I'm trying to show:
Let $X$ and $Y$ be independent Cauchy random variables. Find the density function of $Z=\alpha X+ \beta Y$ assuming that $\alpha\beta\neq 0$.
I suspect $Z$ is also Cauchy: Suppose that $$ X\sim \mathrm{Cauchy}(x_0,\gamma) \mbox{ and } Y\sim \mathrm{Cauchy}(y_0,\delta)$$
Let $Z =\alpha X + \beta Y$ and $W = \beta Y$. Then, we have $$X=\frac{1}{\alpha}Z - \frac{1}{\alpha}W \qquad Y = \frac{1}{\beta}W $$ the Jacobian is $$J =\begin{bmatrix} \frac{1}{\alpha} & -\frac{1}{\alpha} \\ 0 & \frac{1}{\beta} \end{bmatrix}$$ which has determinant $\frac{1}{\alpha\beta}$.
Therefore, we have \begin{eqnarray*} f_{Z,W}(z,w) &=& |\det(J)|f_{X,Y}\left(\frac{1}{\alpha}(z-w),\frac{1}{\beta}w\right)\\ &=& \frac{1}{\alpha\beta}\frac{1}{\pi^2 \gamma\delta \left[ 1+ \left(\frac{\frac{1}{\alpha}(z-w)-x_0}{\gamma} \right)^2 \right]\left[ 1+ \left(\frac{\frac{1}{\beta}w-y_0}{\delta} \right)^2 \right]} \end{eqnarray*} By the Lagrange identity; $$(a^2+b^2)(c^2+d^2) = (ac+bd)^2 + (ad-bc)^2$$ So we have \begin{eqnarray*} \alpha\beta\gamma\delta\left[ 1+ \left(\frac{\frac{1}{\alpha}(z-w)-x_0}{\gamma} \right)^2 \right]\left[ 1+ \left(\frac{\frac{1}{\beta}w-y_0}{\delta} \right)^2 \right] \\ =\alpha\beta\gamma\delta \left[1 + \left(\frac{\frac{1}{\alpha}(z-w)-x_0}{\gamma} \right)\left(\frac{\frac{1}{\beta}w-y_0}{\delta} \right)\right] + \left[\left(\frac{\frac{1}{\beta}w-y_0}{\delta} \right)- \left(\frac{\frac{1}{\alpha}(z-w)-x_0}{\gamma} \right) \right]\\ = \left[\alpha\beta\gamma\delta + \left((z-w)-\alpha x_0 \right)\left(w-\beta y_0\right)\right] + \left[\alpha\gamma \left(w- \beta y_0 \right)- \beta\delta\left((z-w)-\alpha x_0 \right) \right] \end{eqnarray*} Then to get the density of $Z$, we integrate out the w: $$\int_0^\infty f_{Z,W}(z,w) dw = \int_0^\infty \frac{1}{\pi^2}\frac{1}{ \left[\alpha\beta\gamma\delta + \left((z-w)-\alpha x_0 \right)\left(w-\beta y_0\right)\right] + \left[\alpha\gamma \left(w- \beta y_0 \right)- \beta\delta\left((z-w)-\alpha x_0 \right) \right]}dw$$ but I have no clue what to do with this integral. Is there an easier approach?
Following Angryavian's suggestion:
A $\text{Cauchy}(x_0,\gamma)$ random variable has characteristic function $\phi(t)=\exp(itx_0)\exp(-\gamma|t|)$, according to Wikipedia. So the characteristic function of the sum $\alpha X+\beta Y$, where $X\sim\text{Cauchy}(x_0,\gamma)$ and $Y\sim\text{Cauchy}(y_0,\delta)$ are independent, is $$\varphi(t) = e^{it\alpha x_0}e^{-\gamma\alpha|t|}\times e^{it\beta y_0}e^{-\delta\beta|t|}.$$
You can express this in the form $\varphi(t) = e^{itA}e^{-B|t|}$ for $A$ and $B$ you can work out in terms of $x_0$ and your other parametors; then $\alpha X+\beta Y\sim\text{Cauchy}(A,B)$.