Let \(z = a+bi\) be a complex number where \(a,b\in \mathbb{R}\). The complex conjugate of \(z\), denoted by \(\overline{z}\), is given by \(a - bi\). In other words, to obtain the complex conjugate of \(z\), one simply flips the sign of its imaginary part.
\(\overline{4} = 4\) because the imaginary part of \(4\) is \(0\).
\(\overline{1+2i} = 1 - 2i\).
\(\overline{3i} = - 3i\).
One can avoid memorizing the formula for the multiplicative inverse by making use of the complex conjugate. For example, to turn \(\displaystyle\frac{1}{1+i}\) into the form \(p+qi\) where \(p,q \in \mathbb{R}\), multiply the numerator and denominator by the complex conjugate of the denominator. Namely, \[\begin{eqnarray} \frac{1}{1+i} & = & \frac{1}{1+i} \frac{\overline{1+i}}{\overline{1+i}} \\ & = & \frac{1}{1+i} \frac{1-i}{1-i} \\ & = & \frac{(1-i)}{(1+i)(1-i)} \\ & = & \frac{1-i}{1-i^2} \\ & = & \frac{1-i}{2} = \frac{1}{2}-\frac{1}{2}i \end{eqnarray}\]
Hence, dividing the complex number \(z\) by the complex number \(w\) can be performed as \(\displaystyle\frac{z}{w} = \frac{z}{w}\frac{\overline{w}}{\overline{w}}\). For example, \[\frac{3+2i}{2-i} =\frac{3+2i}{2-i}\cdot\frac{2+i}{2+i} =\frac{4+7i}{2^2-i^2} =\frac{4+7i}{4+1} = \frac{4}{5} + \frac{7}{5}i.\]
Give the complex conjugate of each of the following:
\(2i\)
\(\displaystyle -2+\frac{3}{5}i\)
\(\pi\)
Let \(z\) and \(w\) be complex numbers. Show that
\(\overline{\overline{z}} = z\).
\(\overline{z + w} = \overline{z} + \overline{w}\).
\(\overline{z w} = \overline{z}~\overline{w}\).
\(\displaystyle\overline{z^{-1}} = (\overline{z})^{-1}\).