3. Let $A$ and $B$ be the images of the solutions of the equation: $z^{2}-13 z+\mathbf{7 2}+\mathbf{3 0} i=0$.

a) Calculate the area of the triangle $\boldsymbol{A O B}$, where $\boldsymbol{O}$ is the origin of the complex plane.

b) What percentage of the area of triangle $\boldsymbol{A O B}$ is covered by the set of solutions to the inequality $|z| \leq 4$?
3.a) $\Delta=(5-12 \mathrm{i})^{2}$, where $z_{1}=9-6 i, z_{2}=4+6 i$.

Then $A B=\left|z_{1}-z_{2}\right|=13, A O=\left|z_{1}\right|=3 \sqrt{13}, B O=\left|z_{2}\right|=2 \sqrt{13}$

It is observed that $O A^{2}+O B^{2}=A B^{2}$, so $\triangle A O B$ is a right triangle with the right angle at $O$, and the area of the triangle is:

$$
A_{\triangle A O B}=\frac{A O \cdot B O}{2}=39
$$

b) Since $\triangle A O B$ is a right triangle with the right angle at $O$, it follows that one quarter of the circle that represents the graphical solution of the inequality $|z| \leq 4$ covers the triangle.

$$
A_{\text {sect }}=\frac{16 \pi}{4}=4 \pi \cong 12.56 \cong 32.2 \% A_{\triangle A O B}
$$