Suppose that f E H(C) and |f(z)| < ekez for all z. Show that f(z) =ce? for some constant c.

Name: Suppose that f E H(C) and |f(z)| < ekez for all z. Show that f(z) =ce
Uploaded: 2024-03-20T06:38:58.000Z
Channel: Bartleby
Description: Suppose that f E H(C) and |f(z)| < ekez for all z. Show that f(z) =ce? for some constant c.

Consider fz=cez for z=x+i y Take absolute value on both sides: fz=cez=cex+iy=cexeiy=cex…

Answered: Suppose that f e H(C) and |f(2)I < eRez…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Concept explainers

Topic Video

Question

Suppose that f E H(C) and |f(z)| < ekez for all z. Show that f(z) =
ce? for some constant c.

Expert Solution

Solution:

Consider $f (z) = c e^{z}$ for $z = x + i y$

Take absolute value on both sides:

$\begin{array}{rcl} |f (z)| & = & c |e^{z}| \\ = & c |e^{x + i y}| \\ = & c |e^{x}| |e^{i y}| \\ = & c |e^{x}| ∵ |e^{i y}| = 1 \end{array}$

We know, in polar form $x = r \cos θ$ ,

$\begin{array}{rcl} |f (z)| & = & c |e^{r \cos θ}| \\ = & c |e^{r}| |e^{\cos θ}| \\ = & |e^{\cos θ}| if c = \frac{1}{|e^{r}|} \\ \leq & e^{R e (z)} \end{array}$

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Sequence$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Suppose that f e H(C) and |f(2)I < eRez for all z. Show that f(z) ce for some constant c.