Difference between revisions of "Euler's formula"
From specialfunctionswiki
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==Proof== | ==Proof== | ||
| + | Recall the definition of $\cos$ | ||
| + | $$\cos(z) = \dfrac{e^{iz} + e^{-iz}}{2}$$ | ||
| + | and the definition of $\sin$ | ||
| + | $$\sin(z) = \dfrac{e^{iz}-e^{-iz}}{2i}.$$ | ||
| + | Now compute | ||
| + | $$\begin{array}{ll} | ||
| + | \cos(z) + i\sin(z) &= \left( \dfrac{e^{iz}+e^{-iz}}{2} \right) + i \left( \dfrac{e^{iz}-e^{-iz}}{2i} \right) \\ | ||
| + | &= e^{iz}, | ||
| + | \end{array}$$ | ||
| + | as was to be shown. █ | ||
==References== | ==References== | ||
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Proven]] | ||
Latest revision as of 00:20, 23 December 2016
Theorem
The following formula holds: $$e^{iz}=\cos(z)+i\sin(z),$$ where $e^{iz}$ denotes the exponential function, $\cos$ denotes the cosine, $i$ denotes the imaginary number, and $\sin$ denotes the sine.
Proof
Recall the definition of $\cos$ $$\cos(z) = \dfrac{e^{iz} + e^{-iz}}{2}$$ and the definition of $\sin$ $$\sin(z) = \dfrac{e^{iz}-e^{-iz}}{2i}.$$ Now compute $$\begin{array}{ll} \cos(z) + i\sin(z) &= \left( \dfrac{e^{iz}+e^{-iz}}{2} \right) + i \left( \dfrac{e^{iz}-e^{-iz}}{2i} \right) \\ &= e^{iz}, \end{array}$$ as was to be shown. █
