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

Last updated date: 21st Jun 2024
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Hint: We can see $i$ .We see here everything in the form of $a+ib$ where $a,b$ are real numbers. And this is related to complex numbers. The basic thing which we know about complex numbers is the value of $i$. The value of $i=\sqrt{-1}$ . In radical expressions , we basically say that inside the root , we cannot have a negative value . But $i$ let us use it and we have some set of rules that we should abide by while solving complex numbers. For now , we just stick to rationalizing the denominator.

Complete step-by-step solution:
Even when we solve radical expressions or surds , we generally try to rationalize the denominator whenever we find roots in it.
For example, if we have something of the form $\dfrac{1}{\sqrt{a}+\sqrt{b}}$ , then the first thing that strikes our mind is rationalizing the denominator. And it goes like this :
$\Rightarrow \dfrac{1}{\sqrt{a}+\sqrt{b}}\times \dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{a}-\sqrt{b}}$
We know that $\left( x-y \right)\left( x+y \right)={{x}^{2}}-{{y}^{2}}$
$\Rightarrow \dfrac{\sqrt{a}-\sqrt{b}}{a-b}$
Now , we apply the same concept here .
But the most important thing to remember here is that ${{i}^{2}}=-1$ .
Upon rationalizing $\dfrac{2-3i}{5+i}$ , i.e we multiply the numerator and denominator with $5-i$ ,we get the following :
\begin{align} & \Rightarrow \dfrac{2-3i}{5+i}\times \dfrac{5-i}{5-i} \\ & \Rightarrow \dfrac{10-2i-15i-3}{25+1} \\ & \Rightarrow \dfrac{7-17i}{26} \\ \end{align}
Here $5-i$ is called the conjugate of $5+i$ .
$\therefore$ Hence this is how we simplify $\dfrac{2-3i}{5+i}$ and on simplifying it , we get , $\dfrac{7-17i}{26}$.

Note: We should know to rationalize the denominator and we should also know what value should be used to rationalize in order to get the required answer. We should notice that value of ${{i}^{2}}=-1$ . We should keep this mind and solve as this may lead to errors in signs.