tambustqaze

2022-04-01

$i\mathrm{sin}\left(x\right)$ phase shift
By Euler's formula, I can express i in the following way:
$i=\mathrm{cos}\left(\frac{\pi }{2}\right)+i\mathrm{sin}\left(\frac{\pi }{2}\right)=\mathrm{exp}\left(i\frac{\pi }{2}\right)$
I wonder if it is legitimate to write
$i\mathrm{sin}x=\mathrm{exp}\left(i\frac{\pi }{2}\right)\cdot \frac{\mathrm{exp}\left(ix\right)-\mathrm{exp}\left(-ix\right)}{2i}$
$=\frac{\mathrm{exp}\left(ix\right)\mathrm{exp}\left(i\frac{\pi }{2}\right)-\mathrm{exp}\left(-ix\right)\mathrm{exp}\left(i\frac{\pi }{2}\right)}{2i}$
$=\frac{\mathrm{exp}\left(i\left(x+\frac{\pi }{2}\right)\right)-\mathrm{exp}\left(-i\left(x+\frac{\pi }{2}\right)\right)}{2i}$
$=\mathrm{sin}\left(x+\frac{\pi }{2}\right)$
I don't feel like this is right, because it would imply a lot of weird things. So where is my mistake?

mistemePietsffi

In your sleep-deprived delirium, you have said
${e}^{-ix}{e}^{i\frac{\pi }{2}}={e}^{-i\left(x+\frac{\pi }{2}\right)}$
where it is actually
${e}^{-ix}{e}^{i\frac{\pi }{2}}={e}^{-i\left(x-\frac{\pi }{2}\right)}$
It would certainly be strange if $i\mathrm{sin}\left(x\right)=\mathrm{sin}\left(x+\frac{\pi }{2}\right)$ since one is pure imaginary and the other real for real x.

Do you have a similar question?