# Corollary from Cauchy's integral formula

I have doubts about the proof given in Stein's complex analysis textbook from the following Cauchy's integral formula corollary:

Corollary: If ##f## is holomorphic in an open set ##Ω##, then ##f## has infinitely many complex derivatives in ##Ω##. Moreover, if ##C⊂Ω## is a circle whose interior is also contained in ##Ω##, then

##f^{(n)}(z)=\dfrac{n!}{2πi}∫_C \dfrac{f(ζ)}{(ζ−z)^{n+1}}dζ## for all ##z## in the interior of ##C##
I'll write the complete proof copied from Stein's textbook and then I'll ask my question:

Proof The proof is by induction on ##n##, the case ##n=0## being simply the Cauchy integral formula. Suppose that ##f## has up to ##n−1## complex derivatives and that ##f^{(n−1)}(z)=\dfrac{(n−1)!}{2πi}∫_C \dfrac{f(ζ)}{(ζ−z)^n}dζ##.

Now for ##h## small, the difference quotient for ##f^{((n−1)}## takes the form

(1) ##f^{(n−1)}(z+h)−f^{(n−1)}(z)h=\dfrac{(n−1)!}{2πi}∫_C f(ζ)\dfrac{1}{h}[\dfrac{1}{(ζ−z−h)^n}-\dfrac{1}{(ζ−z)^n}]dζ##
We now recall that ##A^n−B^n=(A−B)[A^{n−1}+A^{n−2}B+⋅⋅⋅+AB^{n−2}+B^{n−1}]##
With ##A=\dfrac{1}{(ζ−z−h)}## and ##B=\dfrac{1}{(ζ−z)}##, we see that the term in brackets in equation (1) is equal to

##\dfrac{h}{(ζ−z−h)(ζ−z)}[A^{n−1}+A^{n−2}B+⋅⋅⋅+AB^{n−2}+B^{n−1}]##. But observe that if ##h## is small, then ##z+h## and ##z## stay at a finite distance from the boundary circle ##C##, so in the limit as ##h## tends to ##0##, we find that the quotient converges to

##\dfrac{(n−1)!}{2πi}∫_C f(ζ)[\dfrac{1}{(ζ−z)^2}][\dfrac{n}{(ζ−z)^{n−1}}]dζ=\dfrac{n!}{2πi}∫_C \dfrac{f(ζ)}{(ζ−z)^{n+1}}dζ##,

which completes the induction argument and proves the theorem.

My question is: how is it that ##\dfrac{h}{(ζ−z−h)(ζ−z)}[A^{n−1}+A^{n−2}B+⋅⋅⋅+AB^{n−2}+B^{n−1}]## converges to ##[\dfrac{1}{(ζ−z)^2}][\dfrac{n}{(ζ−z)^{n−1}}]## given the condition of ##h## being sufficiently small?

I've tried to manipulate the first expression in order to get to the latter but I couldn't arrive to anything. I would be greatly appreciate if someone could clearly explain me this.

## Answers and Replies

STEMucator
Homework Helper
I have doubts about the proof given in Stein's complex analysis textbook from the following Cauchy's integral formula corollary:

Corollary: If ##f## is holomorphic in an open set ##Ω##, then ##f## has infinitely many complex derivatives in ##Ω##. Moreover, if ##C⊂Ω## is a circle whose interior is also contained in ##Ω##, then

##f^{(n)}(z)=\dfrac{n!}{2πi}∫_C \dfrac{f(ζ)}{(ζ−z)^{n+1}}dζ## for all ##z## in the interior of ##C##
I'll write the complete proof copied from Stein's textbook and then I'll ask my question:

Proof The proof is by induction on ##n##, the case ##n=0## being simply the Cauchy integral formula. Suppose that ##f## has up to ##n−1## complex derivatives and that ##f^{(n−1)}(z)=\dfrac{(n−1)!}{2πi}∫_C \dfrac{f(ζ)}{(ζ−z)^n}dζ##.

