ but it's not the same thing.