Question

What is the conjugate of a fraction?

Answer 1

Method $1$:

A conjugate of a complex number is another complex number that has the same real part as the original complex number and the imaginary part has the same magnitude but opposite sign.

To find the conjugate of a fraction, multiply the numerator and denominator by the complex conjugate of the denominator.

Let $z=\frac{1}{x+yi}$ is a fraction, then

Rewrite the complex number as $a+ib$ form

$z=\frac{1\left(x-yi\right)}{\left(x+yi\right)\left(x-yi\right)}$

$\Rightarrow$ $z=\frac{x-yi}{x^2-{\left(yi\right)}^2}$ $\left[\because \left(a+b\right)\left(a-b\right)=a^2-b^2\right]$

$\Rightarrow$ $z=\frac{x-yi}{x^2-\left(-y^2\right)}$ $\left[\because i^2=-1\right]$

$\Rightarrow$ $z=\frac{x-yi}{x^2+y^2}$

The conjugate of $z$ is $\overline{z}=\frac{x+yi}{x^2+y^2}$

Method $2$: Conjugate radical function

A radical function is a function that contains a square root.

$\sqrt{a}+\sqrt{b}$ and $\sqrt{a}-\sqrt{b}$ are conjugate radical.

To find the conjugate of a fraction, multiply the numerator and denominator by the conjugate of the denominator.

Let $c=\frac{1}{\sqrt{x}+\sqrt{y}}$ is a fraction, then

Rewrite the radical function

$c=\frac{1\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}$

$\Rightarrow$ $c=\frac{\sqrt{x}-\sqrt{y}}{{\left(\sqrt{x}\right)}^2-{\left(\sqrt{y}\right)}^2}$ $\left[\because \left(a+b\right)\left(a-b\right)=a^2-b^2\right]$

$\Rightarrow$ $c=\frac{\sqrt{x}-\sqrt{y}}{x-y}$

The conjugate of the radical function is $\overline{c}=\frac{\sqrt{x}+\sqrt{y}}{x-y}$.