avewright/proofwiki-math huggingface dataset:
Theorem: Sequence Converges to Within Half Limit/Normed Division Ring

Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring with zero $0$.
Let $\sequence {x_n}$ be a sequence in $R$.
Let $\sequence {x_n}$ be convergent in the norm $\norm {\, \cdot \,}$ to the following limit:
:$\ds \lim_{n \mathop \to \infty} x_n = l \ne 0$
Then:
:$\exists N: \forall n > N: \norm {x_n} > \dfrac {\norm l} 2$

Proof:
Since $l \ne 0$, by {{Norm-axiom-mult|1}}:
:$\norm l > 0$
Let us choose $N$ such that:
:$\forall n > N: \norm {x_n - l} < \dfrac {\norm l} 2$
Then:
{{begin-eqn}}
{{eqn | l = \norm {x_n - l}
      | o = <
      | r = \frac {\norm l} 2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \norm l - \norm {x_n}
      | o = \le
      | r = \norm {x_n - l}
      | c = Reverse Triangle Inequality
}}
{{eqn | o = <
      | r = \frac {\norm l} 2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \norm {x_n}
      | o = >
      | r = \norm l - \frac {\norm l} 2
      | c = 
}}
{{eqn | r = \frac {\norm l} 2
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Sequences
Category:Limits of Sequences
Category:Normed Division Rings
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Theorem: Limit of Subsequence equals Limit of Sequence/Normed Division Ring

Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring with zero: $0$.
Let $\sequence {x_n}$ be a sequence in $R$.
Let $\sequence {x_n}$ be convergent in the norm $\norm {\, \cdot \,}$ to the following limit:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
Let $\sequence {x_{n_r} }$ be a subsequence of $\sequence {x_n}$.
Then:
:$\sequence {x_{n_r} }$ is convergent and $\ds \lim_{r \mathop \to \infty} x_{n_r} = l$

Proof:
Let $d$ denote the metric induced by $\norm {\, \cdot \,}$, that is,
:$d \tuple {x, y} = \norm {x - y}$
By definition of convergence in a normed division ring:
:$\sequence {x_n}$ converges to $l$ in $\struct {R, \norm {\, \cdot \,} }$ {{iff}} $\sequence {x_n}$ converges to $l$ in the metric space $\struct {R, d}$.
We can then apply the result of Limit of Subsequence equals Limit of Sequence for metric spaces to get the result for normed division rings.
{{qed}}
Category:Sequences
Category:Normed Division Rings
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Theorem: Combination Theorem for Sequences/Normed Division Ring/Quotient Rule

Suppose $m \ne 0$.
Then:
:$\exists k \in \N : \forall n \in \N: y_{k + n} \ne 0$
and the sequences:
:$\sequence {x_{k + n} \ {y_{k + n} }^{-1} }$ and $\sequence { {y_{k + n} }^{-1} \ x_{k + n} }$ are well-defined and convergent with:
:$\ds \lim_{n \mathop \to \infty} x_{k + n} \ {y_{k + n} }^{-1} = l m^{-1}$
:$\ds \lim_{n \mathop \to \infty} {y_{k + n} }^{-1} \ x_{k + n} = m^{-1} l$

Proof:
From Inverse Rule for Sequences in Normed Division Ring:
:$\exists k \in \N : \forall n \in \N : y_{k + n} \ne 0$
and the sequence:
:$\sequence { {y_{k + n} }^{-1} }$
is well-defined and convergent with:
:$\ds \lim_{n \mathop \to \infty} {y_{k + n} }^{-1} = m^{-1}$
From Limit of Subsequence equals Limit of Sequence, $\sequence {x_{k + n} }$ is convergent with:
:$\ds \lim_{n \mathop \to \infty} x_{k + n} = l$
From Product Rule for Sequences in Normed Division Ring:
:$\ds \lim_{n \mathop \to \infty} x_{k + n} \ {y_{k + n} }^{-1} = l m^{-1}$
:$\ds \lim_{n \mathop \to \infty} {y_{k + n} }^{-1} \ x_{k + n} = m^{-1} l$
{{qed}}
Category:Combination Theorem for Sequences in Normed Division Rings
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Theorem: Real Sine Function is Bounded

