hcju/proofwikiqa huggingface dataset:
Let $N_f$ and $N_{f + g}$ be the number of zeroes of $f$ and $f + g$ in $D$ respectively.  By the Argument Principle:  :$\displaystyle N_f = \frac 1 {2 \pi i} \oint_D \frac {f' \left({z}\right)} {f \left({z}\right)} \rd z$ Similarly:  :$\displaystyle N_{f + g} = \frac 1 {2 \pi i} \oint_D \frac {\left({f + g}\right)' \left({z}\right)} {\left({f + g}\right) \left({z}\right)} \rd z$ We aim to show that $N_f = N_{f + g}$.  From $\left\vert{g \left({z}\right)}\right\vert < \left\vert{f \left({z}\right)}\right\vert$ we have that $f$ is non-zero on $D$, as $\left\vert{z}\right\vert \ge 0$ for all $z \in \C$.  From the fact that $\left\vert{g \left({z}\right)}\right\vert \ne \left\vert{f \left({z}\right)}\right\vert$ we also have that $g \left({z}\right) \ne - f \left({z}\right)$, so $f + g$ is also non-zero on $D$.  We have:  {{begin-eqn}} {{eqn	| l = N_{f + g} - N_f  	| r = \frac 1 {2 \pi i} \oint_D \frac {\left({f + g}\right)' \left({z}\right)} {\left({f + g}\right) \left({z}\right)} \rd z - \frac 1 {2 \pi i} \oint_D \frac {f' \left({z}\right)} {f \left({z}\right)} \rd z }} {{eqn	| r = \frac 1 {2 \pi i} \oint_D \left({\frac {\left({f + g}\right)' \left({z}\right)} {\left({f + g}\right) \left({z}\right)} - \frac {f' \left({z}\right)} {f \left({z}\right)} }\right) \rd z 	| c = Linear Combination of Contour Integrals }} {{eqn	| r = \frac 1 {2 \pi i} \oint_D \left({\frac {\left({f \left({1 + \frac g f}\right)}\right)' \left({z}\right)} {\left({f \left({1 + \frac g f}\right)}\right) \left({z}\right)} - \frac{f' \left({z}\right)} {f \left({z}\right)} }\right) \rd z }} {{eqn	| r = \frac 1 {2 \pi i} \oint_D \left({\frac {\left({f' \left({1 + \frac g f}\right)}\right) \left({z}\right)} {\left({f \left({1 + \frac g f}\right)}\right) \left({z}\right)} +  \frac {\left({f \left({1 + \frac g f}\right)'}\right)\left({z}\right)} {\left({f \left({1 + \frac g f}\right)}\right) \left({z}\right)} - \frac {f' \left({z}\right)} {f \left({z}\right)} }\right) \mathrm dz 	| c = Product Rule }} {{end-eqn}} As $g \left({z}\right) \ne - f \left({z}\right)$, we have $\dfrac {g \left({z}\right)} {f \left({z}\right)} \ne -1$, so $1 + \dfrac g f$ has no zeroes on $D$. So: {{begin-eqn}} {{eqn	| l = \frac 1 {2 \pi i} \oint_D \left({\frac {\left({f' \left({1 + \frac g f}\right)}\right) \left({z}\right)} {\left({f \left({1 + \frac g f}\right)}\right) \left({z}\right)} +  \frac {\left({f \left({1 + \frac g f}\right)'}\right)\left({z}\right)} {\left({f \left({1 + \frac g f}\right)}\right) \left({z}\right)} - \frac {f' \left({z}\right)} {f \left({z}\right)} }\right) \mathrm dz 	| r = \frac 1 {2 \pi i} \oint_D \left({\frac {f' \left({z}\right)} {f \left({z}\right)} + \frac {\left({1 + \frac g f}\right)' \left({z}\right)} {\left({1 + \frac g f}\right) \left({z}\right)} - \frac {f' \left({z}\right)} {f \left({z}\right)} }\right) \rd z }} {{eqn	| r = \frac 1 {2 \pi i} \oint_D \frac {\left({1 + \frac g f}\right)' \left({z}\right)} {\left({1 + \frac g f}\right) \left({z}\right)} \rd z }} {{eqn	| r = 0 }} {{end-eqn}} Hence $N_{f + g} = N_f$.  {{qed}} {{Namedfor|Eugène Rouché|cat = Rouché}} Category:Complex Analysis t3yui5qtz8w91eu2d6cc4zf4tuxsqkh
