Nombres complexes : point de vue algébrique

Appliquer la formule du binôme de Newton - Exercice 2

10 min
25
Utiliser la formule du binôme pour développer :
Question 1

(4+i)3\left(4+i\right)^{3}

Correction
  • Soient aa et bb deux nombres complexes. Pour tout entier naturel nn, on a : (a+b)n=k=0n(nk)akbnk\left(a+b\right)^{n} =\sum _{k=0}^{n}\left(\begin{array}{c} {n} \\ {k} \end{array}\right) a^{k} b^{n-k}
  • i3=i2×i=(1)×i=ii^{3} =i^{2} \times i=\left(-1\right)\times i=-i
  • (4+i)3=k=03(3k)4ki3k\left(4+i\right)^{3} =\sum _{k=0}^{3}\left(\begin{array}{c} {3} \\ {k} \end{array}\right) 4^{k} i^{3-k} équivaut successivement à :
    (4+i)3=(30)40i30+(31)41i31+(32)42i32+(33)43i33\left(4+i\right)^{3} =\left(\begin{array}{c} {3} \\ {0} \end{array}\right)4^{0} i^{3-0} +\left(\begin{array}{c} {3} \\ {1} \end{array}\right)4^{1} i^{3-1} +\left(\begin{array}{c} {3} \\ {2} \end{array}\right)4^{2} i^{3-2} +\left(\begin{array}{c} {3} \\ {3} \end{array}\right)4^{3} i^{3-3}
    (4+i)3=(30)40i3+(31)41i2+(32)42i1+(33)43i0\left(4+i\right)^{3} =\red{\left(\begin{array}{c} {3} \\ {0} \end{array}\right)}4^{0} i^{3} +\blue{\left(\begin{array}{c} {3} \\ {1} \end{array}\right)}4^{1} i^{2} +\pink{\left(\begin{array}{c} {3} \\ {2} \end{array}\right)}4^{2} i^{1} +\purple{\left(\begin{array}{c} {3} \\ {3} \end{array}\right)}4^{3} i^{0}
    (4+i)3=1×40i3+3×41i2+3×42i1+1×43i0\left(4+i\right)^{3} =\red{1}\times 4^{0} i^{3} +\blue{3}\times 4^{1} i^{2} +\pink{3}\times 4^{2} i^{1} +\purple{1}\times 4^{3} i^{0}
    (4+i)3=1×1×(i)+3×4×(1)+3×16×i+64\left(4+i\right)^{3} =1\times 1\times \left(-i\right)+3\times 4\times \left(-1\right)+3\times 16\times i+64
    (4+i)3=i12+48i+64\left(4+i\right)^{3} =-i-12+48i+64
    Ainsi :
    (4+i)3=52+47i\left(4+i\right)^{3} =52+47i

    Question 2

    (3+i)4\left(3+i\right)^{4}

    Correction
      Formule du binoˆme de Newton\red{\text{Formule du binôme de Newton}}
    Soient aa et bb deux nombres complexes. Pour tout entier naturel nn, on a : (a+b)n=k=0n(nk)akbnk\left(a+b\right)^{n} =\sum _{k=0}^{n}\left(\begin{array}{c} {n} \\ {k} \end{array}\right) a^{k} b^{n-k}
    (3+i)4=k=0n(4k)3ki4k\left(3+i\right)^{4} =\sum _{k=0}^{n}\left(\begin{array}{c} {4} \\ {k} \end{array}\right) 3^{k} i^{4-k} équivaut successivement à :
    (3+i)4=(40)30i4+(41)31i41+(42)32i42+(43)33i43+(44)34i44\left(3+i\right)^{4} =\left(\begin{array}{c} {4} \\ {0} \end{array}\right)3^{0} i^{4} +\left(\begin{array}{c} {4} \\ {1} \end{array}\right)3^{1} i^{4-1} +\left(\begin{array}{c} {4} \\ {2} \end{array}\right)3^{2} i^{4-2} +\left(\begin{array}{c} {4} \\ {3} \end{array}\right)3^{3} i^{4-3} +\left(\begin{array}{c} {4} \\ {4} \end{array}\right)3^{4} i^{4-4}
    (3+i)4=(40)30i4+(41)31i3+(42)32i2+(43)33i1+(44)34i0\left(3+i\right)^{4} =\left(\begin{array}{c} {4} \\ {0} \end{array}\right)3^{0} i^{4} +\left(\begin{array}{c} {4} \\ {1} \end{array}\right)3^{1} i^{3} +\left(\begin{array}{c} {4} \\ {2} \end{array}\right)3^{2} i^{2} +\left(\begin{array}{c} {4} \\ {3} \end{array}\right)3^{3} i^{1} +\left(\begin{array}{c} {4} \\ {4} \end{array}\right)3^{4} i^{0}
    (3+i)4=(40)30i4+(41)31i3+(42)32i2+(43)33i1+(44)34i0\left(3+i\right)^{4} =\red{\left(\begin{array}{c} {4} \\ {0} \end{array}\right)}3^{0} i^{4} +\blue{\left(\begin{array}{c} {4} \\ {1} \end{array}\right)}3^{1} i^{3} +\pink{\left(\begin{array}{c} {4} \\ {2} \end{array}\right)}3^{2} i^{2} +\purple{\left(\begin{array}{c} {4} \\ {3} \end{array}\right)}3^{3} i^{1} +\green{\left(\begin{array}{c} {4} \\ {4} \end{array}\right)}3^{4} i^{0}
    (3+i)4=1×30i4+4×31i3+6×32i2+4×33i1+1×34i0\left(3+i\right)^{4} =\red{1}\times 3^{0} i^{4} +\blue{4}\times 3^{1} i^{3} +\pink{6}\times 3^{2} i^{2} +\purple{4}\times 3^{3} i^{1} +\green{1}\times 3^{4} i^{0}
    (3+i)4=1×1×i4+4×3×i3+6×9×i2+4×27×i1+1×81×i0\left(3+i\right)^{4} =1\times 1\times i^{4} +4\times 3\times i^{3} +6\times 9\times i^{2} +4\times 27\times i^{1} +1\times 81\times i^{0}
    (3+i)4=1×1×i2×i2+4×3×i2×i+6×9×i2+4×27×i1+1×81×i0\left(3+i\right)^{4} =1\times 1\times i^{2}\times i^{2} +4\times 3\times i^{2}\times i +6\times 9\times i^{2} +4\times 27\times i^{1} +1\times 81\times i^{0}
    (3+i)4=1×1×(1)×(1)+4×3×(1)×i+6×9×(1)+4×27×i1+1×81×i0\left(3+i\right)^{4} =1\times 1\times \left(-1\right)\times \left(-1\right) +4\times 3\times \left(-1\right)\times i +6\times 9\times\left(-1\right) +4\times 27\times i^{1} +1\times 81\times i^{0}
    (3+i)4=112i54+108i+81\left(3+i\right)^{4} =1-12i-54+108i+81
    Ainsi :
    (3+i)4=28+96i\left(3+i\right)^{4} =28+96i