:$\size {\sin x} \le 1$

Proof:
From the algebraic definition of the real sine function:
:$\ds \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}$
it follows that $\sin x$ is a real function.
Thus $\sin^2 x \ge 0$.
From Sum of Squares of Sine and Cosine, we have that $\cos^2 x + \sin^2 x = 1$.
Thus it follows that:
:$\sin^2 x = 1 - \cos^2 x \le 1$
From Ordering of Squares in Reals and the definition of absolute value, we have that:
:$x^2 \le 1 \iff \size x \le 1$
The result follows.
{{qed}}
Theorem: Complex Cosine Function is Unbounded

The complex cosine function is unbounded.

Proof:
Let $K \in \R_{>0}$ be an arbitrary real number.
Let $p = \ln {2 K}$.
Let $z = i p$, where $i$ denotes the imaginary unit.
Then:
{{begin-eqn}}
{{eqn | l = \cos z
      | r = \dfrac {\map \exp {i \paren {i p} } + \map \exp {-i \paren {i p} } } 2
      | c = Euler's Cosine Identity
}}
{{eqn | r = \dfrac {\exp p + \map \exp {-p} } 2
      | c = simplifying: $i^2 = -1$
}}
{{eqn | o = >
      | r = \dfrac {\exp p} 2
      | c = as $\map \exp {-p} > 0$
}}
{{eqn | r = \dfrac {\map \exp {\ln {2 K} } } 2
      | c = 
}}
{{eqn | r = K
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \cmod {\map \cos {i p} }
      | o = >
      | r = K
      | c = 
}}
{{end-eqn}}
Thus for any $K$ we can find $z \in \C$ such that $\cmod {\cos z} > K$.
Hence the result by definition of unbounded.
{{qed}}
Theorem: Complex Cosine Function is Unbounded

The complex cosine function is unbounded.

Proof:
By Complex Cosine Function is Entire, we have that $\cos$ is an entire function. 
{{AimForCont}} that $\cos$ is a bounded function.
By Liouville's Theorem, we have that $\cos$ is a constant function.
However, by Cosine of Zero is One: 
:$\cos 0 = 1$
and by Cosine of Right Angle: 
:$\cos \dfrac \pi 2 = 0$
Therefore, $\cos$ is not a constant function.
This contradicts the statement that $\cos$ is a constant function.
We hence conclude, by Proof by Contradiction, that our assumption that $\cos$ is a bounded function is false.
Hence $\cos$ is unbounded.
{{qed}}
Theorem: Real Part of Complex Exponential Function

Let $z = x + i y \in \C$ be a complex number, where $x, y \in \R$.
Let $\exp z$ denote the complex exponential function.
Then:
:$\map \Re {\exp z} = e^x \cos y$
where:
:$\Re z$ denotes the real part of a complex number $z$
:$e^x$ denotes the real exponential function of $x$
:$\cos y$ denotes the real cosine function of $y$.

Proof:
From the definition of the complex exponential function:
:$\exp z := e^x \paren {\cos y + i \sin y}$
The result follows by definition of the real part of a complex number.
{{qed}}
Theorem: Imaginary Part of Complex Exponential Function

Let $z = x + i y \in \C$ be a complex number, where $x, y \in \R$.
Let $\exp z$ denote the complex exponential function.
Then:
:$\map \Im {\exp z} = e^x \sin y$
where:
:$\map \Im z$ denotes the imaginary part of a complex number $z$
:$e^x$ denotes the real exponential function of $x$
:$\sin y$ denotes the real sine function of $y$.

