Question

How do you simplify $\dfrac{2}{4-9i}$ ?

Answer 1

In the given question, we are given a fractional term in which the denominator involves the complex number and the numerator is 2. So, for this firstly we need to remember the property that ${{i}^{2}}=-1$ and also, we need to rationalise in order to remove the complex number iota from the denominator.

Complete step-by-step solution:
We are asked to simplify the given fraction $\dfrac{2}{4-9i}$in the question and we can clearly see that complex number iota is present in the denominator so firstly what we need to do is we need to remove that iota from the denominator. So, for this we will simplify the given fraction by dividing and multiplying by $4+9i$ .
So, after rationalising we get,
$\begin{aligned} & \dfrac{2}{4-9i}\times \dfrac{4+9i}{4+9i} \\\ & \Rightarrow \dfrac{2\left( 4+9i \right)}{16+81} \\\ & \Rightarrow \dfrac{2\left( 4+9i \right)}{97} \\\ \end{aligned}$
So, now we can see that after rationalising iota is removed from the denominator and is present in the numerator. Also, we can clearly see that we can not simplify this fraction further so the simplified form of the given fraction $\dfrac{2}{4-9i}$is $\dfrac{2\left( 4+9i \right)}{97}$.

Note: Most common mistake that we make in this question is that we forget to remove the iota from the denominator after rationalising and continue to simplify this complex fraction which will not give us any output as it is required. Apart from this we need to make the fraction in real and complex just two terms while simplifying which mostly we leave as it is.