Can Complex Function q(z) Be a Contraction Mapping?

AI Thread Summary
The discussion revolves around proving that the complex function q(z) = (z^7 + z^3 - i)/9 is a contraction mapping. The key requirement is to demonstrate the existence of a real number α < 1 such that the inequality |(q(z) - q(w))/|z - w|| < 1 holds. Participants express difficulty in algebraic manipulation to reach this conclusion, questioning whether certain conditions on z and w are necessary. Concerns are raised about the validity of q(z) being a contraction, particularly when evaluating specific values like q(2+0i), which yields a magnitude greater than 2. The thread highlights the need for further exploration of the function's properties to establish contraction mapping criteria.
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Homework Statement



Let z,w be complex numbers.

Homework Equations



Prove there is a real number \alpha &lt; 1 such that
<br /> \left|\frac{z^7 + z^3 - i}{9} - \frac{w^7 + w^3 - i}{9}\right| \leq \alpha<br /> \left|z - w\right|<br />

The goal is to show that \displaystyle q(z) = \frac{z^7 + z^3 - i}{9} is a contraction mapping in a real analysis contraction mapping problem. I am stuck here maybe because of algebraic manipulation.

The Attempt at a Solution



Is this the same as proving the following inequality?
<br /> \begin{align*}<br /> \frac{\left|\frac{z^7 + z^3 - i}{9} - \frac{w^7 + w^3 -<br /> i}{9}\right|}{\left|z - w\right|} &lt; 1<br /> \end{align*}<br />

If so then
<br /> \begin{align*}<br /> &amp; \frac{\left|\frac{z^7 + z^3 - i}{9} - \frac{w^7 + w^3 -<br /> i}{9}\right|}{\left|z - w\right|} = \left|\frac{\frac{z^7 + z^3 - i}{9} -<br /> \frac{w^7 + w^3 - i}{9}}{z - w}\right| = \left|\frac{\frac{(z^7 + z^3) -<br /> (w^7 + w^3)}{9}}{z-w}\right| = \frac{1}{9} \left|\frac{(z^7 + z^3) - (w^7 +<br /> w^3)}{z-w}\right| \\<br /> = &amp; \frac{1}{9} \left|\frac{z^7 - w^7 + z^3 - w^3}{z-w}\right| =<br /> \frac{1}{9} \left| \frac{(z-w)(z^6 + z^5w + \cdots + zw^5 + w^6) +<br /> (z-w)(z^2 + zw + w^2)}{z-w} \right| \\<br /> =&amp; \frac{1}{9} |(z^6 + z^5w + z^4w^2 + z^3w^3 + z^2w^4 + zw^5 +<br /> w^6) + (z^2 + zw + w^2)|<br /> \end{align*}<br />

How can I proceed from here?
 
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complexnumber said:

Homework Statement



Let z,w be complex numbers.

Homework Equations



Prove there is a real number \alpha &lt; 1 such that
<br /> \left|\frac{z^7 + z^3 - i}{9} - \frac{w^7 + w^3 - i}{9}\right| \leq \alpha<br /> \left|z - w\right|<br />

The goal is to show that \displaystyle q(z) = \frac{z^7 + z^3 - i}{9} is a contraction mapping in a real analysis contraction mapping problem. I am stuck here maybe because of algebraic manipulation.

The Attempt at a Solution



Is this the same as proving the following inequality?
<br /> \begin{align*}<br /> \frac{\left|\frac{z^7 + z^3 - i}{9} - \frac{w^7 + w^3 -<br /> i}{9}\right|}{\left|z - w\right|} &lt; 1<br /> \end{align*}<br />

If so then
<br /> \begin{align*}<br /> &amp; \frac{\left|\frac{z^7 + z^3 - i}{9} - \frac{w^7 + w^3 -<br /> i}{9}\right|}{\left|z - w\right|} = \left|\frac{\frac{z^7 + z^3 - i}{9} -<br /> \frac{w^7 + w^3 - i}{9}}{z - w}\right| = \left|\frac{\frac{(z^7 + z^3) -<br /> (w^7 + w^3)}{9}}{z-w}\right| = \frac{1}{9} \left|\frac{(z^7 + z^3) - (w^7 +<br /> w^3)}{z-w}\right| \\<br /> = &amp; \frac{1}{9} \left|\frac{z^7 - w^7 + z^3 - w^3}{z-w}\right| =<br /> \frac{1}{9} \left| \frac{(z-w)(z^6 + z^5w + \cdots + zw^5 + w^6) +<br /> (z-w)(z^2 + zw + w^2)}{z-w} \right| \\<br /> =&amp; \frac{1}{9} |(z^6 + z^5w + z^4w^2 + z^3w^3 + z^2w^4 + zw^5 +<br /> w^6) + (z^2 + zw + w^2)|<br /> \end{align*}<br />

How can I proceed from here?

Are there some conditions on z and w that you don't show and aren't using? It doesn't seem to me that q(z) = (z7 + z3 - i)/9 is a contraction mapping, in general. For example, q(2+0i) = (128 + 8 - i)/9 has a magnitude considerably larger than 2. If q were a contraction mapping, I would expect |q(z)| <= |z|.
 
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