Differentiating a complex function using the definition

Question

I need to differentiate the complex function $f(z)=z^2+z$.

I know that the definition of a derivative is $f'(z)=\frac{f(z)-f(z_0)}{z-z_0}$. In this case, $f'(z)=\frac{(z^2+z)-(z_0^2+z_0)}{z-z_0}$.

According to the solution, the numerator factorises into $(z-z_0)(z+z_0+1)$. I am assuming that $(z-z_0)$ was factored out so that cancellation could be performed with the denominator, but I am struggling to understand how this factorisation was carried out, since this is not a conventional factorisation which I am accustomed to performing.

I understand how to proceed from here: $(z-z_0)$ cancels and we are left with $\lim_{z \to z_0}(z+z_0+1)=2z_0+1$, and thus $f'(z)=2z+1$.

So, the only step which I am struggling to grasp is the factorisation step! How do I perform a factorisation like this?

Answer 1

$$(z^2+z)-(z_0^2+z_0)=(z^2-z_0^2)+(z-z_0)=(z-z_0)(z+z_0)+(z-z_0)=(z-z_0)(z+z_0+1)$$

Answer 2

Just use the fact that\begin{align}(z^2+z)-({z_0}^2+z_0)&=z^2-{z_0}^2+z-z_0\\&=(z-z_0)(z+z_0)+z-z_0\\&=(z-z_0)(z+z_0+1).\end{align}

Answer 3

Whenever $z\mapsto f(z)$ is a polynomial such a factorization is possible. For the proof it is sufficient to observe that for all $n\in{\mathbb N}_{\geq0}$ one has $$z^n-z_0^n=(z-z_0)\bigl(z^{n-1}+z^{n-2}z_0+\ldots+z_0^{n-1}\bigr)\ .$$