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demokratxd
@demokratxd
July 2022
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Решите, плизз, баллы приличные!
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Удачник66
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1) 1 - i = √2*(1/√2 - i*1/√2) = √2*(cos 7pi/4 + i*sin 7pi/4)
(1 - i)^12 = √2^12*(cos 84pi/4 + i*sin 84pi/4) = 64*(cos 21pi + i*sin 21pi) =
= 64(cos pi + i*sin pi) = 64*(-1 + 0*i) = -64
2) 1 + i = √2*(1/√2 + i*1/√2) = √2*(cos pi/4 + i*sin pi/4)
(1 + i)^17 = √2^17*(cos 17pi/4 + i*sin 17pi/4) = 256√2*(cos pi/4 + i*sin pi/4) =
= 256√2*(1/√2 + i/√2) = 256 + 256i
3) (-1 + i*√3*i)/2 = -1/2 + i*√3/2 = 1*(cos 2pi/3 + i*sin 2pi/3)
((-1 + i*√3*i)/2)^3 = 1^3*(cos 2pi + i*sin 2pi) = 1*(1 + i*0) = 1
4) (a + bi)/(a - bi) = (a + bi)^2 / ((a - bi)(a + bi)) = (a^2 + 2abi - b^2)/(a^2 + b^2)
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Answers & Comments
Verified answer
1) 1 - i = √2*(1/√2 - i*1/√2) = √2*(cos 7pi/4 + i*sin 7pi/4)(1 - i)^12 = √2^12*(cos 84pi/4 + i*sin 84pi/4) = 64*(cos 21pi + i*sin 21pi) =
= 64(cos pi + i*sin pi) = 64*(-1 + 0*i) = -64
2) 1 + i = √2*(1/√2 + i*1/√2) = √2*(cos pi/4 + i*sin pi/4)
(1 + i)^17 = √2^17*(cos 17pi/4 + i*sin 17pi/4) = 256√2*(cos pi/4 + i*sin pi/4) =
= 256√2*(1/√2 + i/√2) = 256 + 256i
3) (-1 + i*√3*i)/2 = -1/2 + i*√3/2 = 1*(cos 2pi/3 + i*sin 2pi/3)
((-1 + i*√3*i)/2)^3 = 1^3*(cos 2pi + i*sin 2pi) = 1*(1 + i*0) = 1
4) (a + bi)/(a - bi) = (a + bi)^2 / ((a - bi)(a + bi)) = (a^2 + 2abi - b^2)/(a^2 + b^2)