What does conjugation in the time-domain of a signal mean? I've never been explicitly told what the conjugation of a signal in the time-domain means. I'm mainly asking because in my signals class, my professor stated that for a signal x(t) to be real: x(t) = x(t)*, and I don't understand why.
 A: If 
$$
z\equiv a+ib
$$
is a complex number ($a$ and $b$ are real numbers), its conjugate
is defined to be
$$
\overline{z}\equiv a-ib.
$$
Note that we can write a real number $x$ as
$$
x=x+i0.
$$
The conjugate of $x$ is equal to itself:
$$
\overline{x}=x-i0=x.
$$
Therefore, if a number is real, it is equal to its conjugate.
For the converse, if $z\equiv a+ib$ is equal to its conjugate,
$$
a+ib=a-ib
$$
which implies
$$
ib=-ib.
$$
If $b$ is anything other than zero, we arrive at a contradiction,
since $i\neq-i$. Therefore, if a complex number is equal to its conjugate, it is real.
All that remains to be done is to apply your result pointwise (i.e. for every possible time $t$), and you arrive at the result $x(t)=x(t)^*$ for all $t$ if and only if $x(t)$ is real for all $t$.
A: Let $z$ be some fixed but arbitrary complex number with Cartesian representation $z=\Re{(z)}+i\,\Im{(z)}$ for some real pair $(\Re{(z)},\Im{(z)})\in\mathbb{R}^2$. The complex conjugate of $z$ is then,
$$z^*=\Re{(z)}-i\,\Im{(z)}.$$
Subtracting the conjugate of $z$ from $z$ itself then yields:
$$\begin{align}
z-z^*
&=(\Re{(z)}+i\,\Im{(z)})-(\Re{(z)}-i\,\Im{(z)})\\
&=2i\,\Im{(z)}.
\end{align}$$
Solving for the imaginary component of $z$, we have:
$$\Im{(z)}=\frac{z-z^*}{2i}.$$
Thus, if $z\neq z^*$, or equivalently if $z-z^*\neq 0$, then $\Im{(z)}\neq 0$. Contrapositively, if $\Im{(z)}=0$ then $z=z^*$. $\blacksquare$
