(6+4i)/(1+i) + (1-8i)(1-2i). Created by данясмирнов89. algebra-ru

(1+i) + (1-8i)(1-2i)...

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(6+4i)/(1+i) + (1-8i)(1-2i)

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Dимасuk

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(6 + 4i)/(1 + i) + (1 - 8i)(1 - 2i) = (6 + 4i)(1 - i)/(1 + i)(1 - i) + (1 - 2i - 8i + 16i²) =
(6 - 6i + 4i - 4i²)/(1 - i²) + (1 - 16 - 10i) = (6 + 4 - 2i)/2 - 15 - 10i =
(10 - 2i)/2 - 15 - 10i = 5 - i - 15 - 10i = -10 - 11i.

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