Giải bài 4 trang 138 – SGK môn Giải tích lớp 12
Giải các phương trình sau:
a) \(\left( 3-2i \right)z+\left( 4+5i \right)=7+3i\)
b) \(\left( 1+3i \right)z-\left( 2+5i \right)=\left( 2+i \right)z\);
c) \(\dfrac{z}{4-3i}+\left( 2-3i \right)=5-2i \).
\(\begin{aligned}a)\, & \left( 3-2i \right)z+\left( 4+5i \right)=7+3i \\ \Leftrightarrow & \left( 3-2i \right)z=3-2i \\ \Leftrightarrow & z=1 \\ \end{aligned} \)
\(\begin{aligned}b)\, & \left( 1+3i \right)z-\left( 2+5i \right)=\left( 2+i \right)z \\ \Leftrightarrow &\left( -1+2i \right)z=2+5i \\ \Leftrightarrow &z=\dfrac{2+5i}{-1+2i} \\ \Leftrightarrow& z=\dfrac{\left( 2+5i \right)\left( -1-2i \right)}{5} \\ \Leftrightarrow& z=\dfrac{8}{5}-\dfrac{9}{5}i \\ \end{aligned} \)
\(\begin{aligned}c)\, & \dfrac{z}{4-3i}+\left( 2-3i \right)=5-2i \\ \Leftrightarrow &\dfrac{z}{4-3i}=3+i \\ \Leftrightarrow &z=\left( 3+i \right)\left( 4-3i \right) \\ \Leftrightarrow &z=15-5i \\ \end{aligned} \)
Ghi nhớ: Cho hai số phức \(z=a+bi\) và \(z'=c+di\).
Khi đó \(z.z'=(ac-bd)+(ad+bc)i\).
\(\dfrac{z}{z'}=\dfrac{z.\overline{z'}}{{{\left| z' \right|}^{2}}}=\dfrac{\left( a+bi \right)\left( c-di \right)}{{{c}^{2}}+{{d}^{2}}}\).