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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
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- PopeLv 78 months agoFavorite 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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