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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 ?

Nathuram
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  • It is a division of polynomials, not of complex numbers. Anyway, there's no Euclidean division in $\mathbf C$. – Bernard Sep 17 '17 at 22:13
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    It's equivalent to finding that $(x-2)$ is a factor of $x^2 - 5x + 6$ by polynomial division. There are other methods that don't involve division (e.g. inspection) and when the equivalence is expressed in the form $x^2 - 5x + 6 = (x-2)(x-3)$, there is no division operation in use and therefore the expression is valid for all x (including $x = 2$). – A.M. Sep 17 '17 at 22:24

3 Answers3

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The identity $$4z^3-13z^2+169 = (4z+13)(z^2-4z+13)$$ is true for every complex number $z$.

One can't say the same for $$4z+13 = \frac{4z^3-13z^2+169}{z^2-4z+13},$$ which is true only for $z\neq 2\pm 3i$.

Rodrigo Dias
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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 $\,4z^3 - 3z^2 + 169 = 4\color{green}{(3z-52)}-3\color{blue}{(4z-13)}+169=0\,$.

dxiv
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Slightly abstracted it is: $\,z\!-\!a\mid f(z)\mid g(z)\,\Rightarrow\, g(a) = 0,\,$ i.e. $\,z\!-\!a\,$ is a factor of $f$ so also of its multiple $g$, hence $\,g = (z\!-\!a)h\,$ so $\, g(a) = 0.\ $ No divisions were used, only divisibilities.

Bill Dubuque
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