Difference between revisions of "Euler's formula"

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(Created page with "==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 ima...")
 
 
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==Proof==
 
==Proof==
 +
Recall the definition of $\cos$
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$$\cos(z) = \dfrac{e^{iz} + e^{-iz}}{2}$$
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and the definition of $\sin$
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$$\sin(z) = \dfrac{e^{iz}-e^{-iz}}{2i}.$$
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Now compute
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$$\begin{array}{ll}
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\cos(z) + i\sin(z) &= \left( \dfrac{e^{iz}+e^{-iz}}{2} \right) + i \left( \dfrac{e^{iz}-e^{-iz}}{2i} \right) \\
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&= e^{iz},
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\end{array}$$
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as was to be shown. █
  
 
==References==
 
==References==
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Logarithm at -i|next=e is limit of (1+1/n)^n}}: 4.1.16
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[[Category:Theorem]]
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[[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. █

References