Proving an Inequality with Complex Numbers

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Homework Help Overview

The discussion revolves around proving an inequality involving complex numbers, specifically that if z and w are complex numbers with magnitudes less than or equal to 1, then the inequality |z+w| ≤ |1 + \overline{z} w| must hold. Participants are exploring various approaches to establish this inequality.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • One participant attempts to reduce the problem to a simpler form involving real variables x and y, suggesting that if both are less than or equal to 1, the inequality should hold. Others explore the implications of rewriting the inequality and consider potential connections to known inequalities, such as Jensen's inequality.

Discussion Status

The discussion is ongoing, with participants providing various insights and questioning each other's reasoning. Some have offered alternative perspectives on the inequality, while others seek clarification on specific points, indicating a collaborative exploration of the problem.

Contextual Notes

Participants are working under the constraints of the problem statement and are questioning the validity of certain assumptions and steps taken in their reasoning. There is a focus on understanding the relationships between the variables involved.

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Homework Statement



If z and w are complex numbers such that |z|<=1 and |w|<=1 then prove

\left| z+w \right| \leq \left| 1 + \overline{z} w \right|

The Attempt at a Solution



I have reduced this to essentially

x^2+y^2 <= 1+(xy)^2.

It seems to me if both x and y are less than or equal to 1, then the inequality must hold. I can't think of how to prove this formally, though. Any help on how to do this would be appreciated.
 
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Ok, so if x^2=a and y^2=b then you want to prove a+b<=1+ab, where 0<=a<=1 and 0<=b<=1. Write it as a*(1-b)+1*b<=1. If b is in [0,1] then the left side is a number between 'a' and '1', right? There's a name for this kind of inequality, but I forget what it is. There may also be an easier way to prove this. But I forget that too.
 
Maybe it's a form of a Jensen's inequality, in kind of a vague way.
 
Last edited:
Dick,

Thanks for your post. I am sorry, but I do not see how/why the following is true:

If b is in [0,1] then the left side is a number between 'a' and '1', right?
 
Suppose |\bar{z}| &lt; 1 then<br /> |w|^2 \le \frac{1-|z|^2}{1-|z|^2}<br /><br /> |w|^2-|\overline{z} w|^2 \le 1-|z|^2<br /><br /> |w|^2 + |z|^2 + 2\mbox{ Re}(\overline{z} w) \le 1 + |\overline{z} w|^2 + 2\mbox{ Re}(\overline{z} w)<br /><br /> \left| z+w \right|^2 \leq \left| 1 + \overline{z} w \right|^2<br />
 
Last edited:
mjordan2nd said:
Dick,

Thanks for your post. I am sorry, but I do not see how/why the following is true:

Can you show if t is in [0,1] then f(t)=a*t+b*(1-t) is between a and b? f(0)=b, f(1)=a and f(t) is a linear function.
 

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