The proof strategy is to how that for all $z \in S$: :$\left\{{w \in \C: z = \tanh \left({w}\right)}\right\} = \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$ Note that when $z = -1 + 0 i$: {{begin-eqn}} {{eqn | l = 1 + z       | r = 0 + 0 i       | c =  }} {{eqn | ll= \implies       | l = \frac {1 + z} {1 - z}       | r = 0       | c =  }} {{eqn | ll= \implies       | l = \ln \left({\dfrac {1 + z} {1 - z} }\right)       | o =        | r = \text {is undefined}       | c =  }} {{end-eqn}} Similarly, when $z = 1 + 0 i$: {{begin-eqn}} {{eqn | l = 1 - z       | r = 0 + 0 i       | c =  }} {{eqn | ll= \implies       | l = \frac {1 + z} {1 - z}       | o =        | r = \text {is undefined}       | c =  }} {{end-eqn}} Thus let $z \in \C \setminus \left\{{-1 + 0 i, 1 + 0 i}\right\}$. === Definition 1 implies Definition 2 === It is demonstrated that: :$\left\{{w \in \C: z = \tanh \left({w}\right)}\right\} \subseteq \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$ Let $w \in \left\{{w \in \C: z = \tanh \left({w}\right)}\right\}$. Then: {{begin-eqn}} {{eqn | l = z       | r = \frac {e^w - e^{- w} } {e^w + e^{- w} }       | c = Definition of Hyperbolic Tangent }} {{eqn | l = z       | r = \frac {e^{2 w} - 1} {e^{2 w} + 1}       | c = multiplying top and bottom by $e^w$ }} {{eqn | ll= \implies       | l = e^{2 w}       | r = \frac {1 + z} {1 - z}       | c = solving for $e^{2 w}$ }} {{eqn | ll= \implies       | l = \ln \left({e^{2 w} }\right)       | r = \ln \frac {1 + z} {1 - z} }} {{eqn | ll= \implies       | l = 2 w + 2 k' \pi i: k' \in \Z       | r = \ln \frac {1 + z} {1 - z}       | c = Definition of Complex Natural Logarithm }} {{eqn | ll= \implies       | l = w       | r = \frac 1 2 \ln \frac {1 + z} {1 - z} + k \pi i: k \in \Z       | c = putting $k = -k'$ }} {{end-eqn}} Thus by definition of subset: :$\left\{{w \in \C: z = \tanh \left({w}\right)}\right\} \subseteq \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$ {{qed|lemma}} === Definition 2 implies Definition 1 === It is demonstrated that: :$\left\{{w \in \C: z = \tanh \left({w}\right)}\right\} \supseteq \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$ Let $w \in \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$. Then: {{begin-eqn}} {{eqn | ll= \exists k \in \Z:       | l = w        | r = \dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i       | c =  }} {{eqn | lll= \implies       | ll= \exists k \in \Z:       | l = 2 w + 2 \left({-k}\right) \pi i       | r = \ln \left({\dfrac {1 + z} {1 - z} }\right)       | c =  }} {{eqn | lll= \implies       | l = e^{2 w + 2 \left({-k}\right) \pi i}       | r = \dfrac {1 + z} {1 - z}       | c = Definition of Complex Natural Logarithm }} {{eqn | lll= \implies       | l = e^{2 w}       | r = \dfrac {1 + z} {1 - z}       | c = Complex Exponential Function has Imaginary Period }} {{eqn | lll= \implies       | l = z       | r = \dfrac {e^w - e^{- w} } {e^w + e^{- w} }       | c =  }} {{eqn | lll= \implies       | l = z       | r = \tanh w       | c = Definition of Hyperbolic Tangent }} {{eqn | lll= \implies       | l = w       | o = \in       | r = \left\{ {w \in \C: \tanh \left({w}\right) = z}\right\}       | c =  }} {{end-eqn}} Thus by definition of superset: :$\left\{{w \in \C: z = \tanh \left({w}\right)}\right\} \supseteq \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$ {{qed|lemma}} Thus by definition of set equality: :$\left\{{w \in \C: z = \tanh \left({w}\right)}\right\} = \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$ {{qed}} Category:Inverse Hyperbolic Tangent mqr7r0y3zyb1dfvmowcj3eu6d0s6e0n