Bài 1: Thực hiện các phép chia sau:
a) \( \frac{2+i}{3-2i}\); b) \( \frac{1+i\sqrt{2}}{2+i\sqrt{3}}\); c) \( \frac{5i}{2-3i}\); d) \( \frac{5-2i}{i}\).
a) \( \frac{2+i}{3-2i}\) \( =\frac{(2+i)(3+2i)}{13}=\frac{4}{13}+\frac{7}{13}i\).
b) \( \frac{1+i\sqrt{2}}{2+i\sqrt{3}}\) \( =\frac{(1+i\sqrt{2})(2-i\sqrt{3})}{7}=\frac{2+\sqrt{6}}{7}+\frac{2\sqrt{2}-\sqrt{3}}{7}i\)
c) \( \frac{5i}{2-3i}\) \( =\frac{5i(2+3i)}{13}=-\frac{15}{13}+\frac{10}{13}i\)
d) \( \frac{5-2i}{i}= (5 – 2i)(-i) = -2 – 5i\).
Bài 2: Tìm nghịch đảo \( \frac{1}{z}\) của số phức \(z\), biết:
a) \(z = 1 + 2i\); b) \(z = \sqrt2 – 3i\);
c) \(z = i\); d) \(z = 5 + i\sqrt3\).
a) \( \frac{1}{1+2i}=\frac{1-2i}{5}=\frac{1}{5}-\frac{2}{5}i.\)
Advertisements (Quảng cáo)
b) \( \frac{1}{\sqrt{2}-3i}=\frac{\sqrt{2}+3i}{(\sqrt{2})^{2}+(-3)^{2}}=\frac{\sqrt{2}}{11}+\frac{3}{11}i\)
c) \( \frac{1}{i}=\frac{-i}{1}=-i\)
d) \( \frac{1}{5+i\sqrt{3}}=\frac{5-i\sqrt{3}}{5^{2}+(\sqrt{3})^{2}}=\frac{5}{28}-\frac{\sqrt{3}}{28}i\)
Bài 3: Thực hiện các phép tính sau:
a) \(2i(3 + i)(2 + 4i)\); b) \( \frac{(1+i)^{2}(2i)^{3}}{-2+i}\)
c) \(3 + 2i + (6 + i)(5 + i)\); d) \(4 – 3i + \frac{5+4i}{3+6i}\).
Advertisements (Quảng cáo)
a) \(2i(3 + i)(2 + 4i) = 2i(2 + 14i) = -28 + 4i\)
b) \( \frac{(1+i)^{2}(2i)^{3}}{-2+i}\) \( =\frac{2i(-8i)}{-2+i}=\frac{16(-2-i)}{5}=-\frac{32}{5}-\frac{16}{5}i\)
c) \(3 + 2i + (6 + i)(5 + i) = 3 + 2i + 29 + 11i = 32 + 13i\)
d) \( 4 – 3i + \frac{5+4i}{3+6i}\) = \(4 – 3i + \frac{(5+4i)(3-6i)}{45}\) = \(4 – 3i + \frac{39}{45}-\frac{18}{45}i\)
\(= (4 + \frac{39}{45}\)) \(- (3 + \frac{18}{45}\))i = \( \frac{219}{45}-\frac{153}{45}i\)
Bài 4: Giải các phương trình sau:
a) \((3 – 2i)z + (4 + 5i) = 7 + 3i\);
b) \((1 + 3i)z – (2 + 5i) = (2 + i)z\);
c) \( \frac{z}{4-3i} + (2 – 3i) = 5 – 2i\).
a) Ta có \((3 – 2i)z + (4 + 5i) = 7 + 3i \Leftrightarrow (3 – 2i)z = 7 + 3i – 4 – 5i\)
\(\Leftrightarrow z = \frac{3-2i}{3-2i} \Leftrightarrow z = 1\). Vậy \(z = 1\).
b) Ta có \((1 + 3i)z – (2 + 5i) = (2 + i)z \Leftrightarrow (1 + 3i)z -(2 + i)z = (2 + 5i)\)
\(\Leftrightarrow (1 + 3i – 2 – i)z = 2 + 5i \Leftrightarrow (-1 + 2i)z = 2 + 5i\)
\(\Leftrightarrow z = \frac{2 + 5i}{-1+2i}=\frac{(2+5i)(-1-2i)}{5}=\frac{-2-4i-5i-10i^{2}}{5}=\frac{8-9i}{5}=\frac{8}{5}-\frac{9}{5}i\)
Vậy \(z =\frac{8}{5}-\frac{9}{5}i\)
c) Ta có \( \frac{z}{4-3i} + (2 – 3i) = 5 – 2i \Leftrightarrow \)\( \frac{z}{4-3i}= 5 – 2i – 2 + 3i\)
\(\Leftrightarrow z = (3 + i)(4 – 3i)\Leftrightarrow z = 12 + 3 + (-9 + 4)i \Leftrightarrow z = 15 -5i\)
