answer:(a) We will use the Cauchy integral formula and the maximum modulus principle to show the inequality. First, let's rewrite the integral equation as follows: displaystyle f(z) = lambda int_{|w|=1} frac{e^{f(w)}}{w-z} dw = lambda int_{|w|=1} frac{e^{f(w)}}{w(1-frac{z}{w})} dw Now, let's apply the Cauchy integral formula for the function g(w) = frac{e^{f(w)}}{w} and the point z inside the unit circle: displaystyle f(z) = lambda cdot 2pi i cdot g(z) = lambda cdot 2pi i cdot frac{e^{f(z)}}{z} Taking the absolute value of both sides: displaystyle |f(z)| = |lambda| cdot 2pi cdot frac{e^{text{Re}(f(z))}}{|z|} Since |z| < 1, we have: displaystyle |f(z)| < |lambda| cdot 2pi cdot e^{text{Re}(f(z))} Now, let's apply the maximum modulus principle to the function h(z) = e^{f(z)}. The maximum modulus of h(z) on the unit circle is: displaystyle max_{|z|=1} |h(z)| = max_{|z|=1} |e^{f(z)}| = e^{max_{|z|=1} text{Re}(f(z))} Since |f(z)| < |lambda| cdot 2pi cdot e^{text{Re}(f(z))}, we have: displaystyle max_{|z|=1} text{Re}(f(z)) < ln(2pi |lambda|) Therefore, for any |z| < 1: displaystyle |f(z)| leq |ln(1-|lambda|)| (b) To find the unique solution f(z) for the given values of lambda and f(0), we will use the Schwarz lemma. Let g(z) = frac{f(z)}{lnleft(frac{16}{9}right)}. Then, g(0) = 1 and |g(z)| leq |ln(1-|lambda|)| / lnleft(frac{16}{9}right) for |z| < 1. Applying the Schwarz lemma to the function g(z), we have: displaystyle |g(z)| leq |z| and displaystyle |g'(0)| leq 1 Since g(0) = 1, the equality case of the Schwarz lemma implies that g(z) = z for |z| < 1. Therefore, the unique solution for f(z) is: displaystyle f(z) = lnleft(frac{16}{9}right) cdot z

answer:To solve this problem, we will first rewrite the integral equation using polar coordinates. Let z = e^(iθ) for some θ in [0, 2π]. Then, the integral equation becomes: f(e^(iθ)) = e^(iθ) + (2/π) ∫︀(0)^(2π) ln|e^(iθ) − e^(it)|f(e^(it))dt Now, we want to find the specific value of f(1/2). To do this, we will substitute z = 1/2 into the integral equation. Since |1/2| ≠ 1, we need to find a value of θ such that e^(iθ) = 1/2. We can do this by setting θ = π: f(e^(iπ)) = e^(iπ) + (2/π) ∫︀(0)^(2π) ln|e^(iπ) − e^(it)|f(e^(it))dt Now, we have: f(-1) = -1 + (2/π) ∫︀(0)^(2π) ln|-1 − e^(it)|f(e^(it))dt To evaluate the integral, we will use the property of logarithms that ln|a - b| = ln|b - a|. This allows us to rewrite the integral as: f(-1) = -1 + (2/π) ∫︀(0)^(2π) ln|e^(it) + 1|f(e^(it))dt Now, we will use the substitution u = e^(it) + 1. Then, du = ie^(it)dt, and the integral becomes: f(-1) = -1 + (2/π) ∫︀(0)^(2π) ln|u|f(u - 1)(du/iu) Now, we can use the Cauchy integral formula to evaluate the integral. The Cauchy integral formula states that for a holomorphic function g(z) on a simply connected domain D containing the unit circle, we have: g(z) = (1/2πi) ∫︀(∂D) g(w)/(w - z)dw Applying the Cauchy integral formula to our integral equation, we get: f(-1) = -1 + (2/π) (1/2πi) ∫︀(∂D) ln|u|f(u - 1)/(u - (-1))du Since f(z) is continuous on the unit circle, it is holomorphic on the unit disk, which is a simply connected domain. Therefore, the integral evaluates to 0, and we have: f(-1) = -1 Thus, the specific value of f(1/2) is f(-1) = -1.

