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Solve the following system:

$$ \begin{cases} \text{a}\cdot\text{c}+\text{b}\cdot\text{d}\cdot\epsilon^2=\epsilon\cdot\left(\text{b}\cdot\text{c}-\text{a}\cdot\text{d}\right)\\ \\ \left|\epsilon\right|=\left|-\frac{\text{c}}{\text{d}}\right| \end{cases} $$

I don't know how to proceed?!

With the help of the comments and the answer I got:

$$ \begin{cases} \text{a}\cdot\text{c}+\text{b}\cdot\text{d}\cdot\epsilon^2=\epsilon\cdot\left(\text{b}\cdot\text{c}-\text{a}\cdot\text{d}\right)\\ \\ \epsilon=\pm\frac{\text{c}}{\text{d}} \end{cases}\to\text{a}\cdot\text{c}+\text{b}\cdot\text{d}\cdot\frac{\text{c}^2}{\text{d}^2}=\pm\frac{\text{c}}{\text{d}}\cdot\left(\text{b}\cdot\text{c}-\text{a}\cdot\text{d}\right) $$

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    Are $a,b,c,d$ integers? It looks like a question on $GL(2,\mathbb{Z})$. – 2017-01-02
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    The only thing that I know that those a real numbers, so not only integers – 2017-01-02
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    $(a,b,c,d)$ are integers? Real numbers? – 2017-01-02
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    I'm curious about what is the motivation for this question? Do you come up with this question when proving something? – 2017-01-02
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    @Azzo They are real numbers – 2017-01-02
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    @Jack Yes, it came up proving something. About a property of complex fractions – 2017-01-02
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    Then I think it would be nice to put that background to your post. – 2017-01-02
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    @Jack The problem is I'm not working on my own on this problem and it is holiday here in Holland and the one that knows the background you want to know is not at home :( – 2017-01-02

3 Answers 3

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A case disctinction helps. Assume first that $\epsilon=c/d$. Then we obtain $ac+bc^2/d=c^2b/d-ca$, hence $2ac=0$. Both cases $a=0$ or $c=0$ are easily solved.

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    Can I say that $\epsilon=\pm\frac{\text{c}}{\text{d}}$? – 2017-01-02
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    Yes, we have $|-z|=|z|$. – 2017-01-02
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If $c/d$ is positive, you will have $|-c/d|=c/d$. Thus

$$ ac+bd\frac{c^2}{d^2}=\dfrac{c}{d}(bc-ad), $$

or

$$ ac+\frac{bc^2}{d}=\dfrac{bc^2}{d}-ac, $$

or

$$ac=0.$$

Now if $c/d$ is negative, you will have $|-c/d|=-c/d$. Thus

$$ ac+bd\frac{c^2}{d^2}=-\dfrac{c}{d}(bc-ad), $$

or

$$ ac+\frac{bc^2}{d}=-\dfrac{bc^2}{d}+ac, $$

or

$$bc^2/d=0.$$

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We are given $$ \left\{ \begin{gathered} bd\varepsilon ^{\,2} - \left( {bc - ad} \right)\varepsilon + ac = 0 \hfill \\ \varepsilon = \pm \frac{c} {d} \hfill \\ \end{gathered} \right. $$ So we have the equation of a vertical parabola in $\varepsilon$, whose intercepts with the $x$ axis must be symmetrical (vs. $\varepsilon=0$).
Thus the parabola must be symmetrical as well, i.e. the coefficient of the $\varepsilon$ term must be null, which gives: $$ \begin{gathered} \left\{ \begin{gathered} bd\varepsilon ^{\,2} - \left( {bc - ad} \right)\varepsilon + ac = 0 \hfill \\ \varepsilon = \pm \frac{c} {d} \hfill \\ \end{gathered} \right.\quad \Rightarrow \quad \left\{ \begin{gathered} bc - ad = 0 \hfill \\ b\frac{{c^{\,2} }} {d} + ac = 0 \hfill \\ \end{gathered} \right.\quad \Rightarrow \hfill \\ \Rightarrow \quad \left\{ \begin{gathered} bc - ad = 0 \hfill \\ d \ne 0 \hfill \\ \left( {bc + ad} \right)c = 0 \hfill \\ \end{gathered} \right. \Rightarrow \quad \left\{ \begin{gathered} c = 0 \hfill \\ d \ne 0 \hfill \\ a = 0 \hfill \\ \forall b \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} $$