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2000914
@2000914
August 2022
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Комплексные числа
Решите уравнение:
а)z^2+z+2=0
б) z^2-(3-2i)z+5-5i=0
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ShirokovP
Verified answer
Z^2 + z + 2 = 0
D = 1 - 8 = - 7
z1 = ( - 1 + i √7)/2
z2 = ( - 1 - i √7)/2
z^2 - (3 - 2i)z + 5 - 5i = 0
D = (3 - 2i)^2 - 4*(5 - 5i) = 8i - 15
z1 = (3 - 2i + √(8i - 15))/2 = (3 - 2i + 1 + 4i)/2 = 2 + i
z2 = (3 - 2i - √(8i - 15))/2 = (3 - 2i - 1 - 4i)/2 = 1 - 3i
√(8i - 15) = ± (1 + 4i)
a = ± 1
b = ± 4
10 votes
Thanks 11
2000914
Найдите значение параметра а, при котром уравнение z^2-(3-2i)z+a(1-i)=0 имеет корень z1=2+i
2000914
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2000914
А такое умеешь решать?
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Answers & Comments
Verified answer
Z^2 + z + 2 = 0D = 1 - 8 = - 7
z1 = ( - 1 + i √7)/2
z2 = ( - 1 - i √7)/2
z^2 - (3 - 2i)z + 5 - 5i = 0
D = (3 - 2i)^2 - 4*(5 - 5i) = 8i - 15
z1 = (3 - 2i + √(8i - 15))/2 = (3 - 2i + 1 + 4i)/2 = 2 + i
z2 = (3 - 2i - √(8i - 15))/2 = (3 - 2i - 1 - 4i)/2 = 1 - 3i
√(8i - 15) = ± (1 + 4i)
a = ± 1
b = ± 4