**Exercise 3.13**Let $ V = D(0,1) $. Suppose that $ f \in C(\overline{V}) \cap H(V) $, $ f(0) = 0 $, and $ |f(z)| \leq 1 $ for all $ z \in \overline{V} $. Show that $ |f(z)| \leq |z| $ for all $ z \in V $.*Hint:* Apply two previous exercises to the function $ g(z) = \frac{f(z)}{z} $.

Name: **Exercise 3.13**Let $ V = D(0,1) $. Suppose that $ f \in C(\overl
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Description: **Exercise 3.13**Let \( V = D(0,1) $. Suppose that $ f \in C(\overline{V}) \cap H(V) $, $ f(0) = 0 $, and $ |f(z)| \leq 1 $ for all $ z \in \overline{V} $. Show that $ |f(z)| \leq |z| $ for all $ z \in V $.*Hint:* Apply two previous exe

Answered: 3.13. Let V D(0. 1). Suppose that f e…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

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**Exercise 3.13**

Let $ V = D(0,1) $. Suppose that $ f \in C(\overline{V}) \cap H(V) $, $ f(0) = 0 $, and $ |f(z)| \leq 1 $ for all $ z \in \overline{V} $. Show that $ |f(z)| \leq |z| $ for all $ z \in V $.

*Hint:* Apply two previous exercises to the function $ g(z) = \frac{f(z)}{z} $.

Expert Solution

Step 1

Given $V = D (0, 1)$ which is open disk with radius less than one.

which is $|z| < 1$

Since $f \in C (\bar{V}) \cap H (V)$ means $f$ is analytic for $|z| < 1$ in the given domain, and satisfy the condition $f (0) = 0$ and $|f (z)| \leq 1$ $\forall z \in \bar{V}$ . We have to show $|f (z)| \leq |z| \forall z \in V$ .

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