The proof strategy is to how that for all $z \in \C$: :$\set {w \in \C: \map \tan w = z} = \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$ Note that when $z = 0 + i$: {{begin-eqn}} {{eqn | l = i - z       | r = 0 + 0 i       | c =  }} {{eqn | ll= \leadsto       | l = \frac {i - z} {i + z}       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \map \ln {\dfrac {i - z} {i + z} }       | o =        | r = \text {is undefined}       | c =  }} {{end-eqn}} Similarly, when $z = 0 - i$: {{begin-eqn}} {{eqn | l = i + z       | r = 0 + 0 i       | c =  }} {{eqn | ll= \leadsto       | l = \frac {i - z} {i + z}       | o =        | r = \text {is undefined}       | c =  }} {{end-eqn}} Thus let $z \in \C \setminus \set {0 + i, 0 - i}$. === Definition 1 implies Definition 2 === It is demonstrated that: :$\set {w \in \C: \map \tan w = z} \subseteq \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$ Let $w \in \set {w \in \C: z = \map \tan w}$. Then: {{begin-eqn}} {{eqn | l = z       | r = i \frac {1 - e^{2 i w} } {1 + e^{2 i w} }       | c = Tangent Exponential Formulation }} {{eqn | ll= \leadsto       | l = e^{2 i w}       | r = \frac {1 + i z} {1 - i z}       | c = solving for $e^{2 i w}$ }} {{eqn | r = \frac {i - z} {i + z}       | c = multiplying top and bottom by $i$, noting $i^2 = -1$ }} {{eqn | ll= \leadsto       | l = \map \ln {e^{2 i w} }       | r = \ln \frac {i - z} {i + z} }} {{eqn | ll= \leadsto       | l = 2 i w + 2 k' \pi i: k' \in \Z       | r = \ln \frac {i - z} {i + z}       | c = {{Defof|Complex Natural Logarithm}} }} {{eqn | ll= \leadsto       | l = w       | r = \frac 1 {2 i} \ln \frac {i - z} {i + z} + k \pi: k \in \Z       | c = putting $k = -k'$ }} {{end-eqn}} Thus by definition of subset: :$\set {w \in \C: \map \tan w = z} \subseteq \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$ {{qed|lemma}} === Definition 2 implies Definition 1 === It is demonstrated that: :$\set {w \in \C: \map \tan w = z} \supseteq \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$ Let $w \in \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$. Then: {{begin-eqn}} {{eqn | ll= \exists k \in \Z:       | l = w        | r = \dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi       | c =  }} {{eqn | lll=\leadsto       | ll= \exists k \in \Z:       | l = 2 i w + 2 \paren {-k} \pi i       | r = \map \ln {\dfrac {i - z} {i + z} }       | c =  }} {{eqn | lll=\leadsto       | l = e^{2 i w + 2 \paren {-k} \pi i}       | r = \dfrac {i - z} {i + z}       | c = {{Defof|Complex Natural Logarithm}} }} {{eqn | lll=\leadsto       | l = e^{2 i w}       | r = \dfrac {i - z} {i + z}       | c = Complex Exponential Function has Imaginary Period }} {{eqn | lll=\leadsto       | l = z       | r = i \dfrac {1 - e^{2 i w} } {1 + e^{2 i w} }       | c =  }} {{eqn | lll=\leadsto       | l = z       | r = \tan w       | c = Tangent Exponential Formulation }} {{eqn | lll=\leadsto       | l = w       | o = \in       | r = \set {w \in \C: \map \tan w = z}       | c =  }} {{end-eqn}} Thus by definition of superset: :$\set {w \in \C: \map \tan w = z} \supseteq \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$ {{qed|lemma}} Thus by definition of set equality: :$\set {w \in \C: \map \tan w = z} = \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$ {{qed}} Category:Inverse Tangent s1fxr3hj3h5330i801ydq32gce2lf49