Complex numbers: Calculating with complex numbers
Complex conjugate
In the theory of the quotient we saw that a denominator of the form #a+b\,\ii# can be cleared by multiplying numerator and denominator by #a-b\,\ii#. This approach works because of the notable product #(a+b\,\ii)\cdot(a-b\,\ii)=\left|a+b\,\ii\right|^2# Not only the multiplication by #a-b\,\ii# provides a real number, also the summation does: #(a+b\,\ii) + (a-b\,\ii)=2a#. Reason enough to give #a-b\,\ii# a special name.
The complex conjugate
If #z=a+b\ii# with #a,b\in\mathbb{R}# is a complex number, then #a-b\ii# is called the complex conjugate of #z#.
This number is written as #\overline{z}#.
Interpretation
Complex conjugation is reflection in the real axis.
The conjugate of #\ii# is #-\ii#, the other solution to the equation #x^2=-1#.
The following rules of calculation show how easy it is to express the real and imaginary part in terms of #z# and its complex conjugate #\overline z#.
Rules of calculation of the complex conjugate
For all complex numbers #z# and #w# we have: \[ \begin{array}{rclcrcl}
\Re (\overline{z}) & = & \Re (z)&\phantom{quadquad}&
\Im (\overline{z}) & = & -\,\Im (z)\\
\Re (z) & = & \frac{1}{2}(z+\overline{z})&&
\Im (z) & = & \frac{1}{2\ii}(z-\overline{z}) \\
|\overline{z}| & = & |{z}|&&
\arg (\overline{z}) & = & -\,\arg (z)\phantom{x}\pmod {2\pi}\\
\overline{z+w} & = & \overline{z}+\overline{w}&&
\overline{z\cdot w} & = & \overline{z}\cdot \overline{w}\\
z+\overline{z} & = & 2\,\Re (z)&&
z \cdot \overline{z} & = & |z|^2
\end {array}
\]
Moreover, any complex number #z# satisfies the following quadratic equation with real coefficients #a=\Re(z)# and #r=|z|^2#: \[z^2-2\,a\cdot z+ r=0\]
The inverse of a complex number distinct from #0# can be expressed succinctly by use of the complex conjugate.
Rule for the inverse of a complex number
For any complex number #z# that is different from #0#, we have: \[\frac{1}{z}=\frac{\overline{z}}{|z|^2}\]We discuss two ways to demonstrate this.
First, a direct calculation: because \(\overline{z}=2+5\,\mathrm{i}\), we have \[\frac{1}{2}(z+\overline{z})=\frac{1}{2}\!(2-5\,\mathrm{i}+2+5\,\mathrm{i})=2\tiny.\]
Second, a formula from the theory: #\frac{1}{2}(z+\overline{z})=\Re(z)=\Re(2-5\,\mathrm{i})=2#.
Or visit omptest.org if jou are taking an OMPT exam.