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Let $f$ be analytic on the closed unit disk centered at the origin and $|f(z)| < 1$ for $|z| = 1$. Show that $f$ has exactly one fixed point inside the open unit disk. That is, there exists a unique number $z_0$ with $|z_0| < 1$ such that $f(z_0) = z_0$. We must prove 1 there exists at least on solution, 2 there is at most one solution. 1) Having problems with at least one solution. 2)By definition a is a fixed point if f(a) = a. Assume that f has more then one fixed point a,b and that a

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    The version of MVT applicable to complex-valued functions in one complex variable is as follows: If $U$ is an open subset of $\mathbb{C}$, $f:U\to\mathbb{C}$ analytic, and $a,b\in\mathbb{C}$ are such that the segment $[a,b]\subset U$, then $f(b)-f(a)=\int_a^bf'(z)dz=(b-a)\int_0^1f'\bigl(a+t(b-a)\bigr)dt$.2012-05-14

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Let $g(z)=f(z)-z$ and $h(z)=z$. Then on $\{|z|=1\}$, we have $|g(z)+h(z)|=|f(z)|<|g(z)|+|h(z)|$ This is because $|f(z)|<1$ and $|h(z)|=1$ on the boundary of the disk. Also, $f(z)-z$ and $z$ have no zeros on the boundary. Then Rouche's Theorem applies and $f(z)-z$ and $z$ have the same number of zeros inside the disk. We know $h(z)=z$ has one zero, so $f(z)-z$ has one zero. Hence there exists $z_0$ with $|z_0|<1$ and $f(z_0)=z_0$.

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    I see what your saying now. $T$hanks.2012-05-12