What Is the Sum of the Absolute Values of a and b in This Algebraic Puzzle?

[anemone]

Here is this week's POTW:

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Let $(a,\,b)$ be a pair of real numbers satisfying

$33a-56b=\dfrac{a}{a^2+b^2}$ and $56a+33b=-\dfrac{b}{a^2+b^2}$

Determine the value of $|a|+|b|$.

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[anemone]

Congratulations to the following members for their correct solution::)

1. MarkFL
2. kaliprasad
3. Opalg
4. greg1313

Solution from Opalg:

Spoiler

[Solution using complex numbers.]

Add $i$ times the second equation to the first equation: $$(33 + 56i)(a+ib) = \frac{a-ib}{a^2+b^2} = \frac1{a+ib}.$$ Therefore $$(a+ib)^2 = \frac1{33+56i},$$ $$a+ib = \frac1{\sqrt{33+56i}} = \pm\frac1{7+4i} = \pm\frac{7-4i}{65},$$ and so $|a| + |b| = \dfrac{7}{65} + \dfrac{4}{65} = \dfrac{11}{65}.$

Alternate Solution from kaliprasad:

Spoiler

Given relations are

$33a-56b= \dfrac{a}{a^2+b^2}\cdots(1)$

$56a+33b= \dfrac{-b}{a^2+b^2}\cdots(2)$

Square both and add to get

$(33a-56b)^2 + (56a+33b)^2 = \dfrac{1}{a^2+b^2}$

or $(33^2+56^2)(a^2+b^2) = \dfrac{1}{a^2+b^2}$

or $65^2(a^2+b^2) = \dfrac{1}{a^2+b^2}$

or $(a^2+b^2) = \dfrac{1}{65}\cdots(3) $

so we get

From 1st relation $33a-56b = 65a$

or $32a = - 56b$ or $4a = -7b$

From (3), we get

$a^2 + b^2 = a^2 + \dfrac{16}{49}a^2= \dfrac{65}{49}a^2 = \dfrac{1}{65}$

so $a = \dfrac{7}{65}, b = \dfrac{-4}{65}$ or $a=\dfrac{-7}{65},b= \dfrac{4}{65}$ and hence

$| a | + |b| = \dfrac{11}{65}$