Réponse :
Explications étape par étape :
1a)
[tex]P(i)=2i^3+i^2+2i+1=-2i-1+2i+1=0\\\\P(-i)=2(-i)^3+(-i)^2+2*(-i) +1=-2i^3+i^2-2i+1=2i-1-2i+1=0\\\\P(-\frac{1}{2})=2*(-\frac{1}{2})^3+(-\frac{1}{2})^2+2*(-\frac{1}{2})+1 =-\frac{1}{4}+\frac{1}{4}-1+1=0[/tex]
i, -i et -(1/2) sont des racines du polynôme P(z).
1b)
[tex]P(z)=2(z-i)(z+1)(z+\frac{1}{2})[/tex]
2
[tex]z_1=(2-i)^2-2(1+3i)^2=4-4i+i^2-2(1+6i+9i^2)=4-4i-1-2-12i+18=19-16i\\\\z_2=\frac{1-i}{2+3i} =\frac{(1-i)(2-3i)}{4+9} =\frac{2-3i-2i+3i^2}{13}= -\frac{1}{13}-\frac{5}{13}i\\ \\z_3=(1-2i)^4 =((1-2i)^2)^2=(1-4i+4i^2)^2=(-3-4i)^2=9+24i+16i^2=-7+24i[/tex]
3)
[tex]z=a+ib\\a) \ 3+i(a+ib)+2i=a+ib\\3+ia+i^2b+2i=a+ib\\3-b+(2+a)i =a +ib\\\rightarrow 2+a=b \rightarrow a=b-2\\\rightarrow3-b=a\\\rightarrow b-2=3-b \rightarrow 2b=5 \rightarrow b=\frac{5}{2} \\a=\frac{5}{2}-2=\frac{1}{2}\\\\z=\frac{1}{2}+\frac{5}{2}i \\\\b) \ (1+3i)(a+ib)-(i-1)=i(a-ib)-3\\a+ib+3ia+3i^2b-i+1=ia-i^2b-3\\a-3b+1+i(b+3a-1)=b-3+ia\\\rightarrow a+1-3b=b-3 \rightarrow a=4b-4\\\rightarrow b+3a-1=a \rightarrow 2a+b-1=0\rightarrow 8b-8+b-1=0\rightarrow b=1\\a=4*1-4=0\\\\z=i[/tex]
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Réponse :
Explications étape par étape :
1a)
[tex]P(i)=2i^3+i^2+2i+1=-2i-1+2i+1=0\\\\P(-i)=2(-i)^3+(-i)^2+2*(-i) +1=-2i^3+i^2-2i+1=2i-1-2i+1=0\\\\P(-\frac{1}{2})=2*(-\frac{1}{2})^3+(-\frac{1}{2})^2+2*(-\frac{1}{2})+1 =-\frac{1}{4}+\frac{1}{4}-1+1=0[/tex]
i, -i et -(1/2) sont des racines du polynôme P(z).
1b)
[tex]P(z)=2(z-i)(z+1)(z+\frac{1}{2})[/tex]
2
[tex]z_1=(2-i)^2-2(1+3i)^2=4-4i+i^2-2(1+6i+9i^2)=4-4i-1-2-12i+18=19-16i\\\\z_2=\frac{1-i}{2+3i} =\frac{(1-i)(2-3i)}{4+9} =\frac{2-3i-2i+3i^2}{13}= -\frac{1}{13}-\frac{5}{13}i\\ \\z_3=(1-2i)^4 =((1-2i)^2)^2=(1-4i+4i^2)^2=(-3-4i)^2=9+24i+16i^2=-7+24i[/tex]
3)
[tex]z=a+ib\\a) \ 3+i(a+ib)+2i=a+ib\\3+ia+i^2b+2i=a+ib\\3-b+(2+a)i =a +ib\\\rightarrow 2+a=b \rightarrow a=b-2\\\rightarrow3-b=a\\\rightarrow b-2=3-b \rightarrow 2b=5 \rightarrow b=\frac{5}{2} \\a=\frac{5}{2}-2=\frac{1}{2}\\\\z=\frac{1}{2}+\frac{5}{2}i \\\\b) \ (1+3i)(a+ib)-(i-1)=i(a-ib)-3\\a+ib+3ia+3i^2b-i+1=ia-i^2b-3\\a-3b+1+i(b+3a-1)=b-3+ia\\\rightarrow a+1-3b=b-3 \rightarrow a=4b-4\\\rightarrow b+3a-1=a \rightarrow 2a+b-1=0\rightarrow 8b-8+b-1=0\rightarrow b=1\\a=4*1-4=0\\\\z=i[/tex]