Now for ##h## small, the difference quotient for ##f^{((n−1)}## takes the form

(1) ##f^{(n−1)}(z+h)−f^{(n−1)}(z)h=\dfrac{(n−1)!}{2πi}∫_C f(ζ)\dfrac{1}{h}[\dfrac{1}{(ζ−z−h)^n}-\dfrac{1}{(ζ−z)^n}]dζ##
We now recall that ##A^n−B^n=(A−B)[A^{n−1}+A^{n−2}B+⋅⋅⋅+AB^{n−2}+B^{n−1}]##
With ##A=\dfrac{1}{(ζ−z−h)}## and ##B=\dfrac{1}{(ζ−z)}##, we see that the term in brackets in equation (1) is equal to

##\dfrac{h}{(ζ−z−h)(ζ−z)}[A^{n−1}+A^{n−2}B+⋅⋅⋅+AB^{n−2}+B^{n−1}]##. But observe that if ##h## is small, then ##z+h## and ##z## stay at a finite distance from the boundary circle ##C##, so in the limit as ##h## tends to ##0##, we find that the quotient converges to

##\dfrac{(n−1)!}{2πi}∫_C f(ζ)[\dfrac{1}{(ζ−z)^2}][\dfrac{n}{(ζ−z)^{n−1}}]dζ=\dfrac{n!}{2πi}∫_C \dfrac{f(ζ)}{(ζ−z)^{n+1}}dζ##,

which completes the induction argument and proves the theorem.

My question is: how is it that ##\dfrac{h}{(ζ−z−h)(ζ−z)}[A^{n−1}+A^{n−2}B+⋅⋅⋅+AB^{n−2}+B^{n−1}]## converges to ##[\dfrac{1}{(ζ−z)^2}][\dfrac{n}{(ζ−z)^{n−1}}]## given the condition of ##h## being sufficiently small?

I've tried to manipulate the first expression in order to get to the latter but I couldn't arrive to anything. I would be greatly appreciate if someone could clearly explain me this.

##\frac{h}{(ζ−z−h)(ζ−z)}[A^{n−1}+A^{n−2}B+⋅⋅⋅+AB^{n−2}+B^{n−1}] \space (1)##

From what I can see, the ##h## in the numerator of this expression will first cancel the ##\frac{1}{h}## term in the integrand.

As ##h## gets small, ##(1) → [\dfrac{1}{(ζ−z)(ζ−z)}][\dfrac{n}{(ζ−z)^{n−1}}]##.

If you plug in ##A## and ##B## into the formula for ##A^n - B^n##, what is the result if ##h## is small?

EDIT: As another hint, take a look at the first two terms ##A^{n−1} + A^{n−2}B## and the last two terms ##AB^{n−2} + B^{n−1}##.

If you add the first two terms when ##h## is small, the result is ##\frac{2}{(ζ-z)^{n-1}}##. If you add the last two terms, what do you get? Notice you're really just adding ##\frac{1}{(ζ-z)^{n-1}}## a total of ##n## times.

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##\frac{h}{(ζ−z−h)(ζ−z)}[A^{n−1}+A^{n−2}B+⋅⋅⋅+AB^{n−2}+B^{n−1}] \space (1)##

From what I can see, the ##h## in the numerator of this expression will first cancel the ##\frac{1}{h}## term in the integrand.

As ##h## gets small, ##(1) → [\dfrac{1}{(ζ−z)(ζ−z)}][\dfrac{n}{(ζ−z)^{n−1}}]##.

If you plug in ##A## and ##B## into the formula for ##A^n - B^n##, what is the result if ##h## is small?

EDIT: As another hint, take a look at the first two terms ##A^{n−1} + A^{n−2}B## and the last two terms ##AB^{n−2} + B^{n−1}##.

If you add the first two terms when ##h## is small, the result is ##\frac{2}{(ζ-z)^{n-1}}##. If you add the last two terms, what do you get? Notice you're really just adding ##\frac{1}{(ζ-z)^{n-1}}## a total of ##n## times.

Oh, what a fool of me, I should have put ##h[\dfrac{1}{(ζ−z)^2}][\dfrac{n}{(ζ−z)^{n−1}}]## but, as you've said, then ##h## will cancel with ##\dfrac{1}{h}##. Thanks for your answer, now I see why it converges to that expression.