Let $n_1, n_2, n_3 \ldots n_N$ denote the zeroes of $f$, and $p_1, p_2, p_3 \ldots p_P$ denote its poles. Then, there exists a function $g$ nonzero and holomorphic on the boundary and on the interior of $D$ such that: :$\displaystyle f\left({z}\right) = \frac { \prod_{k \mathop = 1}^N \left( { z - n_k } \right) } { \prod_{j \mathop = 1}^P \left( { z - p_j } \right) } g\left({z}\right)$ Taking the logarithmic derivative: {{begin-eqn}} {{eqn	| l = \mathcal L \left( { f\left( {z} \right) } \right)  	| r = \frac { f'\left({z}\right) } { f\left({z}\right) } 	| c = {{Defof|Logarithmic Derivative of Meromorphic Function}} }} {{eqn	| r = \mathcal L \left( { \prod_{k \mathop = 1}^N \left( { z - n_k } \right) } \right) - \mathcal L \left( { \prod_{j \mathop = 1}^P \left( { z - p_j } \right) } \right) + \mathcal L \left( { g\left({z}\right) } \right) }} {{eqn	| r = \sum_{k \mathop = 1}^N \frac 1 {z - n_k} - \sum_{j \mathop = 1}^P \frac 1 {z - p_j} + \frac { g'\left({z}\right) } { g\left({z}\right) }  }} {{end-eqn}} Then:  {{begin-eqn}} {{eqn	| l = \oint_D \frac { f'\left({z}\right) } { f\left({z}\right) } \, \mathrm d z 	| r = \oint_D \left( \sum_{k \mathop = 1}^N \frac 1 {z - n_k} - \sum_{j \mathop = 1}^P \frac 1 {z - p_j} + \frac { g'\left({z}\right) } { g\left({z}\right) } \right) \, \mathrm d z }} {{eqn	| r = \sum_{k \mathop = 1}^N \oint_D \frac 1 {z - n_k} \, \mathrm d z - \sum_{j \mathop =1}^P \oint_D \frac 1 {z - p_j} \, \mathrm d z + \oint_D \frac { g'\left({z}\right) } { g\left({z}\right) } \, \mathrm d z  	| c = Linear Combination of Contour Integrals }} {{eqn	| r = \sum_{k=1}^N 2\pi i - \sum_{j \mathop = 1}^P 2\pi i + \oint_D \frac{ g'\left({z}\right) } { g\left({z}\right) } \, \mathrm d z }} {{eqn	| r = 2\pi i \left( {N - P} \right) + \oint_D \frac{ g'\left({z}\right) } { g\left({z}\right) } \, \mathrm d z }} {{end-eqn}} By our construction of $g$, $\frac {g'} g$ is holomorphic on and inside $D$ so, by the Cauchy Integral Theorem, the integral is equal to zero. We therefore have: :$\displaystyle \oint_D \frac { f'\left({z}\right) } { f\left({z}\right) } \, \mathrm d z = 2\pi i \left( {N - P} \right)$ Hence: :$\displaystyle N - P = \frac 1 {2\pi i} \oint_D \frac { f'\left({z}\right) } { f\left({z}\right) } \, \mathrm d z$ {{qed}} Category:Complex Analysis 3vebn4wwvo42e2xnediyfj7zdrvz1ta
