Conjugate of a complex Number
For an arbitrary complex number \(z = a + bi\), its conjugate is defined as \(\bar z = a  bi\). For example,
\[\begin{align}&z = 2 + 3i \qquad \Rightarrow \qquad \bar z = 2  3i\\&z =  1  5i \!\!\!\!\!\qquad \Rightarrow\qquad \bar z =  1 + 5i\\&z = i  \frac{1}{2} \qquad \Rightarrow\qquad \bar z =  i  \frac{1}{2}\end{align}\]
Note the last example carefully. When we take the conjugate of a complex number, it is the imaginary part which reverses in sign, and not the real part.
Geometrically, \(\bar z\) will be the mirror image of z in the horizontal axis, as the following figure shows:
We make the following observations:

The real part of \(\bar z\) is the same as the real part of z. That is, \({\mathop{\rm Re}\nolimits} \left( {\bar z} \right) = {\mathop{\rm Re}\nolimits} \left( z \right)\).

The imaginary part of \(\bar z\) is the negative of the imaginary part of z. That is, \({\mathop{\rm Im}\nolimits} \left( {\bar z} \right) =  {\mathop{\rm Im}\nolimits} \left( z \right)\).

If z is purely real, then \(\bar z = z\). For example, if \(z = 3\), then \(\bar z\) is also 3.

If z is purely imaginary, then \(\bar z =  z\). For example, if \(z = 2i\), then \(\bar z =  2i\).

If z is above the horizontal axis, then \(\bar z\) is below the horizontal axis, and viceversa.
Now, let’s see what happens if we add / subtract a complex number and its conjugate. Let \(z = a + bi\). We have:
\[\begin{align}&z + \bar z = \left( {a + bi} \right) + \left( {a  bi} \right) = 2a = 2{\mathop{\rm Re}\nolimits} \left( z \right)\\&z  \bar z = \left( {a + bi} \right)  \left( {a  bi} \right) = 2bi = 2i{\mathop{\rm Im}\nolimits} \left( z \right)\end{align}\]
By virtue of these relations, we can write the real and imaginary parts of a complex number as follows:
\[\begin{align}&{\mathop{\rm Re}\nolimits} \left( z \right) = \frac{{z + \bar z}}{2}\\&{\mathop{\rm Im}\nolimits} \left( z \right) = \frac{{z  \bar z}}{{2i}}\end{align}\]
Next, consider what will happen if we multiply z and \(\bar z\):
\[\begin{align}z\bar z& = \left( {a + bi} \right)\left( {a  bi} \right)\, = {\left( a \right)^2}  {\left( {bi} \right)^2}\\& = {a^2}  {b^2}{i^2} = {a^2}  {b^2}\left( {  1} \right)\\& = {a^2} + {b^2} = {\left z \right^2}\end{align}\]
Thus, the product of a complex number and its conjugate gives the square of its magnitude. This relation is used very frequently, and should be internalized properly.
Example 1: Consider the following two complex numbers:
\[\begin{align}&{z_1} = 2  3i\\&{z_2} =  4  7i\end{align}\]
Find the conjugate of the complex number \(z = 4{z_1}  2i{z_2}\).
Solution: We have:
\[\begin{align}&4{z_1} = 4\left( {2  3i} \right) = 8  12i\\&2i{z_2} = 2i\left( {  4  7i} \right) =  8i  14{i^2} = 14  8i\\& \Rightarrow \,\,\,z = 4{z_1}  2i{z_2} = \left( {8  12i} \right)  \left( {14  8i} \right)\\ &=  6  4i\end{align}\]
Thus,
\[\bar z =  6 + 4i\]