Inequality and complex numbers

  • Thread starter oddiseas
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  • #1
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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.
 
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Answers and Replies

  • #2
LCKurtz
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Homework Statement



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


Homework Equations





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.
 
  • #3
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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.
 

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