answer:To solve the integral equation using complex analysis, we first rewrite the integral in terms of complex exponentials. Recall that cos(nx) = frac{1}{2}(e^{inx} + e^{-inx}) and cos x = frac{1}{2}(e^{ix} + e^{-ix}). Substituting these expressions into the integral, we get: int_{-pi}^{pi} frac{frac{1}{2}(e^{inx} + e^{-inx})}{3-frac{1}{2}(e^{ix} + e^{-ix})}dx = frac{pi}{sqrt{7}}(sqrt{7}-2)^n. Now, let z = e^{ix}. Then, dz = ie^{ix}dx = iz dx, and dx = frac{dz}{iz}. Also, z^{-1} = e^{-ix}. We can rewrite the integral in terms of z: int_{|z|=1} frac{frac{1}{2}(z^n + z^{-n})}{3-frac{1}{2}(z + z^{-1})}frac{dz}{iz} = frac{pi}{sqrt{7}}(sqrt{7}-2)^n. Simplifying the integrand, we have: int_{|z|=1} frac{z^n + z^{-n}}{2iz(3z^2 - 2z^3 - 3 + 2z)}dz = frac{pi}{sqrt{7}}(sqrt{7}-2)^n. Now, we factor the denominator: int_{|z|=1} frac{z^n + z^{-n}}{2iz(z-1)^2(2z-3)}dz = frac{pi}{sqrt{7}}(sqrt{7}-2)^n. We will use the residue theorem to evaluate this integral. The residue theorem states that for a function f(z) with isolated singularities inside a simple closed contour C, oint_C f(z)dz = 2pi i sum_{k=1}^n text{Res}(f, z_k), where text{Res}(f, z_k) is the residue of f at the singularity z_k. The function frac{z^n + z^{-n}}{2iz(z-1)^2(2z-3)} has singularities at z = 1 and z = frac{3}{2}. However, only the singularity at z = 1 is inside the unit circle |z| = 1. Thus, we only need to find the residue at z = 1. Since the singularity at z = 1 is a second-order pole, we can find the residue using the formula: text{Res}(f, z) = lim_{z to 1} frac{d}{dz} left[(z-1)^2 frac{z^n + z^{-n}}{2iz(z-1)^2(2z-3)}right]. Taking the derivative and simplifying, we get: text{Res}(f, z) = lim_{z to 1} frac{z^n + z^{-n} - nz^{n-1} - n z^{-(n+1)}}{2i(2z-3)}. Now, we evaluate the limit as z to 1: text{Res}(f, 1) = frac{1 + 1 - n - n}{2i(2-3)} = frac{2 - 2n}{-2i} = i(1-n). Applying the residue theorem, we have: int_{|z|=1} frac{z^n + z^{-n}}{2iz(z-1)^2(2z-3)}dz = 2pi i cdot i(1-n) = 2pi(1-n). Comparing this result to the given expression, we have: 2pi(1-n) = frac{pi}{sqrt{7}}(sqrt{7}-2)^n. Dividing both sides by pi, we get: 2(1-n) = frac{1}{sqrt{7}}(sqrt{7}-2)^n. Now, we analyze the cases for different values of n: 1. If n = 0, the equation becomes 2 = frac{1}{sqrt{7}}, which is not true. So, n neq 0. 2. If n = 1, the equation becomes 0 = frac{1}{sqrt{7}}(sqrt{7}-2), which is true. So, n = 1 is a solution. 3. If n > 1, the equation becomes 2(1-n) = frac{1}{sqrt{7}}(sqrt{7}-2)^n < 0, which is not true. So, n > 1 is not a solution. 4. If n < 0, the equation becomes 2(1-n) = frac{1}{sqrt{7}}(sqrt{7}-2)^{-n} > 0, which is true. So, n < 0 is a solution. Thus, the given integral equation holds for n = 1 and n < 0.

answer:To solve the integral equation, we first rewrite the integral equation in terms of the convolution of two functions. Let g(t) = frac{t}{t^2 + 4}, then the integral equation can be written as: (f * g)(x) = 3e^{-2x} Now, we take the Fourier transform of both sides of the equation. Recall that the Fourier transform of a convolution is the product of the Fourier transforms of the individual functions: mathcal{F}{(f * g)(x)} = mathcal{F}{f(x)} cdot mathcal{F}{g(x)} We need to find the Fourier transforms of f(x) and g(x). We already know the Fourier transform of the right-hand side of the equation: mathcal{F}{3e^{-2x}} = 3sqrt{2pi} delta(omega + 2) Now, we find the Fourier transform of g(x): mathcal{F}{g(x)} = int_{-infty}^{infty} frac{x}{x^2 + 4} e^{-iomega x} dx This integral can be solved using the residue theorem from complex analysis. The poles of the integrand are at x = pm 2i. Closing the contour in the upper half-plane, we find that the residue at x = 2i is: text{Res}_{x=2i} frac{x}{x^2 + 4} e^{-iomega x} = frac{2i}{4i} e^{-2omega} = frac{1}{2} e^{-2omega} Thus, the Fourier transform of g(x) is: mathcal{F}{g(x)} = 2pi i cdot frac{1}{2} e^{-2omega} = pi i e^{-2omega} Now, we can find the Fourier transform of f(x): mathcal{F}{f(x)} = frac{mathcal{F}{3e^{-2x}}}{mathcal{F}{g(x)}} = frac{3sqrt{2pi} delta(omega + 2)}{pi i e^{-2omega}} Finally, we take the inverse Fourier transform to find f(x): f(x) = mathcal{F}^{-1}{ frac{3sqrt{2pi} delta(omega + 2)}{pi i e^{-2omega}} } Using the inverse Fourier transform properties, we get: f(x) = frac{3sqrt{2pi}}{pi i} e^{2x} u(x) where u(x) is the unit step function. This is the solution to the given integral equation.