How to visualize this statement regarding Conjugate Symmetry A property of real signals states that if $x(t)$ is real then the Fourier series coefficient (frequency spectrum) is given by :
$$
c_{k}=c_{-k}^{*}
$$
In polar form, this can be expressed as :
$$
|c_{k}|=|c_{-k}|\qquad\text{and}\qquad\arg c_{k}=-\arg c_{-k}
$$
It appears here that the magnitude spectrum obeys even symmetry and the phase spectrum obeys odd symmetry. It is said in my textbook :


Due to symmetry in the frequency spectra of real signals, we typically ignore negative frequencies when dealing with such signals.
In the case of signals that are complex but not real, frequency spectra do not possess the above symmetry,
and negative frequencies become important.


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 :
$$
x(t)=\sum_{k=0}^{\infty}c_{k}e^{jk\omega_{0} t}\;???
$$
For $(2)$, is this statement true because the complex exponentials in the frequency domain will scale a given signal to a decaying or growing signal?
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.