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Виолетта15081
@Виолетта15081
July 2022
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Помогите пожалуйста
(1)Z1=3-5i ;Z2=7i-1
a)Z=Z1+Z2
b)Z=Z1-Z2
c)Z=Z1*Z2
(2)Вычислите
a)(1+i)(-1+2i)+1-3i
b)(1-2i)^2+i(4i+2)
c)i^4+7/i^6
(3)
(2x+y)+(x-y)i=18+3i
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fhzttf96
Z1 = - 2 + 5iz2 = 3 + i
z1 + z2 = - 2 + 5i + 3 + i =1 + 6i
z1 - z2 = - 2 + 5i - 3 - i = - 5 + 4i
z1*z2 = ( - 2 + 5i)(3 + i) = - 6 - 2i + 15i + 5i^2 = - 6 + 13i - 5 = - 11 + 13i
z1/z2 = ( - 2 + 5i)/(3 + i) = (( - 2 + 5i)*(3 - i))/((3 + i)*(3 - i)) = = ( - 6 + 2i + 15i - 5i^2) / (9 - i^2) = = ( - 6 + 17i + 5) / (9 + 1) = = ( - 1 + 17i)/10 = = - 0,1 + 1,7i
/z1 = - 2 - 5i
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Answers & Comments
z1 + z2 = - 2 + 5i + 3 + i =1 + 6i
z1 - z2 = - 2 + 5i - 3 - i = - 5 + 4i
z1*z2 = ( - 2 + 5i)(3 + i) = - 6 - 2i + 15i + 5i^2 = - 6 + 13i - 5 = - 11 + 13i
z1/z2 = ( - 2 + 5i)/(3 + i) = (( - 2 + 5i)*(3 - i))/((3 + i)*(3 - i)) = = ( - 6 + 2i + 15i - 5i^2) / (9 - i^2) = = ( - 6 + 17i + 5) / (9 + 1) = = ( - 1 + 17i)/10 = = - 0,1 + 1,7i
/z1 = - 2 - 5i