@jeffcliff The only chance for e^[ik(x)] to equal e^[−ik(x)] is to select x for which k(x) equals either 0 or π.
@chuculate note the bar should be rendering above representing complex conjugacy. it should be valid for all real k(x)
@jeffcliff maybe from their trigonometric expressions
since
conj(e^[ik(x)]) = cos (k(x)) + i · sin (k(x))
and
e^[-ik(x)] = cos (-k(x)) - i · sin(-k(x))
Since cosine is an even function, its argument sign will not matter.
Since sine is an odd function, sin(-x) is equivalent to -sin(x) and thus that part also becomes the same, above and below?
since
conj(e^[ik(x)]) = cos (k(x)) + i · sin (k(x))
and
e^[-ik(x)] = cos (-k(x)) - i · sin(-k(x))
Since cosine is an even function, its argument sign will not matter.
Since sine is an odd function, sin(-x) is equivalent to -sin(x) and thus that part also becomes the same, above and below?
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@chuculate @jeffcliff
\overline{e^{iz}} = \overline{e^{i(a+ib)}} = \overline{e^{i(a+ib)}} = \overline{e^{ia-b}}=e^{-b}\overline{e^{ia}}=e^{-b}e^{-ia} = e^{i(a-ib)} Its not true in general
@chuculate @jeffcliff Then \overline{z} = re^{-i\theta}, the reflection across the real axis.
@mandlebro @chuculate ie your counterexample was k as not real
@jeffcliff @chuculate if what you meant was k(x) is a real valued function then the point wise equality implies equality if functions. So I'm not sure what you're asking.
@chuculate @jeffcliff You can see this from polar form writing the numbers as z = re^{i\theta}