The proof strategy is to how that for all $z \in S$: :$\left\{{w \in \C: \cot \left({w}\right) = z}\right\} = \left\{{\dfrac 1 {2 i} \ln \left({\dfrac {z + i} {z - i}}\right) + k \pi: k \in \Z}\right\}$ Note that when $z = 0 - i$: {{begin-eqn}} {{eqn | l = z + i       | r = 0 + 0 i       | c =  }} {{eqn | ll= \implies       | l = \frac {z + i} {z - i}       | r = 0       | c =  }} {{eqn | ll= \implies       | l = \ln \left({\dfrac {z + i} {z - i} }\right)       | o =        | r = \text {is undefined}       | c =  }} {{end-eqn}} Similarly, when $z = 0 + i$: {{begin-eqn}} {{eqn | l = z - i       | r = 0 + 0 i       | c =  }} {{eqn | ll= \implies       | l = \frac {z + i} {z - i}       | o =        | r = \text {is undefined}       | c =  }} {{end-eqn}} Thus let $z \in \C \setminus \left\{{0 + i, 0 - i}\right\}$. === Definition 1 implies Definition 2 === It is demonstrated that: :$\left\{{w \in \C: \cot \left({w}\right) = z}\right\} \subseteq \left\{{\dfrac 1 {2 i} \ln \left({\dfrac {z + i} {z - i}}\right) + k \pi: k \in \Z}\right\}$ Let $w \in \left\{{w \in \C: z = \cot \left({w}\right)}\right\}$. Then: {{begin-eqn}} {{eqn | l = z       | r = i \frac {e^{i w} + e^{-i w} } {e^{i w} - e^{-i w} }       | c = Cotangent Exponential Formulation }} {{eqn | ll= \implies       | l = e^{2 i w}       | r = \frac {i z - 1} {i z + 1}       | c = solving for $e^{2 i w}$ }} {{eqn | r = \frac {z + i} {z - i}       | c = multiplying top and bottom by $-i$, noting $i^2 = -1$ }} {{eqn | ll= \implies       | l = \ln \left({e^{2 i w} }\right)       | r = \ln \left({\frac {z + i} {z - i} }\right) }} {{eqn | ll= \implies       | l = 2 i w + 2 k' \pi i: k' \in \Z       | r = \ln \left({\frac {z + i} {z - i} }\right)       | c = Definition of Complex Natural Logarithm }} {{eqn | ll= \implies       | l = w       | r = \frac 1 {2 i} \ln \left({\frac {z + i} {z - i} }\right) + k \pi: k \in \Z       | c = putting $k = -k'$ }} {{end-eqn}} Thus by definition of subset: :$\left\{{w \in \C: \cot \left({w}\right) = z}\right\} \subseteq \left\{{\dfrac 1 {2 i} \ln \left({\dfrac {z + i} {z - i}}\right) + k \pi: k \in \Z}\right\}$ {{qed|lemma}} === Definition 2 implies Definition 1 === It is demonstrated that: :$\left\{{w \in \C: \cot \left({w}\right) = z}\right\} \supseteq \left\{{\dfrac 1 {2 i} \ln \left({\dfrac {z + i} {z - i}}\right) + k \pi: k \in \Z}\right\}$ Let $w \in \left\{{\dfrac 1 {2 i} \ln \left({\dfrac {z + i} {z - i}}\right) + k \pi: k \in \Z}\right\}$. Then: {{begin-eqn}} {{eqn | ll= \exists k \in \Z:       | l = w        | r = \dfrac 1 {2 i} \ln \left({\dfrac {z + i} {z - i} }\right) + k \pi       | c =  }} {{eqn | lll= \implies       | ll= \exists k \in \Z:       | l = 2 i w + 2 \left({-k}\right) \pi i       | r = \ln \left({\frac {z + i} {z - i} }\right)       | c =  }} {{eqn | lll= \implies       | l = e^{2 i w + 2 \left({-k}\right) \pi i}       | r = \frac {z + i} {z - i}       | c = Definition of Complex Natural Logarithm }} {{eqn | lll= \implies       | l = e^{2 i w}       | r = \frac {z + i} {z - i}       | c = Complex Exponential Function has Imaginary Period }} {{eqn | lll= \implies       | l = z       | r = i \frac {e^{2 i w} + 1} {e^{2 i w} - 1}       | c =  }} {{eqn | lll= \implies       | l = z       | r = \cot w       | c = Cotangent Exponential Formulation }} {{eqn | lll= \implies       | l = w       | o = \in       | r = \left\{ {w \in \C: \cot \left({w}\right) = z}\right\}       | c =  }} {{end-eqn}} Thus by definition of superset: :$\left\{{w \in \C: \cot \left({w}\right) = z}\right\} \supseteq \left\{{\dfrac 1 {2 i} \ln \left({\dfrac {z + i} {z - i}}\right) + k \pi: k \in \Z}\right\}$ {{qed|lemma}} Thus by definition of set equality: :$\left\{{w \in \C: \cot \left({w}\right) = z}\right\} = \left\{{\dfrac 1 {2 i} \ln \left({\dfrac {z + i} {z - i}}\right) + k \pi: k \in \Z}\right\}$ {{qed}} Category:Inverse Cotangent olaj1niepowkba7m70eh4ocwvpgtk23