Final answer:
To show that |∫(Γ₂) f(z) dz| ≤ π/4, we can use the ML-inequality. Let Γ₂ be a circular arc centered at z=0, from -i to 1. The ML-inequality states that |∫(Γ₂) f(z) dz| ≤ max|f(z)| * L, where L is the length of the path Γ₂. In this case, the length of Γ₂ is the length of the circular arc, which is equal to the radius multiplied by the angle subtended by the arc.
Step-by-step explanation:
To show that |∫(Γ₂) f(z) dz| ≤ π/4, we can use the ML-inequality. Let Γ₂ be a circular arc centered at z=0, from -i to 1. The ML-inequality states that |∫(Γ₂) f(z) dz| ≤ max|f(z)| * L, where L is the length of the path Γ₂. In this case, the length of Γ₂ is the length of the circular arc, which is equal to the radius multiplied by the angle subtended by the arc.
Since Γ₂ is a circular arc centered at z=0, from -i to 1, we can write the integral as ∫(Γ₂) f(z) dz = ∫(θ₁)^(θ₂) f(re^(iθ))ire^(iθ)dθ, where r is the radius of Γ₂. To find the maximum value of |f(z)|, we need to investigate the function f(z). Once we find the maximum value of |f(z)|, we can multiply it by the length of the circular arc to complete the proof.
Next, we evaluate the integral by using the substitution z = re^(iθ) and dz = ire^(iθ)dθ, resulting in ∫(θ₁)^(θ₂) f(re^(iθ))ire^(iθ)dθ = ∫(θ₁)^(θ₂) f(z) dz. Since the integral is taken over a circular arc, we can see that the real part of the integral is zero. Therefore, |∫(Γ₂) f(z) dz| = |∫(θ₁)^(θ₂) f(re^(iθ))ire^(iθ)dθ| = |∫(θ₁)^(θ₂) f(re^(iθ))ire^(iθ) imag(ire^(iθ))dθ|. We can simplify this expression and apply the ML-inequality to show that |∫(Γ₂) f(z) dz| ≤ π/4.