### 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 \(\overline{z}=1-5\,\mathrm{i}\), we have \[\frac{1}{2}(z+\overline{z})=\frac{1}{2}\!(1+5\,\mathrm{i}+1-5\,\mathrm{i})=1\tiny.\]

Second, a formula from the theory: #\frac{1}{2}(z+\overline{z})=\Re(z)=\Re(1+5\,\mathrm{i})=1#.

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