# Is $\frac{1-\alpha}{1+\alpha}=y \Rightarrow \alpha=\frac{1-y}{y+1}$ correct even if $y=-1$?

I was trying to solving this question:

If roots of the equation $$a x^{2}+b x+c=0$$ are $$\alpha$$ and $$\beta$$, find the equation whose roots are $$\frac{1-\alpha}{1+\alpha}, \frac{1-\beta}{1+\beta}$$

I was not able to solve it so I looked to the solution given in the book, it was as follows:

Let $$\frac{1-\alpha}{1+\alpha}=y \Rightarrow \alpha=\frac{1-y}{1+y}$$ .

Now $$\alpha$$ is root of the equation $$a x^{2}+b x+c=0$$ $$\Rightarrow a \alpha^{2}+b \alpha+c=0$$ $$\Rightarrow \quad a\left(\frac{1-y}{1+y}\right)^{2}+b\left(\frac{1-y}{1+y}\right)+c=0$$ . Hence required equation is $$a(1-x)^{2}+b\left(1-x^{2}\right)+c(1+x)^{2}=0$$

In the first step the auther used componendo dividend rule as follows:

\begin{aligned} & \frac{1-\alpha}{1+\alpha}=y \\ \Rightarrow & \frac{1-\alpha+1+\alpha}{1-\alpha-(1+\alpha)}=\frac{y+1}{y-1} \\ \Rightarrow & \frac{2}{-2 \alpha}=\frac{y+1}{y-1} \\ \Rightarrow & \frac{-1}{\alpha}=\frac{y+1}{y-1} \Rightarrow \alpha=\frac{1-y}{y+1} \end{aligned}

But I don't understand how he could use it, as we are not sure whether $$y$$ could be - 1 and hence $$1 +y$$ can be zero. So is this an error? If so then how can we solve this question and if not then why not?

• Just isolate $y=-1$ as a separate case. Here it is a minor flaw in the book's reasoning. Jul 10 at 9:35

Notice that $$y$$ cannot be equal to $$-1$$ in the first place, if $$y = \dfrac{1-\alpha}{1+\alpha}$$

Range 0f $$f(\alpha) = \dfrac{1-\alpha}{1+\alpha}$$ is $$\mathbb{R} \setminus \{-1\} ~\forall ~\alpha \in \mathbb{R}$$

This is because if $$\dfrac{1-\alpha}{1+\alpha} = -1$$ then $$1-\alpha = -1-\alpha$$ $$1=-1$$ Which is clearly meaningless.

In general, $$\dfrac{ax+b}{cx+d}$$ cannot be equal to $$\dfrac{a}{c}$$ if $$\dfrac{a}{c} \ne \dfrac{b}{d}$$

• The last is only true when $ad-bc\ne 0$ Jul 10 at 9:45
• @HagenvonEitzen Yes, indeed. I should have been a bit more careful when making a general comment. Jul 10 at 9:50

Is $$\frac{1-\alpha}{1+\alpha}=y \Rightarrow \alpha=\frac{1-y}{y+1}$$ correct even if $$y=-1$$?

The tag is given. The answer to the question is: No.

But with tag , the answer to the question would be: Yes.
Calculating in the Riemann sphere, we have $$\frac{1-\infty}{1+\infty} = -1\qquad\text{and}\qquad \frac{1-(-1)}{(-1)+1} = \infty .$$