The proof strategy is to show that for all $z \in \C_{\ne 0}$: :$\set {w \in \C: \map \sec w = z} = \set {\dfrac 1 i \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi: k \in \Z}$ Thus let $z \in \C$. === Definition 1 implies Definition 2 === It will be demonstrated that: :$\set {w \in \C: \map \sec w = z} \subseteq \set {\dfrac 1 i \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi: k \in \Z}$ Let $w \in \set {w \in \C: \map \sec w = z}$. From Secant Exponential Formulation: :$(1): \quad z = \dfrac 2 {e^{i w} + e^{-i w}}$ Let $v = e^{i w}$. Then: {{begin-eqn}} {{eqn | l = z \paren {v + \frac 1 v}       | r = 2       | c = multiplying $(1)$ by $v + \dfrac 1 v$ }} {{eqn | ll= \leadsto       | l = z v^2 - 2 v + z       | r = 0       | c = multiplying by $v$ and rearranging }} {{eqn | ll= \leadsto       | l = v       | r = \frac {1 + \paren {1 - z^2}^{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 {\cmod s} \paren {\map \cos {\dfrac {\map \arg s} 2} + i \map \sin {\dfrac {\map \arg s} 2} } } z       | c = {{Defof|Complex Square Root}} }} {{eqn | n = 2       | ll= \leadsto       | l = \ln v       | r = \map \ln {\frac {1 + \sqrt {\cmod s} \paren {\map \cos {\dfrac {\map \arg s} 2} + i \map \sin {\dfrac {\map \arg s} 2} } } z}       | c = where $\ln$ denotes the Complex Natural Logarithm }} {{end-eqn}} We have that: {{begin-eqn}} {{eqn | l = v       | r = e^{i w}       | c =  }} {{eqn | ll= \leadsto       | l = \ln v       | r = \map \ln {e^{i w} }       | c =  }} {{eqn | n = 3       | r = i w + 2 k' \pi i: k' \in \Z       | c = {{Defof|Complex Natural Logarithm}} }} {{end-eqn}} Thus from $(2)$ and $(3)$: {{begin-eqn}} {{eqn | l = i w + 2 k' \pi i       | r = \map \ln {\frac {1 + \sqrt {\cmod s} \paren {\map \cos {\dfrac {\map \arg s} 2} + i \map \sin {\dfrac {\map \arg s} 2} } } z}       | c =  }} {{eqn | ll= \leadsto       | l = w       | r = \frac 1 i \map \ln {\frac {1 + \sqrt {\cmod s} \paren {\map \cos {\dfrac {\map \arg s} 2} + i \map \sin {\dfrac {\map \arg s} 2} } } z} + 2 k \pi       | c = putting $k = -k'$ }} {{eqn | ll= \leadsto       | l = w       | r = \dfrac 1 i \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi       | c = {{Defof|Exponential Form of Complex Number}} }} {{end-eqn}} Thus by definition of subset: :$\set {w \in \C: \map \sec w = z} \subseteq \set {\dfrac 1 i \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi: k \in \Z}$ {{qed|lemma}} === Definition 2 implies Definition 1 === It will be demonstrated that: :$\set {w \in \C: \map \sec w = z} \supseteq \set {\dfrac 1 i \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi: k \in \Z}$ Let $w \in \set {\dfrac 1 i \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi: k \in \Z}$. Then: {{begin-eqn}} {{eqn | lo= \exists k \in \Z:       | l = i w + 2 \paren {-k} \pi i       | r = \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z}       | c =  }} {{eqn | ll= \leadsto       | l = e^{i w + 2 \paren {-k} \pi i}       | r = \dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z       | c = {{Defof|Complex Natural Logarithm}} }} {{eqn | ll= \leadsto       | l = e^{i w}       | r = \dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z       | c = Complex Exponential Function has Imaginary Period }} {{eqn | ll= \leadsto       | l = z e^{i w} - 1       | r = \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} }       | c =  }} {{eqn | ll= \leadsto       | l = \paren {z e^{i w} - 1}^2       | r = \cmod {1 - z^2} e^{i \arg \paren {1 - z^2} }       | c = Roots of Complex Number }} {{eqn | ll= \leadsto       | l = \paren {z e^{i w} - 1}^2       | r = 1 - z^2       | c = {{Defof|Exponential Form of Complex Number}} }} {{eqn | ll= \leadsto       | l = z^2 e^{2 i w} - 2 z e^{i w} + 1       | r = 1 - z^2       | c = Square of Difference }} {{eqn | ll= \leadsto       | l = z^2 e^{2 i w} - 2 z e^{i w}       | r = - z^2       | c =  }} {{eqn | ll= \leadsto       | l = z e^{2 i w} + z        | r = 2 e^{i w}       | c =  }} {{eqn | ll= \leadsto       | l = z \paren {e^{i w} + \frac 1 {e^{i w} } }       | r = 2       | c =  }} {{eqn | ll= \leadsto       | l = z       | r = \frac 2 {e^{i w} + e^{-i w} }       | c =  }} {{eqn | ll= \leadsto       | l = z       | r = \sec w       | c = Secant Exponential Formulation }} {{eqn | ll= \leadsto       | l = w       | o = \in       | r = \set {w \in \C: z = \map \sec w}       | c =  }} {{end-eqn}} Thus by definition of superset: :$\set {w \in \C: \map \sec w = z} \supseteq \set {\dfrac 1 i \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi: k \in \Z}$ {{qed|lemma}} Thus by definition of set equality: :$\set {w \in \C: \map \sec w = z} = \set {\dfrac 1 i \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi: k \in \Z}$ {{qed}} {{Proofread}} Category:Inverse Secant s42xqeaaz292b8xrcy8ryerisevn62b