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kmaltseva22
@kmaltseva22
July 2022
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Выполните деление комплексных чисел
а) 1/i
б) 1/(1-i)
в) (1-i)/(1+i)
г) (3-2i)/(1+3i)
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Shelkovaia
А) 1\i = ((1)(-1i))\((1i)(-1i)) =( -1·1i)\(1·1) = -1i\1 = -1i
б) 1/(1-i)=
(1)(1 + 1i)\
(1 - 1i)(1 + 1i)
= =
1·1 + 1·1i\
1·1 + 1·1
=
1 + 1i\
1 + 1
=
1 + 1i\
2
= 0.5 + 0.5
i
в) (1-i)/(1+i)=
(1 - 1i)(1 - 1i)\
(1 + 1i)(1 - 1i)
=
1·1 - 1·1i - 1·1i + 1·1i^2\
1·1 + 1·1
=
1 - 1i - 1i - 1\
1 + 1
=
-2i\
2
= -1
i
г) (3-2i)/(1+3i)=
(3 - 2i)(1 - 3i)\
(1 + 3i)(1 - 3i)
=
3·1 - 3·3i - 2·1i + 2·3i^2\
1·1 + 3·3
=
3 - 9i - 2i - 6\
1 + 9
=
-3 - 11i\
10
= -0.3 - 1.1
i
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Answers & Comments
б) 1/(1-i)=(1)(1 + 1i)\(1 - 1i)(1 + 1i) = = 1·1 + 1·1i\1·1 + 1·1 = 1 + 1i\1 + 1 = 1 + 1i\2 = 0.5 + 0.5i
в) (1-i)/(1+i)=(1 - 1i)(1 - 1i)\(1 + 1i)(1 - 1i) = 1·1 - 1·1i - 1·1i + 1·1i^2\1·1 + 1·1 = 1 - 1i - 1i - 1\1 + 1 = -2i\2 = -1i
г) (3-2i)/(1+3i)=(3 - 2i)(1 - 3i)\(1 + 3i)(1 - 3i) = 3·1 - 3·3i - 2·1i + 2·3i^2\1·1 + 3·3 = 3 - 9i - 2i - 6\1 + 9 = -3 - 11i\10 = -0.3 - 1.1i