How to visualize this statement regarding Conjugate Symmetry
For $(1)$, how can I visualize this to see why the negative frequency components do not matter anymore$?$, does this mean the Fourier series representation of $x(t)$ becomes :
$$
x(t)=\sum_{k=0}^{\infty}c_{k}e^{jk\omega_{0} t}\;???
$$

No, the Fourier series is still what it was before this conjugate symmetry nonsense reared its ugly head. But what happens is that
\begin{align}
x(t) &amp;= \sum_{k=-\infty}^{\infty}c_{k}e^{jk\omega_{0} t} &amp; {\scriptstyle\text{original definition}}\\
&amp;= \sum_{k=-\infty}^{-1}c_{k}e^{jk\omega_{0} t} + c_0 + \sum_{k=1}^{\infty}c_{k}e^{jk\omega_{0} t}&amp; {\scriptstyle\text{breaking up the sum}}\\
&amp;= \sum_{k=1}^{\infty}c_{-k}e^{-jk\omega_{0} t} + c_0 + \sum_{k=1}^{\infty}c_{k}e^{jk\omega_{0} t}&amp; {\scriptstyle\text{changing indices}}\\
&amp;= c_0 + \sum_{k=1}^{\infty}c_{k}e^{jk\omega_{0} t} + c_{-k}e^{-jk\omega_{0} t} &amp; {\scriptstyle\text{combining sums}}\\
&amp;= c_0 + \sum_{k=1}^{\infty}c_{k}e^{jk\omega_{0} t} + c_{k}^*e^{-jk\omega_{0} t} &amp; {\scriptstyle c_k = c_{-k}^* \implies c_{-k} = c_k^*}\\
&amp;= c_0 + \sum_{k=1}^{\infty}c_{k}e^{jk\omega_{0} t} + \left(c_{k}e^{jk\omega_{0} t}\right)^* &amp; {\scriptstyle a^*b^* = (ab)^*}\\
&amp;= c_0 + 2\sum_{k=1}^{\infty}\operatorname{Re}\left(c_{k}e^{jk\omega_{0} t}\right)
\end{align}
showing that one needs to compute or know the $c_k$ values for nonnegative $k$ only; the values of $c_k$ for $k &lt; 0$ are not needed. Anal-retentives who believe in crossing their eyes and dotting their teas can always determine the values of the $c_{-k}$ by taking the complex conjugates of the known $c_k$'s.