MHB Complex Residue Calculation at a Specific Point

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The discussion centers on correcting the calculation of residues in complex analysis, specifically at the point z=3. The correct residue at this point is derived using the derivative of the function e^(iz) divided by the polynomial z^2 + 4z + 29. The calculated residue at z=3 is (5i-1)e^(3i)/250. Participants emphasize the importance of following proper steps to ensure accurate residue calculations. Accurate residue determination is crucial for complex function analysis and applications in various fields.
Doomknightx9
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My residue is wrong. What is the solutions and the steps to achieve it ?
 

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Doomknightx9 said:
My residue is wrong. What is the solutions and the steps to achieve it ?
I think that you have the correct residues at $z = -2\pm5i$. The residue at $z=3$ is wrong. It should be $$\left.\frac d{dz}\,\frac{e^{iz}}{z^2 + 4z + 29}\right|_{z=3},$$ which I get to be $\dfrac{(5i-1)e^{3i}}{250}.$
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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