Giải bài 4.10 trang 201 - SBT Giải tích lớp 12
Tính các lũy thừa sau :
\(\begin{align} & a){{\left( 3-4i \right)}^{2}} \\ & b){{\left( 2+3i \right)}^{3}} \\ & c){{\left[ \left( 4+5i \right)-\left( 4+3i \right) \right]}^{5}} \\ & d){{\left( \sqrt{2}-i\sqrt{3} \right)}^{2}} \\ \end{align} \)
Áp dụng:
\((a\pm b)^2=a^2\pm 2ab+b^2\\ i^2=-1\)
\(\begin{aligned} & a) \\ & {{\left( 3-4i \right)}^{2}}=9-24i+16{{i}^{2}}=-7-24i \\ & b)\, \\ & {{\left( 2+3i \right)}^{3}}=8+36i+54{{i}^{2}}+27{{i}^{3}}=-46+9i \\ & c) \\ & {{\left[ \left( 4+5i \right)-\left( 4+3i \right) \right]}^{5}}={{\left( 2i \right)}^{5}}=32{{i}^{5}}=32i\,\,\,\left( \,\,\text{Vì}\,\,\,{{i}^{4}}={{\left( {{i}^{2}} \right)}^{2}}=1 \right) \\ & d)\, \\ & {{\left( \sqrt{2}-i\sqrt{3} \right)}^{2}}=2-2\sqrt{6}i+3{{i}^{2}}=-1-2\sqrt{6}i \\ \end{aligned} \)