Let $f(z)$ be a complex function, where $z=x+iy$ and that we can always write $f(z)=u(x,y)+iv(x,y)$, where $u$ and $v$ are real functions involving two independent variables.
Let $\bar{z}=x-iy$. (The complex conjugate of variable $z$)
If $f(z)$ is analytic (holomorphic) then it is obvious that the Cauchy Riemann conditions satisfy and we can derive them from standard definition of differentiation.
$\lim_{\Delta z\rightarrow 0} \frac{f(z+\Delta z)-f(z)}{\Delta z}$
(which is found in most of the books)
Put in another way holomorphic function satisfy $\frac{\partial f}{\partial \bar{z}}=0$ (That is holomorphic function does not involve the variable $\bar{z}$ and it involves only $z$)
In a video lecture I encountered that we can write $\frac{\partial f}{\partial \bar{z}}=$ $\frac{\partial f}{\partial x}$ – $i\frac{\partial f}{\partial y}=0$ and by substituting $f(z)=u(x,y)+iv(x,y)$ and then by equating real and imaginary parts we can derive Cauchy Riemann conditions.
My question is, "Is it correct to write $\frac{\partial f}{\partial \bar{z}}=$ $\frac{\partial f}{\partial x}$ – $i\frac{\partial f}{\partial y}=0$"
If so,
- Please explain $\frac{\partial f}{\partial \bar{z}}=$ $\frac{\partial f}{\partial x}$ – $i\frac{\partial f}{\partial y}$
- How to proceed to derive the Cauchy Riemann Conditions from here?
Also it is said that $z$ and $\bar{z}$ are linearly independent of each other.
3)How they are linearly independent (or) what it precisely means?
Best Answer
I don't know what definitions you know, but I'll assume you know real calculus and linear algebra.
First, note that $z = x+iy = (1,1) \neq (1,-1) = x-iy =\overline{z}$, is direct the linear independence of $z$ with $\overline{z}$ (definition of linear algebra: no exists a scalar such that $\lambda z =\overline{z}$). Even, as vectors of $\mathbb{R}^{2}$ they form orthogonal directions.
By partial derivation of $f(x,y)=u(x,y)+iv(x,y)$, i.e., $$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y}$$ the above equality due to the uniqueness of the limit.
I define \begin{equation*} f'(z)=f'(x,y)=\frac{\partial u}{\partial x}(x,y)+i\frac{\partial v}{\partial x}(x,y) \end{equation*} and
\begin{equation*} \begin{split} \frac{\partial}{\partial z}&=\frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)\\ \frac{\partial}{\partial \overline{z}}&=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right) \end{split} \end{equation*} it is possible to verify that $f'(z)=\frac{\partial f}{\partial z}$ and $\frac{\partial f}{\partial \overline{z}}=0$ by C-R.
Proof: ($\Leftarrow$) Let $f:U\subseteq\mathbb{C}\to\mathbb{C}$, if $\frac{\partial f}{\partial\overline{z}}=0$, then: \begin{equation*} \frac{1}{2}\left(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\right)=0 \end{equation*} then (using $f= u+iv$: \begin{eqnarray*} 0=\frac{\partial }{\partial x}(u+iv)+i\frac{\partial}{\partial y}(u+iv) &=&\frac{\partial u}{\partial x}+i\frac{\partial u}{\partial y}+ i\frac{\partial v}{\partial x}-\frac{\partial v}{\partial y}\\ &=&\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) \end{eqnarray*} Then (separating in real and imaginary parts): $$\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)=0$$ and $$\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)=0$$
concluding C-R.
($\Rightarrow$) By a construction of $f'(z)$, assuming C-R. Then: \begin{eqnarray*} f'(z)=f'(x,y) &=&\frac{\partial u}{\partial x}(x,y)+i\frac{\partial v}{\partial x}(x,y)\\ &=&\frac{1}{2}\left(\frac{\partial u}{\partial x}(x,y)+i\frac{\partial v}{\partial x}(x,y)\right)+\frac{1}{2}\left(\frac{\partial u}{\partial x}(x,y)+i\frac{\partial v}{\partial x}(x,y)\right) \\ &=&\frac{1}{2}\left(\frac{\partial u}{\partial x}(x,y)+i\frac{\partial v}{\partial x}(x,y)\right)+\frac{1}{2}\left(\frac{\partial v}{\partial y}(x,y)-i\frac{\partial u}{\partial y}(x,y)\right) \\ &=&\frac{1}{2}\left(\frac{\partial u}{\partial x}(x,y)+i\frac{\partial v}{\partial x}(x,y)\right)+\frac{1}{2}\left(\frac{\partial v}{\partial y}(x,y)-i\frac{\partial u}{\partial y}(x,y)\right) \\ &=&\dots\text{ sorting out, check it }\dots\\ &=&\frac{\partial f}{\partial z}+\frac{\partial f}{\partial \overline{z}} \end{eqnarray*}
But $f'(z) = \frac{\partial f}{\partial z}$, then $\frac{\partial f}{\partial \overline{z}}=0$.