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vika218
@vika218
August 2022
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помощь! 1- запись в тригоном форме,2-операции над комплексн числами( сложить,вычесть,делить,умнож)3- объяснить почему +/-
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Удачник66
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1) z1 = 3 - 3i = 3√2*(1/√2 - 1/√2*i) = 3√2*(cos(7pi/4) + i*sin(7pi/4))
z2 = -2 - 2i = 2√2*(-1/√2 - 1/√2*i) = 2√2*(cos(5pi/4) + i*sin(5pi/4))
z3 = 4i = 4*(0 + 1*i) = 4*(cos(pi/2) + i*sin(pi/2))
2) z1 = 2 - i; z2 = 3 + i
z1 + z2 = 2 - i + 3 + i = 5
z1 - z2 = 2 - i - 3 - i = -1 - 2i
z1*z2 = (2 - i)(3 + i) = 6 - 3i + 2i - i^2 = 6 - i + 1 = 7 - i
z1 / z2 = (2 - i) / (3 + i) = (2 - i)(3 - i) / (9 + 1) = (6 - 5i - 1)/10 = (1 - i)/2
z2 / z1 = (3 + i) / (2 - i) = (3 + i)(2 + i) / (4 + 1) = (6 + 5i - 1)/5 = 1 + i
3) Извините, не понял задания
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Answers & Comments
Verified answer
1) z1 = 3 - 3i = 3√2*(1/√2 - 1/√2*i) = 3√2*(cos(7pi/4) + i*sin(7pi/4))z2 = -2 - 2i = 2√2*(-1/√2 - 1/√2*i) = 2√2*(cos(5pi/4) + i*sin(5pi/4))
z3 = 4i = 4*(0 + 1*i) = 4*(cos(pi/2) + i*sin(pi/2))
2) z1 = 2 - i; z2 = 3 + i
z1 + z2 = 2 - i + 3 + i = 5
z1 - z2 = 2 - i - 3 - i = -1 - 2i
z1*z2 = (2 - i)(3 + i) = 6 - 3i + 2i - i^2 = 6 - i + 1 = 7 - i
z1 / z2 = (2 - i) / (3 + i) = (2 - i)(3 - i) / (9 + 1) = (6 - 5i - 1)/10 = (1 - i)/2
z2 / z1 = (3 + i) / (2 - i) = (3 + i)(2 + i) / (4 + 1) = (6 + 5i - 1)/5 = 1 + i
3) Извините, не понял задания