Proof:
From the definition of the complex exponential function:
:$\exp z := e^x \paren {\cos y + i \sin y}$
The result follows by definition of the imaginary part of a complex number.
{{qed}}
Theorem: Argument of Exponential is Imaginary Part plus Multiple of 2 Pi

Let $z \in \C$ be a complex number.
Let $\exp z$ denote the complex exponential of $z$.
Let $\arg z$ denote the argument of $z$.
Then:
:$\map \arg {\exp z} = \set {\Im z + 2 k \pi: k \in \Z}$
where $\Im z$ denotes the imaginary part of $z$.

Proof:
Let $z = x + i y$.
Let $\theta \in \map \arg {\exp z}$.
We have:
{{begin-eqn}}
{{eqn | l = \exp z
      | r = e^x \paren {\cos y + i \sin y}
      | c = {{Defof|Exponential Function/Complex|Exponential Function|subdef = Real Functions}}
}}
{{eqn | ll= \leadsto
      | l = y
      | o = \in
      | r = \map \arg {\exp z}
      | c = {{Defof|Polar Form of Complex Number}}
}}
{{eqn | ll= \leadsto
      | l = \map \arg {\exp z}
      | r = \set {y + 2 k \pi: k \in \Z}
      | c = {{Defof|Argument of Complex Number}}
}}
{{eqn | r = \set {\Im z + 2 k \pi: k \in \Z}
      | c = {{Defof|Imaginary Part}}
}}
{{end-eqn}}
{{qed}}
Theorem: Real Part of Sine of Complex Number

Let $z = x + i y \in \C$ be a complex number, where $x, y \in \R$.
Let $\sin z$ denote the complex sine function.
Then:
:$\map \Re {\sin z} = \sin x \cosh y$
where:
:$\map \Re z$ denotes the real part of a complex number $z$
:$\sin$ denotes the sine function (real and complex)
:$\cosh$ denotes the hyperbolic cosine function.

Proof:
From Sine of Complex Number:
:$\map \sin {x + i y} = \sin x \cosh y + i \cos x \sinh y$
The result follows by definition of the real part of a complex number.
{{qed}}
Theorem: Imaginary Part of Sine of Complex Number

Let $z = x + i y \in \C$ be a complex number, where $x, y \in \R$.
Let $\sin z$ denote the complex sine function.
Then:
:$\Im \paren {\sin z} = \cos x \sinh y$
where:
:$\Im z$ denotes the imaginary part of a complex number $z$
:$\sin$ denotes the complex sine function
:$\cos$ denotes the real cosine function
:$\sinh$ denotes the hyperbolic sine function.

Proof:
From Sine of Complex Number:
:$\sin \paren {x + i y} = \sin x \cosh y + i \cos x \sinh y$
The result follows by definition of the imaginary part of a complex number.
{{qed}}
Theorem: Modulus of Sine of Complex Number

Let $z = x + i y \in \C$ be a complex number, where $x, y \in \R$.
Let $\sin z$ denote the complex sine function.
Then:
:$\cmod {\sin z} = \sqrt {\sin^2 x + \sinh^2 y}$
where:
:$\cmod z$ denotes the modulus of a complex number $z$
:$\sin x$ denotes the real sine function
:$\sinh$ denotes the hyperbolic sine function.

Proof:
{{begin-eqn}}
{{eqn | l = \sin \paren {x + i y}
      | r = \sin x \cosh y + i \cos x \sinh y
      | c = Sine of Complex Number
}}
{{eqn | ll= \leadsto
      | l = \cmod {\sin z}^2
      | r = \paren {\sin x \cosh y}^2 + \paren {\cos x \sinh y}^2
      | c = {{Defof|Complex Modulus}}
}}
{{eqn | r = \sin^2 x \paren {1 + \sinh^2 y} + \cos^2 x \sinh^2 y
      | c = Difference of Squares of Hyperbolic Cosine and Sine
}}
{{eqn | r = \sin^2 x + \paren {\sin^2 x + \cos^2 x} \sinh^2 y
      | c = rearranging
}}
{{eqn | r = \sin^2 x + \sinh^2 y
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | ll= \leadsto
      | l = \cmod {\sin z}
      | r = \sqrt {\sin^2 x + \sinh^2 y}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Inverse Tangent of i

The inverse tangent of $i$ is not defined.

