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Does Euler's formula ($e^{ix} = \cos x + i \sin x$) work in base $10$? If it does, how could I express it?

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  • $\begingroup$ One may always write $$10^{ix}=e^{ix\ln(10)}=\cos[ix\ln(10)]+i\sin[ix\ln(10)].$$ $\endgroup$ – Olivier Oloa Jan 2 '18 at 14:26
  • $\begingroup$ @mdewey What ?? $\endgroup$ – Rebellos Jan 2 '18 at 14:40
  • $\begingroup$ If you are ok, you can accept the answer and set as solved. Thanks! $\endgroup$ – gimusi Jan 3 '18 at 17:29
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We can try converting the formula to base $10$.

Let $\log$ be the natural (i.e., base $e$) logarithm. Then $10 = e^{\log10}$, so:

$$ 10^{ix} = e^{ix\log10}=\cos(x\log10) + i\sin(x\log10) $$

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The formula does not hold, you can write:

$$10^{ix}=(e^{\log 10})^{ix} =(e^{ix})^{\log 10} = (\cos x + i \sin x)^{\log 10}$$

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