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Nurayna
@Nurayna
August 2022
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даны два комплексных числа z1=a+bi и z2=c+di. Найти модуль числа z=2z1z2+z1/z2
a4 b1 c1 d-1
буду благодарна)
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allangarsk
Z=2(a+bi)(c+di)+(a+bi)/(c+di)
z=2((a*с - b*d)+(ad + bc)i)+((a*с+b*d)/(c^2 + d^2)+(c*b - d*a)/(c^2 + d^2)i)
z=2((4*1 - 1*(-1))+(4*(-1) + 1*1)i))+((4*1 + 1*(-1))/(1^2 + (-1)^2)+(1*1 - 4*(-1))/(1^2 + (-1)^2)i))
z=2((4 + 1)+(-4 + 1)i))+((4 -1)/(1 + 1)+(1 + 4)/(1 + 1)i))
z=2((5-3i))+((3)/(2)+(5)/(2)i))
z=10-6i + 3/2 + 5/2i
z=(23/2)+(-7/2)i
z= 23/2 -7/2i
z= |23/2 -7/2i|
z= V((23/2)^2 + (7/2)^2)
z= V(132.25 + 12.25)
z= V(132.25 + 12.25)
z= V(144.5)
z=12.02
2 votes
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Answers & Comments
z=2((a*с - b*d)+(ad + bc)i)+((a*с+b*d)/(c^2 + d^2)+(c*b - d*a)/(c^2 + d^2)i)
z=2((4*1 - 1*(-1))+(4*(-1) + 1*1)i))+((4*1 + 1*(-1))/(1^2 + (-1)^2)+(1*1 - 4*(-1))/(1^2 + (-1)^2)i))
z=2((4 + 1)+(-4 + 1)i))+((4 -1)/(1 + 1)+(1 + 4)/(1 + 1)i))
z=2((5-3i))+((3)/(2)+(5)/(2)i))
z=10-6i + 3/2 + 5/2i
z=(23/2)+(-7/2)i
z= 23/2 -7/2i
z= |23/2 -7/2i|
z= V((23/2)^2 + (7/2)^2)
z= V(132.25 + 12.25)
z= V(132.25 + 12.25)
z= V(144.5)
z=12.02