1. The problem statement, all variables and given/known data
I am told to prove a theorem by considering a single case of [itex]z_0=0[/itex]:
Suppose that a function [itex]f=u+iv[/itex] is defined in an open subset [itex]U[/itex] of the complex plane and that the partial derivatives [itex]u_x,u_y,v_x,v_y[/itex] exist everywhere in [itex]U[/itex]. If each of these partial derivatives is continuous at a point [itex]z_0[/itex] of [itex]U[/itex], then [itex]f[/itex] is differentiable in the real sense at [itex]z_0[/itex] and furthermore [itex]f(z)=f(z_0)+f_{z}(z_0)(z-z_0)+f_{\overline{z}}(z_0)(\overline{z}-\overline{z}_{0})+E(z)[/itex] for [itex]z\in{U}[/itex], where [itex]E:U\rightarrow\mathbb{C}[/itex] satisfies [itex]\lim\limits_{z\rightarrow{z_{0}}}\frac{|E(z)|}{|z-z_0|}=0[/itex]
in the manner of a previous proof, which involved using the mean value theorem.
2. Relevant equations
MVT; [itex]z[/itex] and [itex]\overline{z}[/itex] partials
3. The attempt at a solution
I'm pretty sure I have an answer, but I was hoping for an elucidation and perhaps a different way of thinking about.
Having been told to only consider the case of [itex]z_0=0[/itex] I have:
[itex]\lim\limits_{z\rightarrow{z_{0}}}\frac{|E(z)|}{|z-z_0|}=\lim\limits_{z\rightarrow{z_{0}}}\frac{|u_x(a)x+u_y(b)y+i(v_x(c)+ v_y(d))-\alpha\cdot{z}-\beta\cdot{\overline{z}}|}{|z|}[/itex]
I have those unexplained terms from taking [itex]u(x,y)-u(0,y)+u(0,y)-u(0,0)[/itex] and [itex]u(x,y)-u(x,0)+u(x,0)-u(0,0)[/itex] and applying the mean value theorem, yielding [itex]u(x,y)-u(0,0)=u_x(a)x[/itex] and then doing that three more times.
Now it becomes apparent to me that kinds of functions that will allow me to satisfy the condition that this "Error" term (as my book puts it) goes to 0 will be the two functions [itex]\alpha=f_{\overline{z}}(z_0)[/itex] and [itex]\beta=f_{z}(z_0)[/itex], as they cancel some of their mutual terms and leave me with [itex]-u_x(z)x-u_y(z)y-i(v_x(z)x+v_y(z)y)[/itex]
What concerns me, however, is that I do not know if it is always the case that these are the only two functions that will satisfy this condition. Is there some way I might go about demonstrating that this is a unique answer? Or is there some way that I am supposed to arrive at the conclusion that the two complex constants are [itex]z[/itex] and [itex]\overline{z}[/itex] partials?
I am told to prove a theorem by considering a single case of [itex]z_0=0[/itex]:
Suppose that a function [itex]f=u+iv[/itex] is defined in an open subset [itex]U[/itex] of the complex plane and that the partial derivatives [itex]u_x,u_y,v_x,v_y[/itex] exist everywhere in [itex]U[/itex]. If each of these partial derivatives is continuous at a point [itex]z_0[/itex] of [itex]U[/itex], then [itex]f[/itex] is differentiable in the real sense at [itex]z_0[/itex] and furthermore [itex]f(z)=f(z_0)+f_{z}(z_0)(z-z_0)+f_{\overline{z}}(z_0)(\overline{z}-\overline{z}_{0})+E(z)[/itex] for [itex]z\in{U}[/itex], where [itex]E:U\rightarrow\mathbb{C}[/itex] satisfies [itex]\lim\limits_{z\rightarrow{z_{0}}}\frac{|E(z)|}{|z-z_0|}=0[/itex]
in the manner of a previous proof, which involved using the mean value theorem.
2. Relevant equations
MVT; [itex]z[/itex] and [itex]\overline{z}[/itex] partials
3. The attempt at a solution
I'm pretty sure I have an answer, but I was hoping for an elucidation and perhaps a different way of thinking about.
Having been told to only consider the case of [itex]z_0=0[/itex] I have:
[itex]\lim\limits_{z\rightarrow{z_{0}}}\frac{|E(z)|}{|z-z_0|}=\lim\limits_{z\rightarrow{z_{0}}}\frac{|u_x(a)x+u_y(b)y+i(v_x(c)+ v_y(d))-\alpha\cdot{z}-\beta\cdot{\overline{z}}|}{|z|}[/itex]
I have those unexplained terms from taking [itex]u(x,y)-u(0,y)+u(0,y)-u(0,0)[/itex] and [itex]u(x,y)-u(x,0)+u(x,0)-u(0,0)[/itex] and applying the mean value theorem, yielding [itex]u(x,y)-u(0,0)=u_x(a)x[/itex] and then doing that three more times.
Now it becomes apparent to me that kinds of functions that will allow me to satisfy the condition that this "Error" term (as my book puts it) goes to 0 will be the two functions [itex]\alpha=f_{\overline{z}}(z_0)[/itex] and [itex]\beta=f_{z}(z_0)[/itex], as they cancel some of their mutual terms and leave me with [itex]-u_x(z)x-u_y(z)y-i(v_x(z)x+v_y(z)y)[/itex]
What concerns me, however, is that I do not know if it is always the case that these are the only two functions that will satisfy this condition. Is there some way I might go about demonstrating that this is a unique answer? Or is there some way that I am supposed to arrive at the conclusion that the two complex constants are [itex]z[/itex] and [itex]\overline{z}[/itex] partials?
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