Proof:
{{AimForCont}} $\tan z_0 = i$.
{{begin-eqn}}
{{eqn | l = \dfrac {\sin z_0} {\cos z_0}
      | r = i
      | c = {{Defof|Tangent Function}}
}}
{{eqn | ll= \leadsto
      | l = \sin z_0
      | r = i \cos z_0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \sin^2 z_0
      | r = -\cos^2 z_0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \sin^2 z_0 + \cos^2 z_0
      | r = 0
      | c = 
}}
{{end-eqn}}
This contradicts the theorem Sum of Squares of Sine and Cosine:
:$\sin^2 z_0 + \cos^2 z_0 = 1$
Hence the result by Proof by Contradiction.
{{qed}}
Theorem: Cauchy-Hadamard Theorem/Complex Case

Let $\xi \in \C$ be a complex number.
Let $\ds \map S z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ be a (complex) power series about $\xi$.
Then the radius of convergence $R$ of $\map S z$ is given by:
:$\ds \dfrac 1 R = \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n}$
If:
:$\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} = 0$
then the radius of convergence is infinite, and $\map S z$ is absolutely convergent for all $z \in \C$.

Proof:
Let $\epsilon \in \R_{>0}$, and let $z \in \C$.
Suppose that $\cmod {z - \xi} = R - \epsilon$.
By definition of radius of convergence, it follows that $S \paren z$ is absolutely convergent.
From the $n$th Root Test:
:$\ds \limsup_{n \mathop \to \infty} \cmod {a_n \paren {z - \xi}^n}^{1/n} \le 1$
By Multiple Rule for Complex Sequences, this inequality can be rearranged to obtain:
:$\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} \le \dfrac 1 {\cmod {z - \xi} } = \dfrac 1 {R - \epsilon}$
As $\epsilon > 0$ was arbitrary, it follows that:
:$\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} \le \dfrac 1 R$
{{qed|lemma}}
Now, suppose that $\cmod {z - \xi} = R + \epsilon$.
Then $S \paren z$ is divergent, so the $n$th Root Test shows that:
:$\ds \limsup_{n \mathop \to \infty} \cmod {a_n \paren {z - \xi}^n}^{1/n} \ge 1$
which we can rearrange to obtain:
:$\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} \ge \dfrac 1 {\cmod {z - \xi} } = \dfrac 1 {R - \epsilon}$
As $\epsilon > 0$ was arbitrary, it follows that:
:$\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} \ge \dfrac 1 R$
{{qed|lemma}}
Thus:
:$\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} = \dfrac 1 R$
Hence the result.
{{qed}}
Theorem: Cauchy-Hadamard Theorem/Complex Case

Let $\xi \in \C$ be a complex number.
Let $\ds \map S z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ be a (complex) power series about $\xi$.
Then the radius of convergence $R$ of $\map S z$ is given by:
:$\ds \dfrac 1 R = \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n}$
If:
:$\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} = 0$
then the radius of convergence is infinite, and $\map S z$ is absolutely convergent for all $z \in \C$.

