Conjugates (Leaving Cert Mathematics): Revision Notes

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Conjugates

Note

The conjugate of a complex number zz is derived by changing the sign of the imaginary component of the complex number. It is denoted as zˉ\bar{z}.

z=a+bi  ⟹  zˉ=a−biz=a+bi \implies \bar{z}=a-bi

Consider the following examples :

z1=2+i  ⟹  z1ˉ=2−iz_1=2+i \implies \bar{z_1}=2-i z2=−8−2i  ⟹  z2ˉ=−8+2iz_2=-8-2i \implies \bar{z_2}=-8+2i z3=−5i  ⟹  z3ˉ=5iz_3=-5i \implies \bar{z_3}=5i z4=10  ⟹  z4ˉ=10z_4=10 \implies \bar{z_4}=10
Note

The conjugate of a (purely) real number is just a+(0)i‾=a−(0)i=a\overline{a+(0)i}=a-(0)i=a which is just the number itself.

Example

Note

Given that z=2−3iz=2-3i and w=2+4iw=2+4i, find z+w‾\overline{z+w}.

Always evaluate whatever is under the bar first.

z+w‾=(2−3i)+(2+4i)‾=4+i‾=4−i\begin{align*} \overline{z+w}&=\overline{(2-3i)+(2+4i)} \\\\ &=\overline{4+i} \\\\ &=4-i \end{align*}
Note

The conjugate of a conjugate is just the original expression. Given that z=a+biz=a+bi.

z‾‾=a+bi‾‾=a−bi‾=a+bi\begin{align*} \overline{\overline{z}}&=\overline{\overline{a+bi}} \\\\ &= \overline{a-bi} \\\\ &=a+bi \end{align*}

Properties of Conjugates

Try derive the following as an exercise, similar to the proof above.

Given that z,wz,w are complex numbers and a,ba,b are (real) constants :

z+w‾=z‾+w‾\overline{z+w}=\overline{z}+\overline{w} z−w‾=z‾−w‾\overline{z-w}=\overline{z}-\overline{w} z‾‾=z\overline{\overline{z}}=z az+bw‾=az‾+bw‾\overline{az+bw}=a\overline{z}+b\overline{w} zw‾=z‾⋅w‾\overline{zw}=\overline{z} \cdot \overline{w}
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