<s>[INST] If $z = 2 - 3i$, show that $z^2 - 4z + 13 = 0$ and hence find the value of $4z^3 - 3z^2 + 169$ This is an example given in my book, and it seems to involve division by 0, so I'm rather confused 
Here is how it's done in the book:
We have $z = 2 - 3i $
So, $z - 2 = -3i$
Now, ${(z-2)}^2 = {(-3i)}^2$
Or, $z^2 + 4 - 4z = 9i^2$
Or $z^2 -4z + 13 = 0$
Now, $4z^3 - 3z^2 + 169 = 4z(z^2 - 4z + 13) + 13(z^2 -4z + 13)$
In the above step, clearly $4z^3 - 3z^2 + 169$ has been divided by $z^2 - 4z + 13$ to obtain the factors. But $z^2 - 4z + 13 $ is equal to $0$, so wouldn't the division step give an undefined answer ?
 [/INST] Alt. hint:  if $z = 2 - 3i$, then $z+\bar z = 2 \operatorname{Re}(z) = 4$ and $ z \bar z = |z|^2 = 2^2+3^2=13$. It follows by Vieta's formulas that $z$ and $\bar z$ are the roots of $z^2-4z+13=0\,$.
But $z^2-4z+13=0 \iff z^2 = \color{blue}{4z -13}\,$, then multiplying by $z \ne 0\,$ it follows that $z^3 = 4z^2-13 z = 4(4z-13)-13z=\color{green}{3z - 52}\,$
So in the end $\,4z^3 - 3z^2 + 169 = 4\color{green}{(3z-52)}-3\color{blue}{(4z-13)}+169=0\,$.
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