1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ In this case, the fact that $(a+b)(a-b)=a^2-b^2$ is useful. $\endgroup$ – kimchi lover Apr 20 at 14:12
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Thank you, that makes sense! $\endgroup$ – Harman Apr 20 at 21:10
1
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ Thanks so much! $\endgroup$ – Harman Apr 20 at 21:09
1
$\begingroup$

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 $$z^n-z_0^n=(z-z_0)\bigl(z^{n-1}+z^{n-2}z_0+\ldots+z_0^{n-1}\bigr)\ .$$

$\endgroup$
  • $\begingroup$ I didn't know of this result. That's going to help tremendously, thank you! $\endgroup$ – Harman Apr 20 at 21:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.