By Krull's Theorem, $A$ has a maximal ideal. By Maximal Ideal of Commutative and Unitary Ring is Prime Ideal, $A$ has a prime ideal. {{qed}}
Let $A / \mathfrak a$ be the quotient ring. By Proper Ideal iff Quotient Ring is Nontrivial, $A / \mathfrak a$ is non-trivial. By Krull's Theorem, $A / \mathfrak a$ has a maximal ideal. By Correspondence Theorem for Quotient Rings, $A$ has a maximal ideal containing $\mathfrak a$. {{qed}} Category:Commutative Algebra Category:Maximal Ideals of Rings ido4yhyw70by29yomtpi83yvszusm01
As $g$ has a zero of multiplicity $1$ at $a$, $\dfrac f g$ has a simple pole at $a$ by definition. So:  {{begin-eqn}} {{eqn	| l = \Res {\dfrac f g} a 	| r = \lim_{z \mathop \to a} \paren {z - a} \frac {\map f z} {\map g z} 	| c = Residue at Simple Pole }} {{eqn	| r = \lim_{z \mathop \to a} \frac {\map f z} {\frac {\map g z - \map g a} {z - a} } 	| c = $\map g a = 0$ }} {{eqn	| r = \frac {\lim_{z \mathop \to a} \map f z} {\lim_{z \mathop \to a} \frac {\map g z - \map g a} {z - a} }  	| c = Quotient Rule for Limits of Functions }} {{eqn	| r = \frac {\map f a} {\map {g'} a} 	| c = {{Defof|Derivative of Complex Function}} }}  {{end-eqn}} {{qed}} Category:Complex Analysis c1c4bmvfo9s55to5jwmx5asi5hqrex0
Let $\map f x$ be the real function defined on $\openint 0 {2 \pi}$ as: :$\map f x = \begin{cases} \sin \dfrac x 2 & : 0 \le x < \pi \\ -\sin \dfrac x 2 & : \pi < x \le 2 \pi \end{cases}$ From Half-Range Fourier Sine Series: $\sin \dfrac x 2$ over $\closedint 0 \pi$, $-\sin \dfrac x 2$ over $\closedint 0 {2 \pi}$, we have: :$\map f x \sim \displaystyle \frac 8 \pi \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {n \sin n x} {4 n^2 - 1}$ Setting $x = \dfrac {\pi} 2$, we have: {{begin-eqn}} {{eqn | l = \sin \dfrac \pi 4       | r = \frac 8 \pi \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {n \sin \dfrac {n \pi} 2 } {4 n^2 - 1}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\sqrt 2} 2       | r = \frac 8 \pi \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {n \sin \dfrac {n \pi} 2 } {4 n^2 - 1}       | c = Sine of $\dfrac \pi 4$ }} {{eqn | ll= \leadsto       | l = \frac \pi {\sqrt 2}       | r = 8 \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {n \sin \dfrac {n \pi} 2 } {4 n^2 - 1}       | c = rearranging }} {{end-eqn}} When $n$ is even, $\dfrac {n \pi} 2$ is an integer multiple of $\pi$. Hence, in this case, from Sine of Multiple of Pi: :$\sin \dfrac {n \pi} 2 = 0$ When $n$ is odd it can be expressed as $n = 2 r - 1$ for $r \ge 1$. Hence we have: {{begin-eqn}} {{eqn | l = \frac \pi {\sqrt 2}       | r = 8 \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {n \sin \dfrac {n \pi} 2 } {4 n^2 - 1}       | c = from above }} {{eqn | r = 8 \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {n \sin \dfrac {n \pi} 2 } {\paren {2 n + 1} \paren {2 n - 1} }       | c = Difference of Two Squares }} {{eqn | r = 8 \sum_{r \mathop = 1}^\infty \paren {-1}^{\paren {2 r - 1} - 1} \frac {\paren {2 r - 1} \sin \dfrac {\paren {2 r - 1} \pi} 2 } {\paren {2 \paren {2 r - 1} + 1} \paren {2 \paren {2 r - 1} - 1} }       | c = from above: terms in even $n$ vanish }} {{eqn | r = 8 \sum_{r \mathop = 1}^\infty \paren {-1}^{2 r} \frac {\paren {2 r - 1} \map \sin {\paren {r - 1} + \frac 1 2} \pi} {\paren {4 r - 1} \paren {4 r - 3} }       | c = simplification }} {{eqn | r = 8 \sum_{r \mathop = 1}^\infty \frac {\paren {2 r - 1} \map \sin {\paren {r - 1} + \frac 1 2} \pi} {\paren {4 r - 1} \paren {4 r - 3} }       | c = $\paren {-1}^{2 r} = 1$ }} {{eqn | r = 8 \sum_{r \mathop = 1}^\infty \frac {\paren {2 r - 1} \paren {-1}^{r - 1} } {\paren {4 r - 1} \paren {4 r - 3} }       | c = Sine of Half-Integer Multiple of Pi }} {{eqn | r = 8 \sum_{r \mathop = 1}^\infty \paren {-1}^{r - 1} \frac {2 \paren {r - \frac 1 2} } {16 r^2 - 16 r + 3}       | c = multiplying out }} {{eqn | r = \sum_{r \mathop = 1}^\infty \paren {-1}^{r - 1} \frac {r - \frac 1 2} {r^2 - r + \frac 3 {16} }       | c = simplifying }} {{end-eqn}} {{qed}}
We have:  {{begin-eqn}} {{eqn	| l = \frac {r - \frac 1 2} {r^2 - r + \frac 3 {16} } 	| r = \frac {16 r - 8} {16 r^2 - 16 r + 3} }} {{eqn	| r = 2 \paren {\frac {8 r - 4} {\paren {4 r - 1} \paren {4 r - 3} } } 	| c = factorising, extracting a factor of $2$ }} {{eqn	| r = 2 \paren {\frac {\paren {4 r - 3} + \paren {4 r - 1} } {\paren {4 r - 1} \paren {4 r - 3} } } }} {{eqn	| r = 2 \paren {\frac 1 {4 r - 1} + \frac 1 {4 r - 3} } }} {{end-eqn}} So, we have:  {{begin-eqn}} {{eqn	| l = \sum_{r \mathop = 1}^\infty \paren {-1}^r \frac {r - \frac 1 2} {r^2 - r + \frac 3 {16} } 	| r = -2 \sum_{r \mathop = 1}^\infty \paren {-1}^r \paren {\frac 1 {4 r - 1} + \frac 1 {4 r - 3} } }} {{eqn	| r = -2 \sum_{r = 1}^\infty \paren {-1}^r \paren {\int_0^1 x^{4 r - 2} \rd x + \int_0^1 x^{4 r - 4} \rd x} 	| c = Primitive of Power, Fundamental Theorem of Calculus }} {{eqn	| r = -2 \sum_{r \mathop = 1}^\infty \paren {-1}^r \int_0^1 \paren {x^{4 r - 2} + x^{4 r - 4} } \rd x 	| c = Linear Combination of Definite Integrals }} {{eqn	| r = -2 \int_0^1 \paren {\sum_{r \mathop = 1}^\infty \paren {-1}^r \paren {x^{4 r - 2} + x^{4 r - 4} } } \rd x 	| c = Fubini's Theorem }} {{eqn	| r = -2 \int_0^1 \paren {\frac 1 {x^2} \sum_{r \mathop = 1}^\infty \paren {-x^4}^r - \sum_{r \mathop = 0}^\infty \paren {-x^4}^r} \rd x 	| c = shifting the index in the second sum, and extracting a factor of $x^{-4}$ from the first }} {{eqn	| r = -2 \int_0^1 \paren {\frac 1 {x^2} \paren {\sum_{r \mathop = 0}^\infty \paren {-x^4}^r - 1} - \sum_{r \mathop = 0}^\infty \paren {-x^4}^r} \rd x }} {{eqn	| r = -2 \int_0^1 \paren {\paren {\frac 1 {x^2} - 1} \sum_{r \mathop = 0}^\infty \paren {-x^4}^r - \frac 1 {x^2} } \rd x 	| c = grouping terms }} {{eqn	| r = -2 \int_0^1 \paren {\paren {\frac {1 - x^2} {x^2} } \paren {\frac 1 {1 + x^4} } - \frac 1 {x^2} } \rd x 	| c = Sum of Infinite Geometric Sequence }} {{eqn	| r = -2 \int_0^1 \paren {\frac {1 - x^2 - \paren {1 + x^4} } {x^2 \paren {1 + x^4} } } \rd x 	| c = writing as a single fraction }} {{eqn	| r = 2 \int_0^1 \paren {\frac {x^2 \paren {1 + x^2} } {x^2 \paren {1 + x^4} } } \rd x 	| c = extracting a factor of $x^2$ }} {{eqn	| r = 2 \int_0^1 \frac {1 + x^2} {1 + x^4} \rd x 	| c = cancelling the factor of $x^2$ }} {{eqn	| r = \sqrt 2 \intlimits {\map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } } 0 1 	| c = Primitive of $\dfrac {1 + x^2} {1 + x^4}$ }} {{eqn	| r = \sqrt 2 \paren {\arctan 0 - \lim_{x \mathop \to 0^+} \map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } } }} {{eqn	| r = -\sqrt 2 \lim_{u \mathop \to -\infty} \arctan u 	| c = setting $\displaystyle \frac 1 {\sqrt 2} \paren {x - \frac 1 x} = u$, Arctangent of Zero is Zero }} {{eqn	| r = -\sqrt 2 \paren {-\frac \pi 2} 	| c = Limit to Infinity of Arctangent Function }} {{eqn	| r = \frac \pi {\sqrt 2} }} {{end-eqn}} {{qed}}
Let $\map f x$ be the real function defined on $\openint {-\pi} \pi$ as: :$\map f x = \begin{cases} \cos x & : -\pi < x < 0 \\ -\cos x & : 0 < x < \pi \end{cases}$ From Fourier Series: $\cos x$ over $\openint {-\pi} 0$, $-\cos x$ over $\openint 0 \pi$, we have: :$\displaystyle \map f x \sim -\frac 8 \pi \sum_{r \mathop = 1}^\infty \frac {r \sin 2 r x} {4 r^2 - 1}$ Setting $x = \dfrac \pi 4$, we have: {{begin-eqn}} {{eqn | l = -\cos \dfrac \pi 4       | r = -\frac 8 \pi \sum_{r \mathop = 1}^\infty \frac {r \sin 2 r \dfrac \pi 4} {4 r^2 - 1}       | c =  }} {{eqn | ll= \leadsto       | l = \cos \dfrac \pi 4       | r = \frac 8 \pi \sum_{r \mathop = 1}^\infty \frac {r \sin \dfrac {r \pi} 2} {4 r^2 - 1}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\sqrt 2} 2       | r = \frac 8 \pi \sum_{r \mathop = 1}^\infty \frac {r \sin \dfrac {r \pi} 2} {4 r^2 - 1}       | c = Cosine of $\dfrac \pi 4$ }} {{eqn | ll= \leadsto       | l = \frac {\pi} {8 \sqrt 2}       | r = \sum_{r \mathop = 1}^\infty \frac {r \sin \dfrac {r \pi} 2} {4 r^2 - 1}       | c = rearranging }} {{end-eqn}} When $r$ is even, $\dfrac {r \pi} 2$ is an integer multiple of $\pi$. Hence, in this case, from Sine of Multiple of Pi: :$\sin \dfrac {r \pi} 2 = 0$ When $r$ is odd it can be expressed as $r = 2 n - 1$ for $n \ge 1$. Hence we have: {{begin-eqn}} {{eqn | l = \frac {\pi} {8 \sqrt 2}       | r = \sum_{r \mathop = 1}^\infty \frac {r \sin \dfrac {r \pi} 2} {4 r^2 - 1}       | c = from above }} {{eqn | r = \sum_{r \mathop = 1}^\infty \frac {r \sin \dfrac {r \pi} 2} {\paren {2 r + 1} \paren {2 r - 