Determine o quarto vértice do paralelogramo

O problema tem três soluções:

$$\begin{array}{*{20}{l}}
{{z_A}}& = &{{z_2} + \left( {{z_1} – {z_2}} \right) + \left( {{z_3} – {z_2}} \right)} \\
{}& = &{1 + 3i + \left( {(2 – i) – (1 + 3i)} \right) + \left( {( – 3 + 2i) – (1 + 3i)} \right)} \\
{}& = &{1 + 3i + \left( {1 – 4i} \right) + \left( { – 4 – i} \right)} \\
{}& = &{ – 2 – 2i}
\end{array}$$

$$\begin{array}{*{20}{l}}
{{z_B}}& = &{{z_3} + \left( {{z_1} – {z_3}} \right) + \left( {{z_2} – {z_3}} \right)} \\
{}& = &{ – 3 + 2i + \left( {(2 – i) – ( – 3 + 2i)} \right) + \left( {(1 + 3i) – ( – 3 + 2i)} \right)} \\
{}& = &{ – 3 + 2i + \left( {5 – 3i} \right) + \left( {4 + i} \right)} \\
{}& = &6
\end{array}$$

$$\begin{array}{*{20}{l}}
{{z_C}}& = &{{z_1} + \left( {{z_2} – {z_1}} \right) + \left( {{z_3} – {z_1}} \right)} \\
{}& = &{2 – i + \left( {(1 + 3i) – (2 – i)} \right) + \left( {( – 3 + 2i) – (2 – i)} \right)} \\
{}& = &{2 – i + \left( { – 1 + 4i} \right) + \left( { – 5 + 3i} \right)} \\
{}& = &{ – 4 + 6i}
\end{array}$$

(Porquê?)