The proof strategy is to show that for all $z \in \C$: :$\left\{{w \in \C: z = \sin \left({w}\right)}\right\} = \left\{{\dfrac 1 i \ln \left({i z + \sqrt{\left|{1 - z^2}\right|} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)} }\right) + 2 k \pi: k \in \Z}\right\}$ Thus let $z \in \C$. === Definition 1 implies Definition 2 === It will be demonstrated that: :$\left\{{w \in \C: z = \sin \left({w}\right)}\right\} \subseteq \left\{{\dfrac 1 i \ln \left({i z + \sqrt{\left|{1 - z^2}\right|} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)} }\right) + 2 k \pi: k \in \Z}\right\}$ Let $w \in \left\{{w \in \C: z = \sin \left({w}\right)}\right\}$. From Sine Exponential Formulation: :$(1): \quad z = \dfrac {e^{i w} - e^{-i w} } {2 i}$ Let $v = e^{i w}$. Then: {{begin-eqn}} {{eqn | l = 2 i z       | r = v - \frac 1 v       | c = multiplying $(1)$ by $2 i$ }} {{eqn | ll= \implies       | l = v^2 - 2 i z v - 1       | r = 0       | c = multiplying by $v$ and rearranging }} {{eqn | ll= \implies       | l = v       | r = i z + \left({1 - z^2}\right)^{1/2}       | c = Quadratic Formula, and $i^2 = -1$ }} {{end-eqn}} Let $s = 1 - z^2$. Then: {{begin-eqn}} {{eqn | l = v       | r = i z + s^{1/2}       | c =  }} {{eqn | r = i z + \sqrt{\left\vert{s}\right\vert} \left({\cos \left({\frac {\arg \left({s}\right)} 2}\right) + i \sin \left({\frac {\arg \left({s}\right)} 2}\right)}\right)       | c = Definition of Complex Square Root }} {{eqn | n = 2       | ll= \implies       | l = \ln v       | r = \ln \left({i z + \sqrt{\left\vert{s}\right\vert} \left({\cos \left({\frac {\arg \left({s}\right)} 2}\right) + i \sin \left({\frac {\arg \left({s}\right)} 2}\right)}\right)}\right)       | 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= \implies       | l = \ln v       | r = \ln \left({e^{i w} }\right)       | c =  }} {{eqn | n = 3       | r = i 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 = i w + 2 k' \pi i       | r = \ln \left({i z + \sqrt{\left\vert{s}\right\vert} \left({\cos \left({\frac {\arg \left({s}\right)} 2}\right) + i \sin \left({\frac {\arg \left({s}\right)} 2}\right)}\right)}\right)       | c =  }} {{eqn | ll= \implies       | l = w       | r = \frac 1 i \ln \left({i z + \sqrt{\left\vert{s}\right\vert} \left({\cos \left({\frac {\arg \left({s}\right)} 2}\right) + i \sin \left({\frac {\arg \left({s}\right)} 2}\right)}\right)}\right) + 2 k \pi       | c = putting $k = -k'$ }} {{eqn | ll= \implies       | l = w       | r = \frac 1 i \ln \left({i z + \sqrt{\left\vert{1 - z^2}\right\vert} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)} }\right) + 2 k \pi       | c = Definition of Exponential Form of Complex Number }} {{end-eqn}} Thus by definition of subset: :$\left\{{w \in \C: z = \sin \left({w}\right)}\right\} \subseteq \left\{{\frac 1 i \ln \left({i z + \sqrt{\left\vert{1 - z^2}\right\vert} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)} }\right) + 2 k \pi: k \in \Z}\right\}$ {{qed|lemma}} === Definition 2 implies Definition 1 === It will be demonstrated that: :$\left\{{w \in \C: z = \sin \left({w}\right)}\right\} \supseteq \left\{{\dfrac 1 i \ln \left({i z + \sqrt{\left\vert{1 - z^2}\right\vert} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)} }\right) + 2 k \pi: k \in \Z}\right\}$ Let $w \in \left\{{\dfrac 1 i \ln \left({i z + \sqrt{\left\vert{1 - z^2}\right\vert} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)} }\right) + 2 k \pi: k \in \Z}\right\}$. Then: {{begin-eqn}} {{eqn | ll= \exists k \in \Z:       | l = i w + 2 \left({-k}\right) \pi i       | r = \ln \left({i z + \sqrt{\left\vert{1 - z^2}\right\vert} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)} }\right)       | c =  }} {{eqn | lll= \implies       | l = e^{i w + 2 \left({-k}\right) \pi i}       | r = i z + \sqrt{\left\vert{1 - z^2}\right\vert} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)}       | c = Definition of Complex Natural Logarithm }} {{eqn | lll= \implies       | l = e^{i w}       | r = i z + \sqrt{\left\vert{1 - z^2}\right\vert} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)}       | c = Complex Exponential Function has Imaginary Period }} {{eqn | lll= \implies       | l = e^{i w} - i z       | 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({e^{i w} - i z}\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({e^{i w} - i z}\right)^2       | r = 1 - z^2       | c = Definition of Exponential Form of Complex Number }} {{eqn | lll= \implies       | l = e^{2 i w} - 2 i z e^{i w} - z^2       | r = 1 - z^2       | c = Square of Difference and $i^2 = -1$ }} {{eqn | lll= \implies       | l = e^{2 i w}       | r = 1 + 2 i z e^{i w}       | c =  }} {{eqn | lll= \implies       | l = e^{i w} - \frac 1 {e^{i w} }       | r = 2 i z       | c =  }} {{eqn | lll= \implies       | l = z       | r = \frac {e^{i w} - e^{-i w} } {2 i}       | c =  }} {{eqn | lll= \implies       | l = z       | r = \sin w       | c = Sine Exponential Formulation }} {{eqn | lll= \implies       | l = w       | o = \in       | r = \left\{ {w \in \C: z = \sin \left({w}\right)}\right\}       | c =  }} {{end-eqn}} Thus by definition of superset: :$\left\{{w \in \C: z = \sin \left({w}\right)}\right\} \supseteq \left\{{\dfrac 1 i \ln \left({i z + \sqrt{\left\vert{1 - z^2}\right\vert} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)} }\right) + 2 k \pi: k \in \Z}\right\}$ {{qed|lemma}} Thus by definition of set equality: :$\left\{{w \in \C: z = \sin \left({w}\right)}\right\} = \left\{{\dfrac 1 i \ln \left({i z + \sqrt{\left\vert{1 - z^2}\right\vert} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)} }\right) + 2 k \pi: k \in \Z}\right\}$ {{qed}} Category:Inverse Sine 138fjty5uljt4bfupssro53l7dxswc9