1. In the complex plane, consider the set of all points $z$ of the form $(4 t+1)+(3 t+7) i$, where $t$ is a real number. What is that set?

Determine the number from that set which has the smallest absolute value.
1. To the point $(x, y)$ in the coordinate plane, the complex number $z=x+i y$ is assigned. In our case, $x=4 t+1, y=3 t+7$. From this, we have $t=\frac{x-1}{4}, t=\frac{y-7}{3}$, which leads to $\frac{x-1}{4}=\frac{y-7}{3}$ or $3 x-4 y+25=0$. Therefore, the set of points is a line whose equation is this. Now,

$$
|z|=\sqrt{(4 t+1)^{2}+(3 t+7)^{2}}=5 \sqrt{t^{2}+2 t+2}
$$

Since $t^{2}+2 t+2=(t+1)^{2}+1 \geq 1$, then $|z| \geq 5$, and the minimum value is achieved for $t=-1$. In this case, $x=-3, y=4$.

Thus, among all the observed complex numbers, the one with the smallest modulus is $z=-3+4 i$.