Đề bài
Thực hiện phép tính: \(\displaystyle {1 \over {2 - 3i}}\); \(\displaystyle {1 \over {{1 \over 2} - {{\sqrt 3 } \over 2}i}}\); \(\displaystyle {{3 - 2i} \over i}\); \(\displaystyle {{3 - 4i} \over {4 - i}}\)
Phương pháp giải - Xem chi tiết
Chia hai số phức \(\dfrac {{a + bi}}{{c + di}} = \dfrac {{\left( {a + bi} \right)\left( {c - di} \right)}}{{\left( {c + di} \right)\left( {c - di} \right)}}\)
Lời giải chi tiết
\(\displaystyle {1 \over {2 - 3i}} \) \(\displaystyle = \frac{{2 + 3i}}{{\left( {2 - 3i} \right)\left( {2 + 3i} \right)}} \) \(\displaystyle = {{2 + 3i} \over {4 - 9{i^2}}} \) \(\displaystyle = \frac{{2 + 3i}}{{13}}\) \(\displaystyle = {2 \over {13}} + {3 \over {13}}i\)
\(\displaystyle {1 \over {{1 \over 2} - {{\sqrt 3 } \over 2}i}} \) \(\displaystyle = \frac{{\frac{1}{2} + \frac{{\sqrt 3 }}{2}i}}{{\left( {\frac{1}{2} - \frac{{\sqrt 3 }}{2}i} \right)\left( {\frac{1}{2} + \frac{{\sqrt 3 }}{2}i} \right)}}\) \(\displaystyle = {{{1 \over 2} + {{\sqrt 3 } \over 2}i} \over {{1 \over 4} - {{\left( {{{\sqrt 3 } \over 2}i} \right)}^2}}} \) \(\displaystyle = {{{1 \over 2} + {{\sqrt 3 } \over 2}i} \over 1} \) \(\displaystyle = {1 \over 2} + {{\sqrt 3 } \over 2}i\)
\(\displaystyle {{3 - 2i} \over i} \) \(\displaystyle = {{i\left( {3 - 2i} \right)} \over {{i^2}}} \) \(\displaystyle = \frac{{3i - 2{i^2}}}{{ - 1}} \) \(\displaystyle = - 3i + 2{i^2} \) \(\displaystyle = - 3i - 2\)
\(\displaystyle {{3 - 4i} \over {4 - i}} \) \(\displaystyle = \frac{{\left( {3 - 4i} \right)\left( {4 + i} \right)}}{{\left( {4 - i} \right)\left( {4 + i} \right)}}\) \(\displaystyle = \frac{{12 - 13i - 4{i^2}}}{{{4^2} + {1^2}}} \) \(\displaystyle = {{16 - 13i} \over {17}} \) \(\displaystyle = {{16} \over {17}} - {{13} \over {17}}i.\)
