How to prove that my generalisation is true?

Question

I've been working with holomorphic functions, and noticed, that for polynomial functions, like $(z+n)^4$, the differentiated real part $4a^3+6an-12ab^2+12a^2n+4n^3-12b^2n$ was the same as the real part of $(z+n)^3$ $a^3+n^3+3a^2n-3b^2a-3b^2n+3n^2a$ multiplied by 4. I see that the pattern repeats for the degrees 1, 2, 3, 4, yet I don't know how to prove it. It reminds me of the chain rule, but is it enough to just use it as a justification?

Answer 1

If $f(z)=u(a, b)+iv(a, b)$, where $a=\Re(z), b=\Im(z)$, existence of:

$$\frac{df}{dz}(z_0)=\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}$$

($z_0=a_0+ib_0$) implies that the same limit exists when we let "only" $a$ approach $a_0$ (and keep $b=b_0$ constant, basically approaching $z_0$ from the "left/right" direction), i.e.

$$\begin{align}\frac{df}{dz}(z_0)&=\lim_{a\to a_0}\frac{u(a, b_0) + iv(a+b_0)-u(a_0, b_0)-iv(a_0, b_0)}{a-a_0}\\&=\lim_{a\to a_0}\frac{u(a, b_0)-u(a_0,b_0)}{a-a_0} + i\lim_{a\to a_0}\frac{v(a, b_0)-v(a_0,b_0)}{a-a_0}\\&=\frac{\partial u}{\partial a}(a_0, b_0) + i\frac{\partial v}{\partial a}(a_0, b_0)\end{align}$$

This then implies the equality:

$$\Re(\frac{df}{dz}(z_0))=\frac{\partial u}{\partial a}(a_0, b_0)$$

and also:

$$\Im(\frac{df}{dz}(z_0))=\frac{\partial v}{\partial a}(a_0, b_0)$$

In your example, the "differentiated real part" (i.e. $\frac{\partial u}{\partial a}$) is equal to the real part of the differentiated function (i.e. $\Re(\frac{df}{dz})$) where $\frac{df}{dz}=4(z+n)^3$ for $f(z)=(z+n)^4$).

Exercise for you: Apply the same logic, but then vary only $b$ and keep $a=a_0$ constant. Derive two new equalities, and conclude the validity of Cauchy-Riemann equations (https://en.wikipedia.org/wiki/Cauchy%E2%80%93Riemann_equations).