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[Linear Algebra] Why is z̅ᵏ = z̅ ᵏ  ?

Conjugate covering both on one of them, and just z^k on the other

1 Answer

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  • Pope
    Lv 7
    8 months ago
    Favorite Answer

    Let z have form |z|cis(θ).

    conj(z) = |z|cis(-θ)

    conj(z)ᵏ = |z|ᵏcis(-kθ) ... (De Moivre)

    zᵏ = |zᵏ|cis(kθ)

    conj(zᵏ) = |zᵏ|cis(-kθ)

    conj(zᵏ) = |z|ᵏcis(-kθ)

    conj(zᵏ) = conj(z)ᵏ

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