# Inequality and complex numbers

## Homework Statement

b)show that if |z|>=3 then the following inequality holds

## Homework Equations

|z|>=3 now for this to hold b must be <=0 which gives the following:
|e^iz|=e^b(cosa+isina) and taking the magnitude yeilds (e^2b)^1/2
I dont know where to go next.

Last edited:

LCKurtz
Homework Helper
Gold Member

## Homework Statement

Show that |e^iz|<=1 for Im(z)>=0
b)show that if |z|>=3 then the following inequality holds

## The Attempt at a Solution

|e^iz|=e^-b(cosa+isina)
and since the inmaginery part is always one, the magnitude is always less or equal to one' provided b is <=0. Now the second part i am having trouble with.

You are having trouble with the first part. What are a and b? Are you writing
z = a + ib. If so, you should say so.

This equation |e^iz|=e^-b(cosa+isina) can't be correct because you have a real number on the left and a complex number on the right.

The "inmaginery" (that's imaginary) part of what is one? And why is it 1? Do you know how to compute the magnitude of a complex number?

"provided b is <=0". You are given Im(z) >= 0.

Back to the drawing board.

Mark44
Mentor
Please post the exact problem you are supposed to solve. The following makes no sense:
oddiseas said:
b)show that if |z|>=3 then the following inequality holds
What following inequality? It doesn't appear in either your original post or the edited post.