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CONJUGATE OF A COMPLEX NUMBER

When two complex numbers differ only in the sign of $i$, they are said to be conjugates of each other.

Thus $x+iy$ and $x-iy$ are two conjugate complex numbers.

The conjugate of a complex number $z$ is denoted by $\overline{z}$.

PROPERTIES OF CONJUGATES

(I) The conjugate of the conjugate of a complex number is the complex number itself,
i.e. $\overline{\overline{z}}=z$.

Proof. Let
$z=x+iy$, where $x,y\in ℝ$.

$\therefore \overline{z}=x-iy.$

$\therefore \overline{\overline{z}}=x-iy=x+iy=z.$

(II) The sum and product of two conjugate complex numbers are purely real.

Let $z=x+iy$. Then $\overline{z}=x-iy$, where $x,y\in ℝ$.

(i) Sum $=z+\overline{z}=(x+iy)+(x-iy)=2x$, which is purely real.

(ii) Product $=z\cdot \overline{z}=(x+iy)(x-iy)=x^2-i^2{y}^2=x^2-(-1)y^2=x^2+y^2$, which is purely real.

(III) The conjugate of the sum (product) of two complex numbers is the sum (product) of their conjugates,
i.e. $\overline{z_1+z_2}=\overline{z_1}+\overline{z_2}$ and $\overline{z_1{z}_2}=\overline{z_1}\cdot \overline{z_2}$

Proof. Let
$z_1=x_1+i{y}_1$ and $z_2=x_2+i{y}_2$, $(x_1,x_2;y_1,y_2\in ℝ)$.

$\therefore \overline{z_1}=x_1-i{y}_1$ and $\overline{z_2}=x_2-i{y}_2.$

(i) $z_1+z_2=(x_1+i{y}_1)+(x_2+i{y}_2)=(x_1+x_2)+i(y_1+y_2).$

Hence, $\overline{z_1+z_2}=(x_1+x_2)-i(y_1+y_2)=(x_1-i{y}_1)+(x_2-i{y}_2)=\overline{z_1}+\overline{z_2}.$

(ii) $z_1{z}_2=(x_1+i{y}_1)(x_2+i{y}_2)=(x_1{x}_2-y_1{y}_2)+i(x_1{y}_2+y_1{x}_2).$

$\therefore \overline{z_1{z}_2}=(x_1{x}_2-y_1{y}_2)-i(x_1{y}_2+y_1{x}_2).$

Also $\overline{z_1}\cdot \overline{z_2}=(x_1-i{y}_1)(x_2-i{y}_2)=(x_1{x}_2-y_1{y}_2)-i(x_1{y}_2+y_1{x}_2).$

Hence, $\overline{z_1{z}_2}=\overline{z_1}\cdot \overline{z_2}$

(IV)

(i) $\overline{z_1-z_2}=\overline{z_1}-\overline{z_2}$

(ii) $\overline{\left({\displaystyle \frac{z_1}{z_2}}\right)}={\displaystyle \frac{\overline{z_1}}{\overline{z_2}}}$, $z_2\ne 0.$

Proof. Let
$z_1=x_1+i{y}_1$ and $z_2=x_2+i{y}_2$, $(x_1,x_2;y_1,y_2\in ℝ)$.

$\therefore \overline{z_1}=x_1-i{y}_1$ and $\overline{z_2}=x_2-i{y}_2.$

(i) $z_1-z_2=(x_1+i{y}_1)-(x_2+i{y}_2)=(x_1-x_2)+i(y_1-y_2).$

$\therefore \overline{z_1-z_2}=(x_1-x_2)-i(y_1-y_2)=(x_1-i{y}_1)-(x_2-i{y}_2)=\overline{z_1}-\overline{z_2}.$

(ii) ${\displaystyle \frac{z_1}{z_2}}={\displaystyle \frac{x_1+i{y}_1}{x_2+i{y}_2}}={\displaystyle \frac{x_1+i{y}_1}{x_2+i{y}_2}}\times {\displaystyle \frac{x_2-i{y}_2}{x_2-i{y}_2}}$

$={\displaystyle \frac{(x_1{x}_2+y_1{y}_2)+i(x_2{y}_1-x_1{y}_2)}{x_2^2+y_2^2}}={\displaystyle \frac{x_1{x}_2+y_1{y}_2}{x_2^2+y_2^2}}+i{\displaystyle \frac{x_2{y}_1-x_1{y}_2}{x_2^2+y_2^2}}.$

$\therefore \overline{\left({\displaystyle \frac{z_1}{z_2}}\right)}={\displaystyle \frac{x_1{x}_2+y_1{y}_2}{x_2^2+y_2^2}}-i{\displaystyle \frac{x_2{y}_1-x_1{y}_2}{x_2^2+y_2^2}}.$

Also

${\displaystyle \frac{\overline{z_1}}{\overline{z_2}}}={\displaystyle \frac{x_1-i{y}_1}{x_2-i{y}_2}}={\displaystyle \frac{x_1-i{y}_1}{x_2-i{y}_2}}\times {\displaystyle \frac{x_2+i{y}_2}{x_2+i{y}_2}}$

$={\displaystyle \frac{(x_1{x}_2+y_1{y}_2)+i(x_2{y}_2-x_1{y}_1)}{x_2^2+y_2^2}}={\displaystyle \frac{x_1{x}_2+y_1{y}_2}{x_2^2+y_2^2}}-i{\displaystyle \frac{x_2{y}_2-x_1{y}_2}{x_2^2+y_2^2}}.$

Hence,

$\overline{\left({\displaystyle \frac{z_1}{z_2}}\right)}={\displaystyle \frac{\overline{z_1}}{\overline{z_2}}},z_2\ne 0.$

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