Factorizing $z^2-z\frac{a^2 +b^2}{ab}+1 = (z-a/b)(z-b/a)$

Question

The intermediate factorization calculation $z^2-z\frac{a^2 +b^2}{ab}+1 = (z-a/b)(z-b/a)$ was coined straightforward by Riley's Mathematical Methods for Physics and Engeneering (eq.24.66).

I've tried completing the square,

$$ z^2-z\frac{a^2 +b^2}{ab}+1 = 0 \\ \bigg( z-\frac{a^2 +b^2}{2ab} \bigg)^2 +1 -\frac{(a^2+b^2)^2}{(2ab)^2} = 0 \\ \bigg( z-\frac{a^2 +b^2}{2ab} \bigg)^2 = \frac{(a^2+b^2)^2}{(2ab)^2}-1 \\ z= \frac{a^2 +b^2}{2ab} \pm \sqrt{\frac{(a^2+b^2)^2}{(2ab)^2}-1} \ , \\ $$ which doesn't nearly suggest the simple end result.

I'm sure it will roll out after reorganizing terms, but I'm interested in an alternative approach which more naturally suggests the end result of $(z-a/b)(z-b/a)$, reflecting the word-use of straightforward in the text.

Context.

Using contour integration to evaluate $I = \int_{0}^{2\pi} \frac{\cos(2\theta)}{a^2+b^2-2ab\cos(\theta)}d\theta$

Answer 1

it must be $$z_{1,2}=\frac{a^2+b^2}{2ab}\pm\sqrt{\left(\frac{a^2+b^2}{2ab}\right)^2-1}$$ and we can factorize to $$-\frac{(a-bz)(az-b)}{ab}$$ and $$ab\ne 0$$

Answer 2

Indeed, one can see that

$$\frac{a}{b} + \frac{b}{a} = \frac{a^2 + b^2}{ab}$$

Going from the RHS to the LHS is called fraction decomposition.

Your result should be clearer now:

$$z^2 - z\left(\frac{a^2 + b^2}{ab}\right) + 1 = z^2 - z\frac{a}{b} - z\frac{b}{a} + 1 = \left(z - \frac{a}{b}\right)\left(z - \frac{b}{a}\right)$$

Answer 3

Note that $$\frac{a^2+b^2}{ab}=\frac ab+\frac ba$$ and $$\frac ab\cdot\frac ba=1$$ Remember that, in a polynomial of second degree, the coefficient of $z$ is the opposite of sum of the roots and the costant term is their product.

Answer 4

You have one mistake.

$\frac{(a^2+b^2)^2}{(ab)^2}$ this term should be

$\frac{(a^2+b^2)^2}{(2ab)^2}$ then its easily solve.