MHB How does Conway's Theorem 2.29 Derive the Simplified Expression for f(z)?

[Math Amateur]

I am reading John B. Conway's book, "Functions of a Complex Variable I" (Second Edition) ...

I am currently focussed on Chapter III Elementary Properties and Examples of Analytic Functions ... Section 2: Analytic Functions ... ...

I need help in fully understanding aspects of Theorem 2.29 ...

[NOTE: Notice that the statement of Theorem 2.29 follows the proof ... see below ... ]

Theorem 2,29 and its proof read as follows:

View attachment 7402
https://www.physicsforums.com/attachments/7403

In the above text by Conway, we read the following:

" ... ... Using the fact that $$u$$ and $$v$$ satisfy the Cauchy-Riemann equations it is easy to see that

$$\frac{ f( z + s + it ) - f(z) }{ s + it } = u_x(z) + iv_x(z) + \frac{ \phi(s, t) + i \psi(x, t) }{ s + it }$$ ... ... ... "

My question is as follows:

What is the (rigorous) process by which we arrive at

$$\frac{ f( z + s + it ) - f(z) }{ s + it } = u_x(z) + iv_x(z) + \frac{ \phi(s, t) + i \psi(x, t) }{ s + it }$$ ...?

That is ... why/how exactly does this follow ...
Hope someone can help ... I am aiming to have a rigorous understanding of the above proof ...

Peter***EDIT***I have been reflecting on the issue/problem above ...

Part of the answer may well be as follows:

$$\frac{ f(z + s + it) - f(z) }{ s + it } $$$$= \frac{ f(x + iy + s + it) - f(x + iy) }{ s + it }$$ $$= \frac{ [ u( x + s, y + t ) + iv( x + s, y + t ) ] - [ u(x, y) + iv(x, y) ] }{ s + it }$$


$$= \frac{ [ u( x + s, y + t ) - u(x, y) ] + i [ v( x + s, y + t ) - v(x, y) ] }{ s + it }$$$$= \frac{ [ u_x ( x, y )s + u_y (x, y)t + \phi (s, t) }{ s + it } + i \frac{ [v_x ( x, y )s + v_y (x, y)t + \psi (s, t) }{ s + it }$$$$ = \frac{ [ u_x ( x, y )s - v_x (x, y)t + \phi (s, t) }{ s + it } + i \frac{ [v_x ( x, y )s + u_x (x, y)t + \psi (s, t) }{ s + it }$$ $$= \frac{ [ u_x ( x, y )s + i u_x (x, y)t ] }{ s + it } + i \frac{ [v_x ( x, y )s + i v_x (x, y)t ] }{ s + it } + \frac{ [ \phi (s, t) + i \psi (s, t)] }{ s + it } $$$$= u_x(z) + iv_x(z) + \frac{ [ \phi (s, t) + i \psi (s, t)] }{ s + it }$$Is that correct?

======================================================================================It may help readers of the above post to have access to Conway's introduction to the Cauchy-Riemann conditions (necessity case) ... so I am providing the same ... as follows:
View attachment 7404
View attachment 7405

 

[Euge]

Re: Cauchy Riemann Equations - Proof of Sufficinecy ... Conway Theorem 2.29 ... Proof ...

***EDIT***I have been reflecting on the issue/problem above ...

Part of the answer may well be as follows:

$$\frac{ f(z + s + it) - f(z) }{ s + it } $$$$= \frac{ f(x + iy + s + it) - f(x + iy) }{ s + it }$$ $$= \frac{ [ u( x + s, y + t ) + iv( x + s, y + t ) ] - [ u(x, y) + iv(x, y) ] }{ s + it }$$


$$= \frac{ [ u( x + s, y + t ) - u(x, y) ] + i [ v( x + s, y + t ) - v(x, y) ] }{ s + it }$$$$= \frac{ [ u_x ( x, y )s + u_y (x, y)t + \phi (s, t) }{ s + it } + i \frac{ [v_x ( x, y )s + v_y (x, y)t + \psi (s, t) }{ s + it }$$$$ = \frac{ [ u_x ( x, y )s - v_x (x, y)t + \phi (s, t) }{ s + it } + i \frac{ [v_x ( x, y )s + u_x (x, y)t + \psi (s, t) }{ s + it }$$ $$= \frac{ [ u_x ( x, y )s + i u_x (x, y)t ] }{ s + it } + i \frac{ [v_x ( x, y )s + i v_x (x, y)t ] }{ s + it } + \frac{ [ \phi (s, t) + i \psi (s, t)] }{ s + it } $$$$= u_x(z) + iv_x(z) + \frac{ [ \phi (s, t) + i \psi (s, t)] }{ s + it }$$Is that correct?

It's correct! (Yes)

 

[Math Amateur]

Re: Cauchy Riemann Equations - Proof of Sufficinecy ... Conway Theorem 2.29 ... Proof ...

It's correct! (Yes)

Thanks Euge ...

I needed the confirmation!

Peter