If $h(k)$ preserves the complex conjugation property, does $ikh(k)$ preserve it too?

My question is this: if $h(k)$ preserves the complex conjugation property (in other words, $h(k) = h(-k)$, $k$ can be just $-n$, $-n+1$, ..., $0$, $1$, ..., $n-1$),then $ikh(k)$ also preserves complex conjugation property which is easy to derive. But actually $ikh(k)$ does not preserve the complex conjugation property. Because we know $h(-n)$ must be real value for $h(k)$ is conjugation. Then $ikh(-n)$ must be image value which destroy the complex conjugation property. How to explain this?

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What do you mean by complex conjugation property? Conjugation invariance would be h(a+ib) = h(a-ib) for a,b real. This doesn't imply that h(k) = h(-k) for k an integer! –  Piotr Pstrągowski Dec 3 '12 at 16:05
I'm sorry.I mean the conjugate symmetric property.For example,the Fourier transform of a real function,f(x,y),is conjugate symmetric:F*(u,v)=F(-u,-v). –  user1859053 Dec 4 '12 at 0:51
I don't see why $h(-n)$ should be real. With the notation $g(k)=ikh(k)$, you want to show that $\overline{g(k)}=g(-k)$. But $\overline{g(k)} = \bar i \bar k \overline{h(k)} = -ik\overline{h(k)}$. –  jathd Dec 4 '12 at 2:34
h(-n) must be real value.Because h(n) is periodic(Sorry I forget to point out that).h(n) is a DFT conjugate symmetric sequence.So h(-n) and h(0) must be real value.Not just h(-n) = h*(n). –  user1859053 Dec 4 '12 at 2:43