Proof:
Let $L = \limsup \cmod {a_n}^{1/n}$.
We will consider only the case $0 < L < \infty$, as the cases $L = 0$ and $L = \infty$ follow quite simply from this one.
We have that:
:$\forall r \in \closedint 0 {\dfrac 1 L}: L < \dfrac 1 r$
Thus there exists $\epsilon \in \R_{> 0}$ such that:
:$L + \epsilon < \dfrac 1 R$
and so:
:$r < \dfrac 1 {L + \epsilon}$
Let $\tilde r = r \paren {L + \epsilon}$.
Then:
:$0 \le \tilde r < 1$
This means that the geometric series $\ds \sum_{n \mathop = 0}^\infty$ is convergent.
From the properties of the limit superior, we know that:
:$\forall n \gg 1: \cmod {a_n}^{1/n} < L + \epsilon$
{{explain|In this context what $n \gg 1$ means}}
Thus:
:$\forall n \gg 1: \cmod {a_n r^n} < \paren {L + \epsilon}^n r^n = \tilde r^n$
This means that the series $\ds \sum_{n \mathop = 0}^\infty \cmod {a_n r^n}$ is also convergent.
Thus $\ds \sum_{n \mathop = 0}^\infty a_n r^n$ is absolutely convergent and therefore convergent.
Thus, by the definition of radius of convergence, we have:
:$r \le R$
which holds for all $r \in \closedint 0 {\dfrac 1 L}$.
Hence:
:$\dfrac 1 L \le R$
Let now $\epsilon \in \openint 0 L$ be fixed.
Because of the properties of the limit superior, we can extract from the sequence of the formula a subsequence such that:
:$\cmod {a_{n_k} }^{1 / n_k} > L - \epsilon$
for all $k \in \N$.
Thus:
:$\cmod {a_{n_k} \cdot \dfrac 1 {\paren {L + \epsilon}^{n_k} } } > 1$
for all $k \in \N$.
So the series:
:$\ds \sum_{n \mathop = 0}^\infty a_n \paren {\frac 1 {L + \epsilon} }^n$
is not convergent.
Thus:
:$R \le \dfrac 1 {L + \epsilon}$
This holds for all $\epsilon \in \R_{>0}$.
So:
:$R \le \dfrac 1 L$
Hence:
:$\dfrac 1 L \le R \le \dfrac 1 L$
which means:
:$R = \dfrac 1 L$
as required.
{{qed}}
Theorem: Cauchy-Hadamard Theorem/Real Case

Let $\xi \in \R$ be a real number.
Let $\ds \map S x = \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ be a power series about $\xi$.
Then the radius of convergence $R$ of $S \paren x$ is given by:
:$\ds \frac 1 R = \limsup_{n \mathop \to \infty} \size {a_n}^{1/n}$
If:
:$\ds \frac 1 R = \limsup_{n \mathop \to \infty} \size {a_n}^{1/n} = 0$
then the radius of convergence is infinite and therefore the interval of convergence is $\R$.

Proof:
From the $n$th root test, $S \paren x$ is convergent if $\ds \limsup_{n \mathop \to \infty} \size {a_n \paren {x - \xi}^n}^{1/n} < 1$.
Thus:
{{begin-eqn}}
{{eqn | l = \size {a_n \paren {x - \xi}^n}^{1/n}
      | o = <
      | r = 1
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \size {a_n}^{1/n} \size {x - \xi}
      | o = <
      | r = 1
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \size {a_n}^{1/n}
      | o = <
      | r = \frac 1 {\size {x - \xi} }
      | c = 
}}
{{end-eqn}}
The result follows from the definition of radius of convergence.
{{qed}}
{{Namedfor|Augustin Louis Cauchy|name2 = Jacques Salomon Hadamard}}
Theorem: Comparison Test for Convergence of Power Series

Let $A = \ds \sum_{n \mathop \ge 0} a_n z^n$ and $B = \ds \sum_{n \mathop \ge 0} b_n z^n$ be power series in $\C$.
Let $R_A$ and $R_B$ be the radii of convergence of $A$ and $B$ respectively.
Let $\cmod {b_n} \le \cmod {a_n}$ for all $n \in \N$.
Then $R_A \le R_B$.

