Question

Punctul d numere complexe​

Answer 1

${(1 + i)}^{10} + {(1 - i)}^{6} + {(1 + i)}^{12} + {(2 - i)}^{12}$

$= {[ {(1 + i)}^{2} ]}^{5} + {[ {(1 - i)}^{2} ]}^{3} + {[ {(1 + i)}^{2} ]}^{6} + {[ {(2 - i)}^{2} ]}^{6}$

$= {( {1}^{2} + 2 \times 1 \times i + {i}^{2}) }^{5} + {( {1}^{2} - 2 \times 1 \times i + {i}^{2} )}^{3} + {( {1}^{2} + 2 \times 1 \times i + {i}^{2}) }^{6} + {( {2}^{2} - 2 \times 2 \times i + {i}^{2}) }^{6}$

$= {(1 + 2i - 1)}^{5} + {(1 - 2i - 1) }^{3} + {(1 + 2i - 1)}^{6} + {(4 - 4i - 1)}^{6}$

$= {(2i)}^{5} + {( - 2i)}^{3} + {(2i)}^{6} + {(3 - 4i)}^{6}$

$= {2}^{5} \times {i}^{5} + {( - 2)}^{3} \times {i}^{3} + {2}^{6} \times {i}^{6} + {[ {(3 -4i)}^{2} ]}^{3}$

$= 32 \times {i}^{4} \times i - 8 \times ( - i) + 64 \times { ({i}^{2}) }^{3} + {( {3}^{2} - 2 \times 3 \times 4i + {(4i)}^{2} )}^{3}$

$= 32 \times 1 \times i + 8i + 64 \times {( - 1)}^{3} + {(9 - 24i + {4}^{2} \times {i}^{2} ) }^{3}$

$= 32i + 8i + 64 \times ( - 1) + {[9 - 24i + 16 \times ( - 1)]}^{3}$

$= 40i - 64 + {(9 - 24i - 16)}^{3}$

$= 40i - 64 + {( - 24i - 7)}^{3}$

$= 40i - 64 + {( - 24i - 7)}^{2} \times ( - 24i - 7)$

$= 40i - 64 + [ {( - 24i)}^{2} + 2 \times 24i \times 7 + {7}^{2} ]( - 24i - 7)$

$= 40i - 64 + [ {( - 24)}^{2} \times {i}^{2} + 336i + 49]( - 24i - 7)$

$= 40i - 64 + [576 \times ( - 1) + 336i + 49]( - 24i - 7)$

$= 40i - 64 + ( - 576 + 336i + 49)( - 24i - 7)$

$= 40i - 64 + (336i - 527)( - 24i - 7)$

$= 40i - 64 + ( - 8064 {i}^{2} - 2352i + 12648i + 3689)$

$= 40i - 64 + [ - 8064 \div ( - 1) + 10296i + 3689]$

$= 40i - 64 + 8064 + 10296i + 3689$

$= 10336i + 11689$