1} }       | c = Difference of Two Squares }} {{eqn | r = \sum_{n \mathop = 1}^\infty \frac {\paren {2 n - 1} \sin \dfrac {\paren {2 n - 1} \pi} 2} {\paren {2 \paren {2 n - 1} + 1} \paren {2 \paren {2 n - 1} - 1} }       | c = from above: terms in even $r$ vanish }} {{eqn | r = \sum_{n \mathop = 1}^\infty \frac {\paren {2 n - 1} \sin \paren {\paren {n - 1} + \frac 1 2} \pi} {\paren {4 n - 1} \paren {4 n - 3} }       | c =  }} {{eqn | r = \sum_{n \mathop = 1}^\infty \frac {\paren {2 n - 1} \paren {-1}^{n - 1} } {\paren {4 n - 1} \paren {4 n - 3} }       | c = Sine of Half-Integer Multiple of Pi }} {{eqn | r = \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {2 n - 1} {\paren {4 n - 1} \paren {4 n - 3} }       | c = rearranging }} {{end-eqn}} {{qed}}
Let $z \in D$. Then:  {{begin-eqn}} {{eqn	| l = \map f z 	| r = \frac 1 {2 \pi i} \int_{\partial D} \frac {\map f t} {t - z} \rd t 	| c = Cauchy's Integral Formula }} {{eqn	| r = \frac 1 {2 \pi i} \int_{\partial D} \frac {\map f t} {\paren {t - a} \paren {1 - \frac {z - a} {t - a} } } \rd t }} {{end-eqn}} Note that for all $t \in \partial D$, we have: :$\cmod {z - a} < \cmod {t - a} = r$. Therefore: :$\cmod {\dfrac {z - a} {t - a} } < 1$ so Sum of Infinite Geometric Sequence may be applied. This gives: {{begin-eqn}} {{eqn	| l = \map f z 	| r = \frac 1 {2 \pi i} \int_{\partial D} \frac {\map f t} {t - a} \sum_{n \mathop = 0}^\infty \paren {\frac {z - a} {t - a} }^n \rd t }} {{eqn	| r = \frac 1 {2 \pi i} \int_{\partial D} \sum_{n \mathop = 0}^\infty \map f t \frac {\paren {z - a}^n} {\paren {t - a}^{n + 1} } \rd t }} {{end-eqn}} From Continuous Function on Compact Space is Bounded, there exists a real $M \ge 0$, such that: :$\forall t \in \partial D: \cmod {\map f t} \le M$ As $\cmod {\dfrac {z - a} {t - a} } < 1$, there exists a real $0 \le N < 1$ such that: :$\cmod {\dfrac {z - a} {t - a} } \le N$ Then:  :$\displaystyle \cmod {\dfrac {\paren {z - a}^n} {\paren {t - a}^{n + 1} } \map f t} = \cmod {\frac 1 {t - a} } \cmod {\dfrac {\paren {z - a}^n} {\paren {t - a}^n} } \cmod {\map f t} \le \frac 1 r M N^n$  As $N < 1$, we have that $\displaystyle \frac 1 r \sum_{n \mathop = 0}^\infty M N^n$ converges by Sum of Infinite Geometric Sequence.  Therefore by the Weierstrass M-Test, we have that:  :$\displaystyle \sum_{n \mathop = 0}^\infty \map f t \frac {\paren {z - a}^n} {\paren {t - a}^{n + 1} }$  converges uniformly on $D$. Therefore:  :$\displaystyle \frac 1 {2 \pi i} \int_{\partial D} \sum_{n \mathop = 0}^\infty \map f t \frac {\paren {z - a}^n} {\paren {t - a}^{n + 1} } \rd t = \sum_{n \mathop = 0}^\infty \paren {z - a}^n \paren {\frac 1 {2 \pi i} \int_{\partial D} \frac {\map f t} {\paren {t - a}^{n + 1} } \rd t}$ Let:  :$\displaystyle c_n = \frac 1 {2 \pi i} \int_{\partial D} \frac {\map f t} {\paren {t - a}^{n + 1} } \rd t$ Then:  :$\displaystyle \map f z = \sum_{n \mathop = 0}^\infty c_n \paren {z - a}^n$  so $f$ is complex analytic on $D$.  {{qed}} Category:Complex Analysis hcrmtz533h0gj5nushcwfbcwj2c1rid
