How to show that the random variables are independent?

The question is about a particular exercise.

Let X,Y be independent random variables with exponential distribution where both have the same parameter $$\lambda$$. Define $$Z_1:=X_1+X_2$$. Calculate the density of $$Z_1$$. Also define $$Z_2:=\frac{X_1 X_2}{X_1+X_2}$$.

$$Z_3:=\frac{X_1^2}{X_1+X_2}$$

$$Z_4:=\frac{X_2}{X_1+X_2}$$

Are $$Z_1,Z_2$$ independent?

Are $$Z_1,Z_3$$independent?

Are $$Z_1,Z_4$$ independent?

My Approach: I calculated the densitiy of Z$$_1$$by using the convolution formula, but now I'm struggling to show that $$Z_1$$ and $$Z_4$$ are independent. (I got a hint that these two are independent.)

Note $$f_{X_1X_2}(x_1,x_2)=\lambda^2 e^{-(x_1+x_2)} \cdot 1_{[0,\infty)^2}$$. Also note $$Z_4$$ is supported on $$(0,1)$$. If $$z_4\in (0,1)$$ is fixed then $$\begin{eqnarray*} F_{Z_4}(z_4)&=&P(Z_4 \leq z_4) \\&=& P\Bigg(X_2\leq\frac{z_4X_1}{1-z_4}\Bigg) \\ &=& \int_0^{\infty} \int_0^{\frac{z_4x}{1-z_4}}\lambda^2e^{-\lambda(x+y)}\mathrm{d}y\mathrm{d}x \\ &=& z_4\end{eqnarray*}$$ This means $$Z_4\sim \mathcal{U}(0,1)$$. You should have gotten with convolution that $$f_{Z_1}(z_1)=\lambda^2 z_1e^{-\lambda z_1}\cdot 1_{(0,\infty)}$$. Using multivariate change of variables we get for $$(z_1,z_4)\in (0,\infty)\times (0,1)$$ that $$\begin{eqnarray*} f_{Z_1Z_4}(z_1,z_4)&=&f_{X_1X_2}\Big(z_1-z_1z_4,z_1z_4\Big)\Bigg|\frac{\partial(z_1-z_1z_4,z_1z_4)}{\partial(z_1,z_4)}\Bigg| \\ &=& \lambda^2 z_1e^{-\lambda z_1} \\ &=& f_{Z_1}(z_1)f_{Z_4}(z_4)\end{eqnarray*}$$ So you get independence.