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Prove that if $g(z)$ is holomorphic everywhere inside and on a simple closed
contour $C$, taken in a positive sense, and $z$ is any point interior to $C$, then
$$g'(z)=dfrac12piiint_Cdfracg(u)du(u-z)^2$$

So far, I know this: If you let $f$ be holomorphic everywhere inside and on a simple closed contour $C$ taken in the positive sense, and if $z_0$ is any point interior to $C$, then

$$f(z_0)=frac12piiint_Cfracf(z)dzz-z_0$$

However, I do not know how to attempt this proof using the knowledge I currently have.

For another approach (without power series), we use the definition of derivative. You want to compute

$f'(z_0)= lim_z to z_0 frac1z - z_0 left [ frac12pi i int_gamma fracf(w)w - z : dw - frac12pi i int_gamma fracf(w)w - z_0 : dw right ].$

The right hand side of this simplifies to

$lim_zto z_0 frac12pi i int_gamma fracf(w)(w - z)(w - z_0) : dw.$

Now the issue is the interchange of the integral with the limit, for if this is valid, then you get

$lim_zto z_0 frac12pi i int_gamma fracf(w)(w - z)(w - z_0) : dw=frac12pi i int_gamma lim_zto z_0fracf(w)(w - z)(w - z_0) : dw=int_gammafracf(w)(w - z_0)^2 : dw$, as desired.

All that remains therefore is to justify the limit/integral switch: a fancy way would be to use the dominated convergence theorem, which applies since $f$ is bounded on $gamma$ and there is a number $delta$ such that $d(z_0,gamma)=delta>0$ and we may also assume that $d(z,gamma)>delta$ for all $z$ close enough to $z_0.$

But you can also argue directly: Let $Mmax_win gamma |, delta$ as in the above remark and $ell(gamma)$ the length of $gamma.$ Then,

$beginalignleft |int_gamma fracf(w)(w - z)(w - z_0) : dw-int_gammafracf(w)(w - z_0)^2 : dwright |le fracMdeltacdot left |int_gamma frac1(w - z) : dw-int_gammafrac1(w - z_0) : dwright |le fracMdeltacdot int_gammaleft|fracz-z_0(w-z)(w-z_0)right|dwle fracMdelta^3int_gamma|z-z_0|dwle fracMcdot ell((gamma)delta^3|z-z_0|endalign$

To finish, let $zto z_0.$

For the $kth$ derivative, this calculation goes through virtually unchanged.

To prove the general case you can take limits of difference quotients and use induction (or expand the Cauchy integral in a power series as I do below). In this case, since you only want the formula for the first derivative, you can simply pass the derivative through the integral; namely,
$$f^prime(z) = fracddzleft(frac12pi iint_C fracf(zeta)zeta - z dzeta right) = frac12pi iint_C fracddzleft(fracf(zeta)zeta - zright) dzeta.$$

Edit: The final step is to compute
$$fracddzleft(fracf(zeta)zeta-zright) = f(zeta)fracddzleft(frac1zeta-zright) = fracf(zeta)(zeta-z)^2.$$

Edit 2: I will actually not do the inductive proof, but rather a proof based on expanding a Cauchy integral as a power series. If you want to see the inductive proof, there are plenty of websites with said proof.

First, note that we can write
$$f(z) = frac12pi iint_gamma fracf(zeta)zeta - z dzeta,$$

Fix a point $z_0$ off $gamma$, set $R = mathrmdist(z_0,gamma)$ and let $z$ be a point satisfying $lvert z - z_0 rvert < R$. Notice that
$$frac1zeta-z = frac1(zeta - z_0) - (z - z_0) = frac1zeta - z_0left(frac11-fracz-z_0zeta-z_0right) = sum_n=0^infty frac(z-z_0)^n(zeta-z_0)^n+1,$$
where the power series converges locally uniformly wrt $zeta$ in $bigglvertfracz-z_0zeta-z_0biggrvert<1$ and so converges uniformly on $gamma$. Since $f$ is bounded on $gamma$, the series
$$frac12pi isum_n=0^infty fracf(zeta)(z-z_0)^n(zeta-z_0)^n+1$$
converges uniformly on $gamma$ to $fracf(zeta)zeta - z$. Therefore, we can integrate the power series term-by-term over $gamma$:
$$f(z) = sum_n=0^infty left(frac12pi iint_gamma fracf(zeta)(zeta-z_0)^n+1 dzetaright) (z-z_0)^n.$$
Therefore, the terms $frac12pi iint_gamma fracf(zeta)(zeta-z_0)^n+1 dzeta$ are the coefficients in the power series for $f$ centered about $z=z_0$ and therefore
$$frac12pi iint_gamma fracf(zeta)(zeta-z_0)^n+1 dzeta = fracf^(n)(z_0)n!.$$

Edit 3: This is a bit of a tangent, so feel free to skip it if you aren't in the mood to indulge me. One reason why I like (and chose to present) this proof is that, in addition to proving the Cauchy integral formula (for derivatives), it actually can (very clearly) tell us something about Cauchy integrals in general. Let $gamma:[a,b] to mathbbC$ be a piecewise-$C^1$ closed curve and let $phi: G supset gamma to mathbbC$ (slightly abusing notation) be continuous on $gamma$. Define
$$F(z) = int_gamma fracphi(zeta)zeta - z dzeta.$$
Then, the (essentially) same proof as above shows that $F$ is analytic on $mathbbCsetminusgamma$. So, we can start with a function which is merely continuous, define its Cauchy integral and get back an analytic function. It also, of course, tells us that all holomorphic functions are analytic.