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Tribekk
@Tribekk
August 2022
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(4+i/1-i)(3-i)+2
Комплексные числа
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Мартын05
(4+i/1-i)(3-i)+2
1) (4+i/1-i)=
(4+i/1-i)*((1+i)/(1+i))=((4+i)(1+i))/((1-i)(1+i))=(
(4+i)(1+i))/2=
=
(
4+4i+i-1)/2=(3+5i)/2;
((3+5i)/2)(3-i)+2=((3+5i)
(3-i))/2 +2=(9-3i+15i+5)/2 +2=(14+12i)/2 +2=7+6i+2=
=9+6i;
вроде так как-то
1 votes
Thanks 1
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Answers & Comments
1) (4+i/1-i)=(4+i/1-i)*((1+i)/(1+i))=((4+i)(1+i))/((1-i)(1+i))=((4+i)(1+i))/2=
=(4+4i+i-1)/2=(3+5i)/2;
((3+5i)/2)(3-i)+2=((3+5i)(3-i))/2 +2=(9-3i+15i+5)/2 +2=(14+12i)/2 +2=7+6i+2=
=9+6i;
вроде так как-то