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# How do you simplify $\dfrac{2-3i}{1-2i}$?

Last updated date: 13th Jun 2024
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Hint:
Any number in the form of $a+bi$ is known as a complex number. Its conjugate complex number is $a-bi$. Product of a complex and its conjugate complex number is the sum of squares of a and b, that is ${{(a)}^{2}}+{{(b)}^{2}}$. Most importantly, the square of $i$ is equal to -1.

In the given question, the complex number is $\dfrac{2-3i}{1-2i}$. While simplifying such types of complex numbers, we will eliminate the complex number from the denominator by multiplying both the numerator and denominator with the conjugate complex number of the denominator.
Here the denominator complex number is $(1-2i)$. The conjugate complex number of the denominator $(1-2i)$ is $\left( 1+2i \right)$. So, on multiplying both the numerator and denominator with $\left( 1+2i \right)$ we get
$\Rightarrow \dfrac{(2-3i)(1+2i)}{(1-2i)(1+2i)}$
We know that $\left( a+bi \right)(a-bi)={{(a)}^{2}}+{{(b)}^{2}}$. On multiplying two constants we get a constant; on multiplying a constant (say c) with iota $i$ gives $ci$ and on multiplying two iota terms (say $ci$, $di$ ) we get ${{i}^{2}}$ term with its coefficient $c\times d$. So, we can write
$\Rightarrow \dfrac{(2+4i-3i-6{{i}^{2}})}{({{1}^{2}}-{{(2i)}^{2}})}$
We know that square of $i$ is equal to -1. On substituting -1 and subtraction of $3i$ from $4i$ we get
\begin{align} & \Rightarrow \dfrac{(2+i-6(-1))}{(1-4(-1))} \\ & \Rightarrow \dfrac{(2+i+6)}{(1+4)} \\ \end{align}
$\Rightarrow \dfrac{(8+i)}{5}$
Hence the simplified form of $\dfrac{2-3i}{1-2i}$ is $\dfrac{8+i}{5}$.
If we want to simplify an expression, it is always important to keep in mind what we mean when we say ’simplify’. Typically, in the case of complex numbers, we aim to never have a complex number in the denominator of any term. To accomplish this, we multiply $a+bi$ by its complex conjugate $a-bi$ , where we end up with a real number ${{(a)}^{2}}+{{(b)}^{2}}$. This will allow us to simplify the complex nature out of a denominator.