Fifth Derivative Limit 02Ca30

1. **State the problem:** Calculate the value of $$\frac{1}{5!}\lim_{z\to1}\frac{d^5 }{dz^5}\left[(z-1)^5 \ln(z)+i(\arg(z)+2k\pi) \right].$$ 2. **Understand the components:** - The term $(z-1)^5 \ln(z)$ is a product of a polynomial and a logarithm. - The term $i(\arg(z)+2k\pi)$ is a constant with respect to $z$ because $\arg(z)$ is the argument of $z$ and $k$ is an integer; its derivative with respect to $z$ is zero. 3. **Focus on the derivative:** Since $i(\arg(z)+2k\pi)$ is constant in $z$, its derivatives vanish: $$\frac{d^5}{dz^5} i(\arg(z)+2k\pi) = 0.$$ 4. **Rewrite the problem:** We only need to compute $$\frac{1}{5!}\lim_{z\to1}\frac{d^5}{dz^5} \left[(z-1)^5 \ln(z)\right].$$ 5. **Use the Leibniz rule for the 5th derivative of a product:** Let $f(z) = (z-1)^5$ and $g(z) = \ln(z)$. Then $$\frac{d^5}{dz^5}[f(z)g(z)] = \sum_{m=0}^5 \binom{5}{m} f^{(m)}(z) g^{(5-m)}(z).$$ 6. **Calculate derivatives of $f(z)$:** - $f(z) = (z-1)^5$ - $f^{(0)}(z) = (z-1)^5$ - $f^{(1)}(z) = 5(z-1)^4$ - $f^{(2)}(z) = 20(z-1)^3$ - $f^{(3)}(z) = 60(z-1)^2$ - $f^{(4)}(z) = 120(z-1)$ - $f^{(5)}(z) = 120$ 7. **Calculate derivatives of $g(z) = \ln(z)$:** Recall: $$g^{(n)}(z) = (-1)^{n-1} (n-1)! / z^n$$ for $n \geq 1$ and $g^{(0)}(z) = \ln(z)$. So: - $g^{(0)}(z) = \ln(z)$ - $g^{(1)}(z) = \frac{1}{z}$ - $g^{(2)}(z) = -\frac{1}{z^2}$ - $g^{(3)}(z) = \frac{2}{z^3}$ - $g^{(4)}(z) = -\frac{6}{z^4}$ - $g^{(5)}(z) = \frac{24}{z^5}$ 8. **Evaluate each term at $z=1$:** Since $(z-1)^k$ for $k>0$ is zero at $z=1$, all terms with $f^{(m)}(1)$ where $m<5$ vanish because they contain $(z-1)$ factors. Specifically: - For $m=0$ to $4$, $f^{(m)}(1) = 0$ because each contains a factor $(z-1)$ to some positive power. - For $m=5$, $f^{(5)}(1) = 120$ (constant). 9. **Therefore, the only surviving term in the sum is for $m=5$:** $$\binom{5}{5} f^{(5)}(1) g^{(0)}(1) = 1 \times 120 \times \ln(1) = 120 \times 0 = 0.$$ 10. **Conclusion:** The 5th derivative at $z=1$ of $(z-1)^5 \ln(z)$ is zero. 11. **Final value:** $$\frac{1}{5!} \times 0 = 0.$$