Evaluate $\dfrac{1}{3+4i}$

Trick for removing complex form from denominator

There is a complex number $3+4i$ in the denominator of the fraction. The form of complex number should be removed from the denominator and it is possible when we use a mathematical trick in this problem.

$=\,\,\,$ $\dfrac{1}{3+4i}$ $\times$ $1$

The form of complex number in the denominator is removed when the complex number $3+4i$ is multiplied by a complex number $3-4i$.

$=\,\,\,$ $\dfrac{1}{3+4i}$ $\times$ $\dfrac{3-4i}{3-4i}$

Evaluate the Product by Multiplying the fractions

The fractions should be multiplied to find their product as per the multiplication rule of the fractions. So, multiply the numerator of first fraction by the numerator of second fraction and also multiply the denominator of first fraction by the denominator of second fraction to obtain the product of them mathematically.

$=\,\,\,$ $\dfrac{1 \times (3-4i)}{(3+4i) \times (3-4i)}$

Firstly, let’s find the product of the number $1$ and the complex number $3-4i$ in the numerator, and their product is equal to $3-4i$.

$=\,\,\,$ $\dfrac{3-4i}{(3+4i) \times (3-4i)}$

Now, let us focus on finding the multiplication of the complex numbers $3+4i$ and $3-4i$ in the denominator, and their product can be obtained as per the sum and difference basis binomial product identity.

$=\,\,\,$ $\dfrac{3-4i}{(3)^2-(4i)^2}$

Find the value of expression by the simplification

The expression in the numerator is already in simplified form. So, let us try to simplify the mathematical expression in the denominator.

$=\,\,\,$ $\dfrac{3-4i}{3^2-(4^2 \times i^2)}$

According to the complex numbers, the square of imaginary unit is negative one.

$=\,\,\,$ $\dfrac{3-4i}{9-\big(16 \times (-1)\big)}$

It is time to simplify the arithmetic expression in denominator of the mathematical expression.

$=\,\,\,$ $\dfrac{3-4i}{9-(-16)}$

$=\,\,\,$ $\dfrac{3-4i}{9+16}$

$=\,\,\,$ $\dfrac{3-4i}{25}$