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# How do you simplify $\dfrac{10+i}{4-i}$ ?

Last updated date: 26th Feb 2024
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Hint: These types of problems are pretty straight forward and are very easy to solve. We need to have a fair knowledge of complex numbers and the different equations and formulae that involves it. The general form of a complex number is $a+ib$ , where the first part is the real term and the second term is the complex part. Here $i$ represents iorta and is defined as,
\begin{align} & i=\sqrt{-1} \\ & \Rightarrow {{i}^{2}}=-1 \\ \end{align}
The easiest way to solve the given problem is to do rationalization of the denominator and then to separate the real and imaginary terms.

\begin{align} & \dfrac{10+i}{4-i} \\ & \Rightarrow \dfrac{\left( 10+i \right)\left( 4+i \right)}{\left( 4-i \right)\left( 4+i \right)} \\ \end{align}
$\Rightarrow \dfrac{40+10i+4i+{{i}^{2}}}{16-{{i}^{2}}}$
Now, using the relation ${{i}^{2}}=-1$ we write,
\begin{align} & \Rightarrow \dfrac{40+10i+4i+\left( -1 \right)}{16-\left( -1 \right)} \\ & \Rightarrow \dfrac{40+10i+4i-1}{16+1} \\ & \Rightarrow \dfrac{39+14i}{17} \\ \end{align}
$\dfrac{39}{17}+\dfrac{14}{17}i$
Thus the answer to our problem is $\dfrac{39}{17}+\dfrac{14}{17}i$.
Note: For these type of problems, we need to remember and keep in mind of the general form of complex numbers. The given problem is solved by a simple rationalization of the denominator, followed by simple multiplication of two complex numbers and replacing the relation ${{i}^{2}}=-1$ . In rationalization, what we do is multiply both the numerator and denominator by the conjugate of the denominator. After all these things we separate the real part and imaginary part and write the hence formed answer.