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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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    Do you know the [contraction mapping theorem](http://en.wikipedia.org/wiki/Banach_fixed_point_theorem)?2012-05-12
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    What about Rouche's theorem?2012-05-12
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    ssume that f has more then one fixed point a,b and that a2012-05-12
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    @JoeStevenson: This version of MVT applies to real-valued functions, and does not generalize to complex-valued functions. We certainly can't say "$a" and have it make any sense, as the complex numbers cannot be ordered in a way compatible with field operations. We *can* however talk about $(a,b)$ for any distinct points $a,b$, but in this case it simply denotes the line segment from $a$ to $b$ with the endpoints removed.2012-05-14
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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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