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I am reading Complex analysis By Stein and Shakarchi.
In that I came across one proposition which I do not understand I thought for some point but Do not know I correct or wrong .

If f is holomorphic function at $z_0$,then $$\frac{\partial f}{\partial \bar{z}}(z_0)=0$$ and $$f'(z_0)=\frac{\partial f}{\partial {z}}(z_0)=2 \frac{\partial u}{\partial {z}}(z_0)$$
Also if we write F(x,y)=f(z) then F is differentible in sense of real varible and $$detJ_F(x_0,y_0)=|f'(z_0)|^2$$


Proof: Before this proof There is mention of 2 operator as enter image description here

I think as follows for this as
$f(x,y)=z$ and $x=\frac{z+\bar{z}}{2}$ and $y=\frac{z-\bar{z}}{2i}$
$\frac{\partial f}{\partial z}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial z}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial z}$
=$\frac{1}{2}(\frac{\partial f}{\partial x}+\frac{\partial f}{i\partial y})$
Is this right? Other is also similarly.
By Cauchy Riemann Equation $u_x=v_y$ and $u_y=-v_x$
. In book it says that by CR equation $\frac{\partial f}{\partial \bar{z}}=0$
$\frac{\partial f}{\partial \bar{z}}=\frac{1}{2}(\frac{\partial f}{\partial x}-\frac{\partial f}{i\partial y})$
Now ,$f'(z_0)=\frac{1}{2}(\frac{\partial f}{\partial x}+\frac{\partial f}{i\partial y})(z_0)=\frac{\partial f}{\partial z }$
$f(z)=u(x,y)+iv(x,y) $
so $f'(z)=\frac{1}{2}(\frac{\partial u}{\partial x}+\frac{\partial u}{i\partial y})$+$i(\frac{\partial v}{\partial x}+\frac{\partial v}{i\partial y)}=u_x-iu_y$ By using CR
enter image description here

I know that Jacobian derivative is matrix but Here they associate some complex function How?
I think that Matrix of Jacobian if we replace terms of v by CR equation and then we can associate complex number as we know that there is isomorphism between Special Type of 2*2 matrix with complex number (I don't know how to write matrix that why not giving explicit isomorphism)
At last det$J_F(x_0,y_0)=u_xv_y-u_yv_x=u_x^2+u_y^2=|2u_z|^2$ Any help will be appreciated

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If $f(x,y)=u(x,y)+v(x,y)i$, then\begin{align}\frac{\partial f}{\partial \overline z}(x+yi)&=\frac12\left(\frac{\partial\bigl(u(x,y)+v(x,y)i\bigr)}{\partial x}-\frac1i\frac{\partial\bigl(u(x,y)+v(x,y)i\bigr)}{\partial y}\right)\\&=\frac12\bigl(u_x(x,y)+v_x(x,y)i+u_y(x,y)i-v_y(x,y)\bigr)\\&=\frac{u_x(x,y)-v_y(x,y)}2+\frac{u_y(x,y)+v_x(x,y)}2i\end{align}and therefore$$\frac{\partial f}{\partial \overline z}(x+yi)=0\iff(x,y)\text{ is a solution of the Cauchy-Riemann equations.}$$

On the other hand, if $z_0=x_0+y_0i$, then\begin{align}f'(z_0)&=u_x(x_0,y_0)+v_x(x_0,y_0)i\\&=\frac{u_x(x_0,y_0)+v_y(x_0,y_0)}2+\frac{v_x(x_0,y_0)-u_y(x_0,y_0)}2i\\&=\frac{u_x(x_0,y_0)+v_x(x_0,y_0)i}2+\frac1i\frac{u_y(x_0,y_0)+v_y(x_0,y_0)i}2\\&=\frac{\partial f}{\partial z}(x_0+y_0i).\end{align}

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