Conjugate of a Complex Number

Let $a$ and $b$ be real numbers and let \[z=a+ib\] be a complex number. Then, conjugate of a complex number $z$ denoted by $\bar{z}$ (read “bar z”) is defined as \[\bar{z}=a-ib\]

Here, we observe that two complex numbers are said to be conjugate of one another if their real parts are same and the imaginary parts differ in sign only.

Thus, $2+3i$ and $2-3i$ are conjugate complex numbers. Each is the conjugate of the other.

Properties of Conjugates

Let $z=a+ib$ and $w=c+id$ be two complex numbers.

P1: \[\begin{array}{l} & \frac{1}{2}(z+\bar{z}) &=\text{Re}(z) \\ \text{and} & \frac{1}{2i}(z-\bar{z}) &=\text{Im}(z)\end{array}\]

\[\begin{array}{l}\frac{1}{2}(z+\bar{z})&=\frac{1}{2}(a+ib+a-ib)\\ &=\frac{1}{2}×2a \\ &=a \\ &=\text{Re}(z)\end{array}\] and, \[\begin{array}{l}\frac{1}{2i}(z-\bar{z}) &=\frac{1}{2i}(a+ib-a+ib) \\ &=\frac{1}{2i}×2ib \\ &= b \\ &=\text{Im}(z)\end{array}\]


P2: \[\bar{\bar{z}}=z\]

Here, \[\begin{array}{l} z &= a+ib \\ \bar{z} &= a-ib \\ \bar{\bar{z}} &= a+ib &= z \end{array}\]


P3: \[z\cdot\bar{z}=\text{a pure real number}\]

\[\begin{array}{l} z\cdot\bar{z} &= (a+ib) (a-ib) \\ &= a^2-i^2b^2 \\ &= a^2+b^2 \\ &=\text{a pure real number}\end{array}\]


P4: \[(\bar{z})^2=\overline{z^2}\]

Here, \[\begin{array}{l} & z &= a+ib \\ & z^2 &= (a+ib)^2 \\ & z^2 &= a^2+i^2b^2+(2ab)i \\ & z^2 &= (a^2-b^2)+(2ab)i \\ \therefore & \overline{z^2} &= (a^2-b^2)-(2ab)i \end{array}\]

Also, \[\begin{array}{l} (\overline{z})^2 &= (a-ib)^2 \\ &= a^2+i^2b^2-(2ab)i \\ &= (a^2-b^2)-(2ab)i \\ &= \overline{z^2} \\ \therefore (\overline{z})^2 &=\overline{z^2}\end{array}\]


P5: \[\overline{z+w}=\overline{z}+\overline{w}\]

Here, \[\begin{array}{l} \overline{z+w} &= a+ib+c+id \\ &= (a+c)+(b+d)i \\ \therefore \overline{z+w} &= (a+c)-(b+d)i \end{array}\]

And, \[\begin{array}{l} \overline{z}+\overline{w} &= a-ib+c-id \\ &= (a+c)-(b+d)i \\ &= \overline{z+w} \end{array}\] \[\therefore \overline{z+w}=\overline{z}+\overline{w}\]


P6: \[\overline{z\cdot w}=\overline{z}\cdot\overline{w}\]

Here, \[\begin{array}{l} z\cdot w &=(a+ib)(c+id) \\ &= ac+iad+ibc+i^2bd \\ &= (ac-bd)+(ad+bc)i \\ \therefore\overline{z\cdot w} &= (ac-bd)-(ad+bc)i\end{array}\]

And, \[\begin{array}{l}\overline{z}\cdot\overline{w} &= (a-ib)(c-id) \\ &= ac-iad-ibc+i^2bd \\ &= (ac-bd)-(ad+bc)i \\ &= \overline{z\cdot w}\end{array}\] \[\therefore\overline{z\cdot w}=\overline{z}\cdot\overline{w}\]



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