胡算 $f(z)=\oint_{C} \frac{4 z}{(z+1)(z-1)(z-3)} d z$,
(a) 若封閉曲缐 $\mathrm{C}$ 分別為: $|z|=2$, 則 $f(z)=$
(b) 若封閉曲缐 C 分別為: $|z|=4$, 則 $f(z)=$

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∻Solution by Steps∻ ‖ step 1 ⋮ We start by identifying the poles of the integrand $f(z) = \frac{4z}{(z+1)(z-1)(z-3)}$. The poles are at $z = -1$, $z = 1$, and $z = 3$. ‖ ‖ step 2 ⋮ For part (a), the path of integration is $|z| = 2$. We need to check which poles lie inside this circle. The poles $z = -1$ and $z = 1$ lie inside $|z| = 2$, but $z = 3$ does not. ‖ ‖ step 3 ⋮ We use the residue theorem to evaluate the integral. The residue theorem states that $\oint_{C} f(z) \, dz = 2\pi i \sum ext{Res}(f, z_k)$, where $z_k$ are the poles inside $C$. ‖ ‖ step 4 ⋮ Calculate the residues at $z = -1$ and $z = 1$. For $z = -1$, the residue is $ ext{Res}(f, -1) = \lim_{z o -1} (z + 1) \frac{4z}{(z+1)(z-1)(z-3)} = \frac{4(-1)}{(-2)(-4)} = \frac{1}{2}$. ‖ ‖ step 5 ⋮ For $z = 1$, the residue is $ ext{Res}(f, 1) = \lim_{z o 1} (z - 1) \frac{4z}{(z+1)(z-1)(z-3)} = \frac{4(1)}{(2)(-2)} = -1$. ‖ ‖ step 6 ⋮ Sum the residues: $ ext{Res}(f, -1) + ext{Res}(f, 1) = \frac{1}{2} - 1 = -\frac{1}{2}$. ‖ ‖ step 7 ⋮ Apply the residue theorem: $f(z) = 2\pi i \left( -\frac{1}{2} ight) = -\pi i$. ‖ ‖ step 8 ⋮ For part (b), the path of integration is $|z| = 4$. All three poles $z = -1$, $z = 1$, and $z = 3$ lie inside this circle. ‖ ‖ step 9 ⋮ Calculate the residue at $z = 3$. The residue is $ ext{Res}(f, 3) = \lim_{z o 3} (z - 3) \frac{4z}{(z+1)(z-1)(z-3)} = \frac{4(3)}{(4)(2)} = 1.5$. ‖ ‖ step 10 ⋮ Sum the residues: $ ext{Res}(f, -1) + ext{Res}(f, 1) + ext{Res}(f, 3) = \frac{1}{2} - 1 + 1.5 = 1$. ‖ ‖ step 11 ⋮ Apply the residue theorem: $f(z) = 2\pi i (1) = 2\pi i$. ‖ ∻Answer∻ ⚹ (a) $f(z) = -\pi i$ ⚹ ⚹ (b) $f(z) = 2\pi i$ ⚹ ∻Key Concept∻ ⚹Residue Theorem⚹ ∻Explanation∻ ⚹The residue theorem allows us to evaluate complex integrals by summing the residues of the integrand's poles inside the contour of integration.⚹◊What is the formula for calculating the value of a complex integral over a closed curve?⍭ Generate me a similar question◊

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