# Thread: Error in Complex Analysis Proof

1. ## Error in Complex Analysis Proof

My Complex Analysis book states the following interesting theorem:

Let $N(c,r)$ be a open disk in $\mathbb{C}$. Let $f$ be a complex function whose domain contains the open disk. Let $f(x+iy) = u(x,y)+iv(x,y) \mbox{ for }x+iy \in N(c,r)$. If the parital derivaties $u_x,u_y,v_x,v_y$ all exist and are continous on $N(c,r)$ and at $c \in N(c,r)$ the Cauchy-Riemann equations are satisfied then $f$ is differenciable at $c$.

This is a partial converse for the necessay Cauchy-Riemann condition.

Here is how the proof begins.

Let $c=a+bi$ and let $h,k$ be so that $|h+ik| (that is they lie in the open disk).
Then,
$u(a+b,b+k)-u(a,b) = [u(a+h,b+k)-u(a,b+k)] + [u(a,b+k)-u(a,b)]$
By the Mean Value Theorem,
$=hu_x(a+x_0h,b+k) + kv_y(a,b+y_0k)$ where $x_0,y_0\in (0,1)$.

And it goes on....

My problem is that, they did not use the MVT properly. Do you agree? (The problem is that this is one of those books which likes to skip steps).

2. Originally Posted by ThePerfectHacker
My Complex Analysis book states the following interesting theorem:

Let $N(c,r)$ be a open disk in $\mathbb{C}$. Let $f$ be a complex function whose domain contains the open disk. Let $f(x+iy) = u(x,y)+iv(x,y) \mbox{ for }x+iy \in N(c,r)$. If the parital derivaties $u_x,u_y,v_x,v_y$ all exist and are continous on $N(c,r)$ and at $c \in N(c,r)$ the Cauchy-Riemann equations are satisfied then $f$ is differenciable at $c$.

This is a partial converse for the necessay Cauchy-Riemann condition.

Here is how the proof begins.

Let $c=a+bi$ and let $h,k$ be so that $|h+ik| (that is they lie in the open disk).
Then,
$u(a+b,b+k)-u(a,b) = [u(a+h,b+k)-u(a,b+k)] + [u(a,b+k)-u(a,b)]$
By the Mean Value Theorem,
$=hu_x(a+x_0h,b+k) + kv_y(a,b+y_0k)$ where $x_0,y_0\in (0,1)$.

And it goes on....

My problem is that, they did not use the MVT properly. Do you agree? (The problem is that this is one of those books which likes to skip steps).
What exactly is your objection, it looks OK to me.

RonL

3. Originally Posted by CaptainBlank
What exactly is your objection, it looks OK to me.
Tell me then then how you got it?

I defined a function of $f(x)=u(a+hx,b+k): [0,1]\to \mathbb{R}$. We can see that this is a well-defined function on this interval and differenciable on $(0,1)$. By mean value theorem there exists $x_0\in (0,1)$ so that, $\frac{f(1)-f(0)}{1-0} = f'(x_0)$. Thus, $f(1)-f(0)=f'(x_0)$. Write out it out fully, $u(a+h,b+k)-u(a,b+k)=u_x(a+x_0h,b+k)$.

4. Originally Posted by ThePerfectHacker
Tell me then then how you got it?

I defined a function of $f(x)=u(a+hx,b+k): [0,1]\to \mathbb{R}$. We can see that this is a well-defined function on this interval and differenciable on $(0,1)$. By mean value theorem there exists $x_0\in (0,1)$ so that, $\frac{f(1)-f(0)}{1-0} = f'(x_0)$. Thus, $f(1)-f(0)=f'(x_0)$. Write out it out fully, $u(a+h,b+k)-u(a,b+k)=u_x(a+x_0h,b+k)$.
$
f(a+h) - f(a) = h f'(u_0)
$

$u_0$ in $(0,h)$, or:

$
f(a+h) - f(a) = h f'(x_0 h)
$

$x_0$ in $(0,1)$

RonL