Question

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

Answer 1

Hint: To simplify the given complex number $\dfrac{5}{2-i}$, we are going to multiply and divide by $2+i$ and then simplify the numerator and denominator of the above expression. The power of (iota) “i” that you require in solving the above problem is ${{i}^{2}}=-1$.

Complete step by step answer:
The complex number given in the above problem that we are asked to simplify is as follows:
$\dfrac{5}{2-i}$
Now, to simplify the above complex number, we are going to rationalize the above number. For rationalization, we are going to multiply and divide by $\left( 2+i \right)$.
$\implies \dfrac{5}{2-i}\times \dfrac{2+i}{2+i}$
In the denominator, we are multiplying $\left( 2+i \right)$ by $\left( 2-i \right)$ and also in the numerator, we are multiplying $\left( 2+i \right)$ by 5 we get,
$\implies \dfrac{5\left( 2+i \right)}{\left( 2-i \right)\left( 2+i \right)}$
In the denominator of the above complex number, we are going to use the identity
i.e.: $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$
$\begin{align}
  & \dfrac{10+5i}{{{\left( 2 \right)}^{2}}-{{i}^{2}}} \\
 & =\dfrac{10+5i}{4-{{i}^{2}}} \\
\end{align}$
In the above, we are going to use the value of ${{i}^{2}}=-1$ in the above equation and we get,
$\begin{align}
  & \dfrac{10+5i}{4-\left( -1 \right)} \\
 & =\dfrac{10+5i}{4+1} \\
 & =\dfrac{10+5i}{5} \\
\end{align}$
Dividing 5 by 10 we get 2 and dividing $5i$ by 5 we get $i$.
\[2+i\]
Hence, we have simplified the given complex number to $2+i$.

Note: The principle of rationalization is that we are going to multiply and divide the conjugate of the denominator and then simplify. The conjugate of $a+ib$ is $a-ib$. Also, whenever you see any complex number in the denominator either with positive or negative sign then rationalization is the method to simplify the complex number.
Also, the possible mistake that could happen in the above problem is that you might write the value of ${{i}^{2}}$ as 1 instead of -1 so make sure you won’t make this mistake in the examination.