The proof strategy is to show that for all $z \in \C_{\ne 0}$: :$\left\{{w \in \C: z = \operatorname{sech} \left({w}\right)}\right\} = \left\{{\ln \left({\dfrac {1 + \sqrt{\left|{1 - z^2}\right|} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)}} z}\right) + 2 k \pi i: k \in \Z}\right\}$ Thus let $z \in \C_{\ne 0}$. === Definition 1 implies Definition 2 === It will be demonstrated that: :$\left\{{w \in \C: z = \operatorname{sech} \left({w}\right)}\right\} \subseteq \left\{{\ln \left({\dfrac {1 + \sqrt{\left|{1 - z^2}\right|} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)}} z}\right) + 2 k \pi i: k \in \Z}\right\}$ Let $w \in \left\{{w \in \C: z = \operatorname{sech} \left({w}\right)}\right\}$. From the definition of hyperbolic secant: :$(1): \quad z = \dfrac 2 {e^w + e^{- w}}$ Let $v = e^w$. Then: {{begin-eqn}} {{eqn | l = z \left({v + \frac 1 v}\right)       | r = 2       | c = multiplying $(1)$ by $v + \dfrac 1 v$ }} {{eqn | ll= \implies       | l = z v^2 - 2 v + z       | r = 0       | c = multiplying by $v$ and rearranging }} {{eqn | ll= \implies       | l = v       | r = \frac {1 + \left({1 - z^2}\right)^{1/2} } z       | c = Quadratic Formula }} {{end-eqn}} Let $s = 1 - z^2$. Then: {{begin-eqn}} {{eqn | l = v       | r = \frac {1 + s^{1/2} } z       | c =  }} {{eqn | r = \frac {1 + \sqrt{\left\vert{s}\right\vert} \left({\cos \left({\dfrac {\arg \left({s}\right)} 2}\right) + i \sin \left({\dfrac {\arg \left({s}\right)} 2}\right)}\right) } z       | c = Definition of Complex Square Root }} {{eqn | n = 2       | ll= \implies       | l = \ln v       | r = \ln \left({\frac {1 + \sqrt{\left\vert{s}\right\vert} \left({\cos \left({\dfrac {\arg \left({s}\right)} 2}\right) + i \sin \left({\dfrac {\arg \left({s}\right)} 2}\right)}\right) } z}\right)       | c = where $\ln$ denotes the Complex Natural Logarithm }} {{end-eqn}} We have that: {{begin-eqn}} {{eqn | l = v       | r = e^w       | c =  }} {{eqn | ll= \implies       | l = \ln v       | r = \ln \left({e^w}\right)       | c =  }} {{eqn | n = 3       | r = w + 2 k' \pi i: k' \in \Z       | c = Definition of Complex Natural Logarithm }} {{end-eqn}} Thus from $(2)$ and $(3)$: {{begin-eqn}} {{eqn | l = w + 2 k' \pi i       | r = \ln \left({\frac {1 + \sqrt{\left\vert{s}\right\vert} \left({\cos \left({\dfrac {\arg \left({s}\right)} 2}\right) + i \sin \left({\dfrac {\arg \left({s}\right)} 2}\right)}\right) } z}\right)       | c =  }} {{eqn | ll= \implies       | l = w       | r = \ln \left({\frac {1 + \sqrt{\left\vert{s}\right\vert} \left({\cos \left({\dfrac {\arg \left({s}\right)} 2}\right) + i \sin \left({\dfrac {\arg \left({s}\right)} 2}\right)}\right) } z}\right) + 2 k \pi i       | c = putting $k = -k'$ }} {{eqn | ll= \implies       | l = w       | r = \ln \left({\frac {1 + \sqrt{\left\vert{1 - z^2}\right\vert} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)} } z}\right) + 2 k \pi i       | c = Definition of Exponential Form of Complex Number }} {{end-eqn}} Thus by definition of subset: :$\left\{{w \in \C: z = \operatorname{sech} \left({w}\right)}\right\} \subseteq \left\{{\ln \left({\dfrac {1 + \sqrt{\left|{1 - z^2}\right|} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)}} z}\right) + 2 k \pi i: k \in \Z}\right\}$ {{qed|lemma}} === Definition 2 implies Definition 1 === It will be demonstrated that: :$\left\{{w \in \C: z = \operatorname{sech} \left({w}\right)}\right\} \supseteq \left\{{\ln \left({\dfrac {1 + \sqrt{\left|{1 - z^2}\right|} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)}} z}\right) + 2 k \pi i: k \in \Z}\right\}$ Let $w \in \left\{{\ln \left({\dfrac {1 + \sqrt{\left|{1 - z^2}\right|} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)}} z}\right) + 2 k \pi i: k \in \Z}\right\}$. Then: {{begin-eqn}} {{eqn | ll= \exists k \in \Z:       | l = w + 2 \left({-k}\right) \pi i       | r = \ln \left({\dfrac {1 + \sqrt{\left\vert{1 - z^2}\right\vert} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)} } z}\right)       | c =  }} {{eqn | lll= \implies       | l = e^{w + 2 \left({-k}\right) \pi i}       | r = \dfrac {1 + \sqrt{\left\vert{1 - z^2}\right\vert} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)} } z       | c = Definition of Complex Natural Logarithm }} {{eqn | lll= \implies       | l = e^w       | r = \dfrac {1 + \sqrt{\left\vert{1 - z^2}\right\vert} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)} } z       | c = Complex Exponential Function has Imaginary Period }} {{eqn | lll= \implies       | l = z e^w - 1       | r = \sqrt{\left\vert{1 - z^2}\right\vert} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)}       | c =  }} {{eqn | lll= \implies       | l = \left({z e^w - 1}\right)^2       | r = \left\vert{1 - z^2}\right\vert e^{i \arg \left({1 - z^2}\right)}       | c = Roots of Complex Number }} {{eqn | lll= \implies       | l = \left({z e^w - 1}\right)^2       | r = 1 - z^2       | c = Definition of Exponential Form of Complex Number }} {{eqn | lll= \implies       | l = z^2 e^{2 w} - 2 z e^w + 1       | r = 1 - z^2       | c = Square of Difference }} {{eqn | lll= \implies       | l = z^2 e^{2 w} - 2 z e^w       | r = - z^2       | c =  }} {{eqn | lll= \implies       | l = z e^{2 w} + z        | r = 2 e^w       | c =  }} {{eqn | lll= \implies       | l = z \left({e^w + \frac 1 {e^w} }\right)       | r = 2       | c =  }} {{eqn | lll= \implies       | l = z       | r = \frac 2 {e^w + e^{- w} }       | c =  }} {{eqn | lll= \implies       | l = z       | r = \operatorname{sech} \left({w}\right)       | c = Definition of Hyperbolic Secant }} {{eqn | lll= \implies       | l = w       | o = \in       | r = \left\{ {w \in \C: z = \operatorname{sech} \left({w}\right)}\right\}       | c =  }} {{end-eqn}} Thus by definition of superset: :$\left\{{w \in \C: z = \operatorname{sech} \left({w}\right)}\right\} \supseteq \left\{{\ln \left({\dfrac {1 + \sqrt{\left|{1 - z^2}\right|} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)}} z}\right) + 2 k \pi i: k \in \Z}\right\}$ {{qed|lemma}} Thus by definition of set equality: :$\left\{{w \in \C: z = \operatorname{sech} \left({w}\right)}\right\} = \left\{{\ln \left({\dfrac {1 + \sqrt{\left|{1 - z^2}\right|} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)}} z}\right) + 2 k \pi i: k \in \Z}\right\}$ {{qed}} Category:Inverse Hyperbolic Secant mstl9zxo120js348aobqoh5ycx22du6