Question

How do you simplify $\dfrac{5-3i}{1+2i}$?

Answer 1

The given expression is of the form $\dfrac{a+bi}{c+di}$, here $i$ is an imaginary number equals $\sqrt{-1}$. To simplify these types of expression means we need to get rid of the imaginary number in the denominator. To do this, we need to multiply and divide the given expression with the conjugate of the imaginary number in the denominator. The conjugate of the number $c+di$ is $c-di$.

Complete step by step solution:
We are asked to simplify $\dfrac{5-3i}{1+2i}$. This expression is of the form of $\dfrac{a+bi}{c+di}$, here the values of $a,b,c,d$ as $5,-3,1,2$ respectively. As we know that to simplify these types of expressions, we need to multiply it by the conjugate of the complex number in the denominator. The complex number in the denominator is $1+2i$, so its complex conjugate will be $1-2i$. By multiplying and dividing the given expression by this conjugate, we get
$\Rightarrow \dfrac{5-3i}{1+2i}\times \dfrac{1-2i}{1-2i}$
First, let’s simplify the numerator $\left( 5-3i \right)\left( 1-2i \right)$. To simplify it, we need to expand the bracket. Thus, the expansion of $\left( 5-3i \right)\left( 1-2i \right)$ is done as,

& \Rightarrow \left( 5-3i \right)\left( 1-2i \right) \\\ & \Rightarrow 5\left( 1-2i \right)-3i\left( 1-2i \right) \\\ & \Rightarrow 5-10i-3i+3i\times 2i \\\ & \Rightarrow 5-13i+6{{\left( \sqrt{-1} \right)}^{2}} \\\ & \Rightarrow 5-13i-6 \\\ & \Rightarrow -1-13i \\\ \end{aligned}$$ The denominator is $$\left( 1+2i \right)\left( 1-2i \right)$$. Its expansion will be $$\begin{aligned} & \Rightarrow \left( 1+2i \right)\left( 1-2i \right) \\\ & \Rightarrow {{\left( 1 \right)}^{2}}-{{\left( 2i \right)}^{2}} \\\ & \Rightarrow 1-4(-1) \\\ & \Rightarrow 5 \\\ \end{aligned}$$ $$\Rightarrow \dfrac{5-3i}{1+2i}\times \dfrac{1-2i}{1-2i}$$ Using the expansions of the numerator and the denominator, we can simplify the expression as $$\Rightarrow \dfrac{5-3i}{1+2i}\times \dfrac{1-2i}{1-2i}$$ $$\begin{aligned} & \Rightarrow \dfrac{\left( 5-3i \right)\left( 1-2i \right)}{\left( 1+2i \right)\left( 1-2i \right)} \\\ & \Rightarrow \dfrac{-1-13i}{5} \\\ \end{aligned}$$ Separating the denominator, we get $$\Rightarrow -\dfrac{1}{5}-\dfrac{13}{5}i$$ This is the simplified form of the given expression. **Note:** To solve these types of problems, it is required to know the properties of the complex number. Some of the important properties that one should know are as follows, $$i=\sqrt{-1}\And {{i}^{2}}=-1$$. We should also know the expansion for $$\left( a+bi \right)\left( a-bi \right)$$, this is simplified as $${{a}^{2}}+{{b}^{2}}$$.