Proof:
{{AimForCont}} $R_A > R_B$.
Let $z_0 \in \C$ such that $R_B < \cmod {z_0} < R_A$.
Then $A$ is convergent at $z_0$ but $B$ is divergent at $z_0$.
But by the Comparison Test, if $A$ is convergent at $z_0$ then $B$ is also convergent at $z_0$.
From this contradiction it follows that there can be no such $z_0$.
That is:
:$R_A \le R_B$
{{qed}}
Theorem: Bounds for Modulus of e^z on Circle x^2 + y^2 - 2x - 2y - 2 = 0

Consider the circle $C$ embedded in the complex plane defined by the equation:
:$x^2 + y^2 - 2 x - 2 y - 2 = 0$
Let $z = x + i y \in \C$ be a point lying on $C$.
Then:
:$e^{-1} \le \cmod {e^z} \le e^3$

Proof:
{{begin-eqn}}
{{eqn | l = x^2 + y^2 - 2 x - 2 y - 2
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {\paren {x - 1}^2 - 1} + \paren {\paren {y - 1}^2 - 1} - 2
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {x - 1}^2 + \paren {y - 1}^2
      | r = 4
      | c = 
}}
{{end-eqn}}
This defines a circle whose center is at $1 + i$ and whose radius is $2$.
From Modulus of Exponential is Exponential of Real Part:
:$\cmod {e^z} = e^x$
If $z \in C$ then from the geometry of the circle $C$:
:$-1 \le x \le 3$
Then from Exponential is Strictly Increasing:
:$e^{-1} \le e^x \le e^3$
Hence the result.
{{qed}}
Theorem: Hyperbolic Cosine of Complex Number

Let $a$ and $b$ be real numbers.
Let $i$ be the imaginary unit.
Then:
:$\map \cosh {a + b i} = \cosh a \cos b + i \sinh a \sin b$
where:
:$\cos$ denotes the real cosine function
:$\sin$ denotes the real sine function
:$\sinh$ denotes the hyperbolic sine function
:$\cosh$ denotes the hyperbolic cosine function

Proof:
{{begin-eqn}}
{{eqn | l = \map \cosh {a + b i}
      | r = \cosh a \map \cosh {b i} + \sinh a \map \sinh {b i}
      | c = Hyperbolic Cosine of Sum
}}
{{eqn | r = \cosh a \cos b + \sinh a \map \sinh {b i}
      | c = Cosine in terms of Hyperbolic Cosine
}}
{{eqn | r = \cosh a \cos b + i \sinh a \sin b
      | c = Sine in terms of Hyperbolic Sine
}}
{{end-eqn}}
{{qed}}
Theorem: Hyperbolic Cosine of Complex Number

Let $a$ and $b$ be real numbers.
Let $i$ be the imaginary unit.
Then:
:$\map \cosh {a + b i} = \cosh a \cos b + i \sinh a \sin b$
where:
:$\cos$ denotes the real cosine function
:$\sin$ denotes the real sine function
:$\sinh$ denotes the hyperbolic sine function
:$\cosh$ denotes the hyperbolic cosine function

Proof:
{{begin-eqn}}
{{eqn | l = \cosh a \cos b - i \sinh a \sin b
      | r = \frac {e^a + e^{-a} } 2 \frac {e^{i b} + e^{-i b} } 2 + i \frac {e^a - e^{-a} } {2 i} \frac {e^{i b} - e^{-i b} } 2
      | c = {{Defof|Hyperbolic Cosine}}, Euler's Cosine Identity, {{Defof|Hyperbolic Sine}}, Euler's Sine Identity
}}
{{eqn | r = \frac {e^{a + i b} + e^{-a + i b} + e^{a - i b} + e^{-a - i b} + e^{a + i b} - e^{-a + i b} - e^{a - i b} + e^{-a - i b} } 4
      | c = simplifying
}}
{{eqn | r = \frac {e^{a + i b} + e^{-\paren {a + i b} } } 2
      | c = simplifying
}}
{{eqn | r = \cosh \paren {a + b i}
      | c = {{Defof|Hyperbolic Cosine}}
}}
{{end-eqn}}
{{qed}}
Theorem: Hyperbolic Tangent of Complex Number/Formulation 1