Let $g : D \to \C$ be a complex function with:  :$\map g z = \begin{cases}\frac {\map f z} z & z \ne 0 \\ \map {f'} 0 & z = 0\end{cases}$ By Schwarz's Lemma: Lemma, $g$ is holomorphic on $D$.  Let $r$ be a real number with $0 < r < 1$.  Let $C_r$ be the closed disk of radius $r$ centred at $0$. Since $C_r$ is compact, by Continuous Function on Compact Space is Bounded, $g$ attains its maximum on $C_r$. Let $\alpha \in C_r$ be a point where this maximum is attained. If $g$ is constant, then we can pick $\alpha$ to be any point on $C_r$.  If $g$ is non-constant, then by the maximum modulus principle, $\alpha$ necessarily lies on the boundary of $C_r$. That is, $\cmod \alpha = r$ if $g$ is non-constant.  If $g$ is constant, pick $\alpha$ such that $\cmod \alpha = r$ to avoid splitting into cases. With that, for all $z \in C_r$ we have: :$\cmod {\map g z} \le \cmod {\map g \alpha}$ Since $0$ lies in the interior of $C_r$, we must have $\alpha \ne 0$, so: {{begin-eqn}} {{eqn	| l = \cmod {\map g \alpha} 	| r = \cmod {\frac {\map f \alpha} \alpha} }} {{eqn	| r = \frac {\cmod {\map f \alpha} } {\cmod \alpha} }} {{eqn	| o = \le 	| r = \frac 1 r 	| c = since $\cmod {\map f z} \le 1$ on $D$, and $\cmod \alpha = r$ }} {{end-eqn}} So, for all $z \in C_r$ we have:  :$\cmod {\map g z} \le \dfrac 1 r$ Suppose, aiming for a contradiction, that there exists $\omega \in D$ with:  :$\cmod \omega < 1$ :$\cmod {\map g \omega} = 1 + \varepsilon > 1$ Note that we then have:  :$0 < \dfrac 1 {1 + \varepsilon} < 1$ Note that on $C_R$ with $R > \map \max {\dfrac 1 {1 + \varepsilon}, \cmod \omega}$, we should have:  :$\cmod {\map g z} \le \dfrac 1 R < 1 + \varepsilon$ However, we have $\omega \in C_R$ and:  :$\cmod {\map g \omega} = 1 + \varepsilon$ a contradiction.  So, it must be the case that:  :$\cmod {\map g z} \le 1$ for all $z \in D$. We have:  :$\cmod {\map {f'} 0} = \cmod {\map g 0} \le 1$ Note that:  :$\cmod {\map f 0} = 0$ so certainly:  :$\cmod {\map f z} \le \cmod z$ for $z = 0$.  Take $z \in D \setminus \set 0$, from:  :$\cmod {\map g z} \le 1$ we have:  :$\cmod {\dfrac {\map f z} z} \le 1$ that is:  :$\cmod {\map f z} \le \cmod z$ hence the result. {{qed}} {{Namedfor|Karl Hermann Amandus Schwarz|cat = Schwarz}} Category:Complex Analysis Category:Schwarz's Lemma swppkjrzg84jlasrupawb8iilml5ik3
