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Is this true? |z1+z2|^2 = (z1+z2)^2, zEC?

1 Answer

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  • Pope
    Lv 7
    8 months ago
    Favorite Answer

    I do not get the zEC part, but I think I get the rest of it.

    The expression |z₁ + z₂|² is real in all cases, but (z₁ + z₂)² is not necessarily real at all. Consider this counterexample:

    z₁ = -1

    z₂ = i

    |z₁ + z₂|²

    |-1 + i|²

    = √[1² + (-1)²]²

    = √(2)²

    = 2

    (z₁ + z₂)²

    = (-1 + i)²

    = (-1)² + i² - 2i

    = 1 - 1 - 2i

    = -2i

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