:$\tanh \paren {a + b i} = \dfrac {\sinh a \cos b + i \cosh a \sin b} {\cosh a \cos b + i \sinh a \sin b}$

Proof:
{{begin-eqn}}
{{eqn | l = \tanh \paren {a + b i}
      | r = \frac {\sinh \paren {a + b i} } {\cosh \paren {a + b i} }
      | c = {{Defof|Hyperbolic Tangent}}
}}
{{eqn | r = \dfrac {\sinh a \cos b + i \cosh a \sin b} {\cosh a \cos b + i \sinh a \sin b}
      | c = Hyperbolic Sine of Complex Number and Hyperbolic Cosine of Complex Number
}}
{{end-eqn}}
{{qed}}
Theorem: Hyperbolic Tangent of Complex Number/Formulation 2

:$\tanh \paren {a + b i} = \dfrac {\tanh a + i \tan b} {1 + i \tanh a \tan b}$

Proof:
{{begin-eqn}}
{{eqn | l = \tanh \paren {a + b i}
      | r = \dfrac {\sinh a \cos b + i \cosh a \sin b} {\cosh a \cos b + i \sinh a \sin b}
      | c = Hyperbolic Tangent of Complex Number: Formulation 1
}}
{{eqn | r = \dfrac {\tanh a \cos b + i \sin b} {\cos b + i \tanh a \sin b}
      | c = multiplying denominator and numerator by $\dfrac 1 {\cosh a}$
}}
{{eqn | r = \frac {\tanh a + i \tan b} {1 + i \tanh a \tan b}
      | c = multiplying denominator and numerator by $\dfrac 1 {\cos b}$
}}
{{end-eqn}}
{{qed}}
Theorem: Hyperbolic Tangent of Complex Number/Formulation 3

:$\tanh \paren {a + b i} = \dfrac {\tanh a + \tanh a \tan^2 b} {1 + \tanh^2 a \tan^2 b} + \dfrac {\tan b - \tanh^2 a \tan b} {1 + \tanh^2 a \tan^2 b} i$

Proof:
{{begin-eqn}}
{{eqn | l = \tan \paren {a + b i}
      | r = \dfrac {\tanh a + i \tan b} {1 + i \tanh a \tan b}
      | c = Hyperbolic Tangent of Complex Number: Formulation 2
}}
{{eqn | r = \frac {\paren {\tanh a + i \tan b} \paren {1 - i \tanh a \tan b} } {1 + \tanh^2 a \tan^2 b}
      | c = multiplying denominator and numerator by $1 - i \tanh a \tan b$
}}
{{eqn | r = \frac {\tanh a + i \tan b - i \tanh^2 a \tan b + \tanh a \tan^2 b} {1 + \tanh^2 a \tan^2 b}
      | c = 
}}
{{eqn | r = \frac {\tanh a + \tanh a \tan^2 b} {1 + \tanh^2 a \tan^2 b} + \frac {\tan b - \tanh^2 a \tan b} {1 + \tanh^2 a \tan^2 b} i
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Hyperbolic Cosecant of Complex Number

Let $a$ and $b$ be real numbers.
Let $i$ be the imaginary unit.
Then:
:$\map \csch {a + b i} = \dfrac {\sinh a \cos b - i \cosh a \sin b} {\sinh^2 a \cos^2 b + \cosh^2 a \sin^2 b}$
where:
:$\csch$ denotes the hyperbolic cosecant function.
:$\sin$ denotes the real sine function
:$\cos$ denotes the real cosine function
:$\sinh$ denotes the hyperbolic sine function
:$\cosh$ denotes the hyperbolic cosine function

