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