For what values of z is log z^2 = 2log z if principal branch of the logarithm is used on both sides of the equation?
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Let r = |z| and θ = Arg(z). Then log z = log r + iθ, -π < θ ≤ π. Thus 2log z = log r + i(2θ). If -π < θ ≤ -π/2, then -2π < 2θ ≤ -π, so 0 < 2θ + 2π ≤ π. This implies Arg(z^2) = 2θ + 2π. Thus log z^2 = log r^2 + i(2θ + 2π) = 2log r + i(2θ) + i(2π) ≠ 2log z. If -π/2 < θ ≤ π/2, then -π < 2θ ≤ π. So Arg(z^2) = 2θ and log z^2 = log r^2 + i(2θ) = 2log r + i(2θ) = 2log z. If π/2 < θ ≤ π, then π < 2θ ≤ 2π, so -π < 2θ - 2π ≤ π. This implies Arg(z^2) = 2θ - 2π. Thus log z^2 = log r^2 + i(2θ - 2π) = 2log r + i(2θ) - i(2π) ≠ 2log z. Therefore, log z^2 = 2log z holds only in the region -π/2 < θ ≤ π/2 (which is contained in the half - plane Re(z) ≥ 0).