Proof:
{{begin-eqn}}
{{eqn | l = \map \csch {a + b i}
      | r = \frac 1 {\map \sinh {a + b i} }
      | c = {{Defof|Hyperbolic Cosecant}}
}}
{{eqn | r = \dfrac 1 {\sinh a \cos b + i \cosh a \sin b}
      | c = Hyperbolic Sine of Complex Number
}}
{{eqn | r = \dfrac {\sinh a \cos b - i \cosh a \sin b} {\paren {\sinh a \cos b + i \cosh a \sin b} \paren {\sinh a \cos b - i \cosh a \sin b} }
      | c = multiplying denominator and numerator by $\sinh a \cos b - i \cosh a \sin b$
}}
{{eqn | r = \dfrac {\sinh a \cos b - i \cosh a \sin b} {\sinh^2 a \cos^2 b - i^2 \cosh^2 a \sin^2 b}
      | c = Difference of Two Squares
}}
{{eqn | r = \dfrac {\sinh a \cos b - i \cosh a \sin b} {\sinh^2 a \cos^2 b + \cosh^2 a \sin^2 b}
      | c = {{Defof|Imaginary Unit}}
}}
{{end-eqn}}
{{qed}}
Theorem: Hyperbolic Secant of Complex Number

Let $a$ and $b$ be real numbers.
Let $i$ be the imaginary unit.
Then:
:$\sech \paren {a + b i} = \dfrac {\cosh a \cos b - i \sinh a \sin b} {\cosh^2 a \cos^2 b + \sinh^2 a \sin^2 b}$
where:
:$\sech$ denotes the hyperbolic secant function.
:$\sin$ denotes the real sine function
:$\cos$ denotes the real cosine function
:$\sinh$ denotes the hyperbolic sine function
:$\cosh$ denotes the hyperbolic cosine function

Proof:
{{begin-eqn}}
{{eqn | l = \sech \paren {a + b i}
      | r = \frac 1 {\cosh \paren {a + b i} }
      | c = {{Defof|Hyperbolic Secant}}
}}
{{eqn | r = \dfrac 1 {\cosh a \cos b + i \sinh a \sin b}
      | c = Hyperbolic Cosine of Complex Number
}}
{{eqn | r = \dfrac {\cosh a \cos b - i \sinh a \sin b} {\paren {\cosh a \cos b + i \sinh a \sin b} \paren {\cosh a \cos b - i \sinh a \sin b} }
      | c = multiplying denominator and numerator by $\cosh a \cos b - i \sinh a \sin b$
}}
{{eqn | r = \dfrac {\cosh a \cos b - i \sinh a \sin b} {\cosh^2 a \cos^2 b - i^2 \sinh^2 a \sin^2 b}
      | c = Difference of Two Squares
}}
{{eqn | r = \dfrac {\cosh a \cos b - i \sinh a \sin b} {\cosh^2 a \cos^2 b + \sinh^2 a \sin^2 b}
      | c = {{Defof|Imaginary Unit}}
}}
{{end-eqn}}
{{qed}}
Theorem: Hyperbolic Cotangent of Complex Number/Formulation 2

:$\map \coth {a + b i} = \dfrac {1 - i \coth a \cot b} {\coth a - i \cot b}$

Proof:
{{begin-eqn}}
{{eqn | l = \map \coth {a + b i}
      | r = \dfrac {\cosh a \cos b + i \sinh a \sin b} {\sinh a \cos b + i \cosh a \sin b}
      | c = Hyperbolic Cotangent of Complex Number: Formulation 1
}}
{{eqn | r = \dfrac {\coth a \cos b + i \sin b} {\cos b + i \coth a \sin b}
      | c = multiplying denominator and numerator by $\dfrac 1 {\sinh a}$
}}
{{eqn | r = \dfrac {\coth a \cot b + i} {\cot b + i \coth a}
      | c = multiplying denominator and numerator by $\dfrac 1 {\sin b}$
}}
{{eqn | r = \dfrac {1 - i \coth a \cot b} {\coth a - i \cot b}
      | c = multiplying denominator and numerator by $-i$
}}
{{end-eqn}}
{{qed}}