Giải bài 1 trang 138 – SGK môn Giải tích lớp 12
Thực hiện các phép chia sau:
a) \(\dfrac{2+i}{3-2i}\);
b) \(\dfrac{1+i\sqrt{2}}{2+i\sqrt{3}}\);
c) \(\dfrac{5i}{2-3i}\);
d) \(\dfrac{5-2i}{i}\)
Hướng dẫn:
Cho hai số phức \(z=a+bi\) và \(z'=c+di\). Khi đó \(\dfrac{z}{z'}=\dfrac{z.\overline{z'}}{{{\left| z' \right|}^{2}}}=\dfrac{\left( a+bi \right)\left( c-di \right)}{{{c}^{2}}+{{d}^{2}}}\).
\(\begin{aligned} a)\, \dfrac{2+i}{3-2i} &=\dfrac{\left( 2+i \right)\left( 3+2i \right)}{13} \\ & =\dfrac{6-2+\left( 4+3 \right)i}{13} \\ & =\dfrac{4}{13}+\dfrac{7}{13}i \\ \end{aligned} \)
\(\begin{aligned}b)\, \dfrac{1+i\sqrt{2}}{2+i\sqrt{3}} &= \dfrac{\left( 1+i\sqrt{2} \right)\left( 2-i\sqrt{3} \right)}{7} \\ & =\dfrac{2+\sqrt{6}+\left(- \sqrt{3}+2\sqrt{2} \right)i} {7} \\ & =\dfrac{2+\sqrt{6}}{7} +\dfrac{- \sqrt{3}+2\sqrt{2}}{7}i \\ \end{aligned} \)
\(\begin{aligned} c)\,\dfrac{5i}{2-3i} &=\dfrac{5i\left( 2+3i \right)}{13} \\ & =-\dfrac{15}{13}+\dfrac{10}{13}i \\ \end{aligned} \)
\(\begin{aligned}d)\, \dfrac{5-2i}{i} &=\dfrac{\left( 5-2i \right)\left( -i \right)}{1} \\ & =-2-5i \\ \end{aligned} \)