# Error in Complex Analysis Proof

• May 24th 2007, 07:20 PM
ThePerfectHacker
Error in Complex Analysis Proof
My Complex Analysis book states the following interesting theorem:

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

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

Here is how the proof begins.

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

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).
• May 24th 2007, 08:48 PM
CaptainBlack
Quote:

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

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

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

Here is how the proof begins.

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

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
• May 25th 2007, 06:20 AM
ThePerfectHacker
Quote:

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 $\displaystyle f(x)=u(a+hx,b+k): [0,1]\to \mathbb{R}$. We can see that this is a well-defined function on this interval and differenciable on $\displaystyle (0,1)$. By mean value theorem there exists $\displaystyle x_0\in (0,1)$ so that, $\displaystyle \frac{f(1)-f(0)}{1-0} = f'(x_0)$. Thus, $\displaystyle f(1)-f(0)=f'(x_0)$. Write out it out fully, $\displaystyle u(a+h,b+k)-u(a,b+k)=u_x(a+x_0h,b+k)$.
• May 25th 2007, 08:24 AM
CaptainBlack
Quote:

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

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

$\displaystyle f(a+h) - f(a) = h f'(u_0)$

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

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

$\displaystyle x_0$ in $\displaystyle (0,1)$

RonL