Question

(4a) (4 points) Consider the following poorly justified proof of the fact that $\lim_{z \to \infty} f(z) = a_0$. 1. For $n < 0$, we know that $\lim_{z \to \infty} a_n (z - z_0)^n = 0$. 2. We have that $\lim_{z \to \infty} \sum_{n=-\infty}^0 a_n (z - z_0)^n = \sum_{n=-\infty}^0 \lim_{z \to \infty} a_n (z - z_0)^n$. 3. Therefore, $\lim_{z \to \infty} f(z) = a_0 + \sum_{n=-\infty}^{-1} a_n (z - z_0)^n = a_0$. This proof is essentially correct, but one line is completely unjustified. Identify the unjustified line and explain why it is unjustified. (4b) (6 points) Assume the claim in part (a) is true. Prove that the function $g(z) = \begin{cases} f(\frac{1}{z} + z_0) & 0 < |z| < \frac{1}{R} \\ a_0 & z = 0 \end{cases}$ is analytic. (Hint: think in terms of removable singularities.)

          (4a) (4 points) Consider the following poorly justified proof of the fact that $\lim_{z \to \infty} f(z) = a_0$.
1. For $n < 0$, we know that $\lim_{z \to \infty} a_n (z - z_0)^n = 0$.
2. We have that
$\lim_{z \to \infty} \sum_{n=-\infty}^0 a_n (z - z_0)^n = \sum_{n=-\infty}^0 \lim_{z \to \infty} a_n (z - z_0)^n$.
3. Therefore, $\lim_{z \to \infty} f(z) = a_0 + \sum_{n=-\infty}^{-1} a_n (z - z_0)^n = a_0$.
This proof is essentially correct, but one line is completely unjustified. Identify the unjustified line and
explain why it is unjustified.
(4b) (6 points) Assume the claim in part (a) is true. Prove that the function $g(z) = \begin{cases} f(\frac{1}{z} + z_0) & 0 < |z| < \frac{1}{R} \\ a_0 & z = 0 \end{cases}$ is analytic. (Hint: think in terms of removable singularities.)

(4a) (4 points) Consider the following poorly justified proof of the fact that limz →∞ f(z) = a0.
1. For n < 0, we know that limz →∞ an (z - z0)^n = 0.
2. We have that
limz →∞∑n=-∞^0 an (z - z0)^n = ∑n=-∞^0 limz →∞ an (z - z0)^n.
3. Therefore, limz →∞ f(z) = a0 + ∑n=-∞^-1 an (z - z0)^n = a0.
This proof is essentially correct, but one line is completely unjustified. Identify the unjustified line and
explain why it is unjustified.
(4b) (6 points) Assume the claim in part (a) is true. Prove that the function g(z) = f((1)/(z) + z0) 0 < |z| < (1)/(R)
a0 z = 0 is analytic. (Hint: think in terms of removable singularities.)

Added by Lance G.

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