Let $\map f x$ be the real function defined on $\hointl {-\pi} \pi$ as: :$\map f x = e^x$ From Fourier Series: $e^x$ over $\openint {-\pi} \pi$, we have: :$(1): \quad \displaystyle \map f x \sim \map S x = \frac {\sinh \pi} \pi \paren {1 + 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac {\cos n x} {1 + n^2} - \frac {n \sin n x} {1 + n^2} } }$ Let $x = \pi$. By Fourier's Theorem, we have: {{begin-eqn}} {{eqn | l = \map S \pi       | r = \frac 1 2 \paren {\lim_{x \mathop \to \pi^+} \map f x + \lim_{x \mathop \to \pi^-} \map f x}       | c =  }} {{eqn | r = \dfrac {e^\pi + e^{-\pi} } 2       | c =  }} {{eqn | r = \cosh \pi       | c = {{Defof|Hyperbolic Cosine}} }} {{end-eqn}} Thus setting $x = \pi$ in $(1)$, we have: {{begin-eqn}} {{eqn | l = \cosh \pi       | r = \frac {\sinh \pi} \pi \paren {1 + 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac {\cos \pi} {1 + n^2} - \frac {n \sin \pi} {1 + n^2} } }       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\pi \cosh \pi} {\sinh \pi}       | r = 1 + 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac {\paren {-1}^n} {1 + n^2} - \frac {n \sin \pi} {1 + n^2} }       | c = Cosine of Multiple of Pi and multiplying both sides by $\dfrac \pi {\sinh \pi}$ }} {{eqn | ll= \leadsto       | l = \dfrac {\pi \coth \pi - 1} 2       | r = \sum_{n \mathop = 1}^\infty \frac 1 {1 + n^2}       | c = {{Defof|Hyperbolic Cotangent|index = 2}}, Sine of Multiple of Pi and simplifying }} {{end-eqn}} {{qed}}
Let $\map f x$ be the real function defined on $\openint 0 \pi$ as: :$\map f x = \cos \lambda x$ From Half-Range Fourier Cosine Series: $\cos \lambda x$ over $\openint 0 \pi$ its Fourier series can be expressed as: :$\displaystyle \cos \lambda x \sim \frac {2 \lambda \sin \lambda \pi} \pi \paren {\frac 1 {2 \lambda^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos n x} {\lambda^2 - n^2} }$ Because of the nature of this expansion, we have that: :$\map f \pi = \map f {-\pi}$ and so the expansion holds for $x = \pi$. Also note that because $\lambda$ is not an integer, $\sin \lambda \pi \ne 0$ and so $\cot \pi \lambda$ is defined. So, setting $x = \pi$: {{begin-eqn}} {{eqn | l = \cos \lambda \pi       | r = \frac {2 \lambda \sin \lambda \pi} \pi \paren {\frac 1 {2 \lambda^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos n \pi} {\lambda^2 - n^2} }       | c =  }} {{eqn | ll= \leadsto       | l = \pi \cot \pi \lambda       | r = 2 \lambda \paren {\frac 1 {2 \lambda^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos n \pi} {\lambda^2 - n^2} }       | c = {{Defof|Real Cotangent Function}} }} {{eqn | r = \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {2 \lambda \cos n \pi} {\lambda^2 - n^2}       | c = simplification }} {{eqn | r = \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {2 \lambda \paren {-1}^n} {\lambda^2 - n^2}       | c = Cosine of Multiple of Pi }} {{eqn | r = \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \frac {2 \lambda} {\lambda^2 - n^2}       | c = simplification }} {{end-eqn}} {{qed}}