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Proposition: Inverse Tangent and Complex Exponential Function

Let $x\in\mathbb R$ be real number and let $\phi=\arctan(x)$ (where $\arctan$ denotes the inverse tangent of $x$). Then the complex exponential function of $2 i\phi$ can be calculated by

$$\exp(2 i \phi)=\frac{1-ix}{1+ix}.$$

Table of Contents

Proofs: 1

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References

Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983