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Express $\sin e^i$ in $a + ib$ form.

$$\sin e^i = \frac{e^{ie^i}-e^{-ie^i}}{2i}$$

I feel like I can express $ie^i$ in a simpler way, but I'm not sure how.

Thanks for your attention.

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    $\begingroup$ It depends on what your criteria for "simple" are. $\endgroup$ – fleablood Jan 30 '17 at 23:56
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$$\displaylines{ e^i=\cos1+i\sin1\cr ie^i=-\sin1+i\cos1\cr e^{ie^i}=e^{-\sin1}(\cos(\cos1)+i\sin(\cos1))\cr e^{-ie^i}=e^{\sin1}(\cos(\cos1)-i\sin(\cos1))\cr e^{ie^i}-e^{-ie^i}=-2\sinh(\sin1)\cos(\cos1)+2i\cosh(\sin1)\sin(\cos1)\cr \sin(e^i)=\cosh(\sin1)\sin(\cos1)+i\sinh(\sin1)\cos(\cos1)\cr }$$

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  • $\begingroup$ Please, can you elaborate on $e^{ie^i} - e^{-ie^i} = -2 \sinh (\sin1)\cos(\cos1) + 2i\cosh(\sin1)\sin(\cos1)$ ? Thanks a lot! $\endgroup$ – paranoidhominid Jan 31 '17 at 2:32
  • $\begingroup$ Just collect the terms involving $\cos(\cos1)$ and use the definition of hyperbolic functions. (Have you met hyperbolic functions?) $\endgroup$ – David Jan 31 '17 at 2:46
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Use the most beautiful equation in mathematics:

$e^{i \pi} = -1$

So $e^i = (-1)^{1/\pi} = (i^2)^{1/\pi} = i^{2/\pi}$.

Multiply by $i$ to get:

$i e^i = i^{(\pi + 2)/\pi} = -0.841471 + 0.540302 i$.

as one "simple" solution (among others). "Simple"? You be the judge.

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  • $\begingroup$ Nice idea, but you have to be careful because complex powers have multiple values and the usual rules of real exponents sometimes need to be modified. For example$$i^{2/\pi}=\exp\Bigl(\frac2\pi\log i\Bigr)=\exp\Bigl(\frac2\pi\,i\Bigl(\frac\pi2+2k\pi\Bigr)\Bigr)=e^{(1+4k)i}\ ,\quad k\in{\Bbb Z}\ ,$$and $e^i$ is just one of many possible values. Not sure if this answer can easily be made more rigorous. Some information. $\endgroup$ – David Jan 31 '17 at 0:38
  • $\begingroup$ The manipulations you do run in a whole number of issues that you seem to ignore. For instance, there is no way to define exponentiation of complex numbers such that the identity $(a^b)^c=a^{bc}$ holds, therefore $x^\pi=-1\not\Rightarrow x=(-1)^{1/\pi}$. All around, this answer seems pointless to me, because it introduces uncertainity in an otherwise straight-forward calculation. $\endgroup$ – user228113 Jan 31 '17 at 0:45

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