I have a sum:
$$\sum _{k=1}^n i \log \left(e^{i k}\right)$$ then using identities: I have:
$$i \log \left(\prod _{k=1}^n e^{i k}\right)=i \log \left(e^{\frac{1}{2} i n (1+n)}\right)$$
,but this identities not hold for my function. I put numericall value $n=1..10$ to Product:
{-1., -3., 0.283185, 2.56637, -2.43363, -2.15044, -2.86726, 1.69911, -1.0177, 1.54867}
In sum I have:
{-1., -3., -6., -3.71681, -2.43363, -2.15044, -2.86726, -4.58407,-7.30089, -4.73452}
values are different in the third, fourth, eighth, ninth, and tenth.
$$\sum _{k=1}^n \log (f(k))\neq \log \left(\prod _{k=1}^n f(k)\right)$$
Why this identities not hold for my function?
It basically comes down to what branch of the natural logarithm you are using. Note that:
$$\ln(a)+\ln(b)=\ln(ab)$$
generally holds only when $a,b>0$. For a quick example of when they don't hold, note that:
$$2\ln(-1)=\ln(-1)+\ln(-1)\ne\ln(1)=0$$
Clearly, $\ln(-1)\ne0$.