
How do you simplify $\dfrac{2+i}{7+5i}$ ?
Answer
548.4k+ views
Hint: In this question we have the fraction given to us in which both the numerator and the denominator is in the complex form. Simplifying in this question implies to remove the complex part from the denominator and to write the fraction in the general format of a complex number which is denoted as $a+bi$ , where $a$ and $b$ are real numbers.
Complete step by step solution:
We have the expression as $\dfrac{2+i}{7+5i}$
Now to simplify the expression, we will multiply both the numerator and denominator of the fraction by the complex conjugate of the denominator. The complex conjugate is a complex number which on multiplying with the original complex number, removes the imaginary part from the number.
The complex conjugate of a complex number $a+bi$ is $a-bi$ therefore, the complex conjugate of $7+5i$ is $7-5i$ which we will multiply with the numerator and the denominator.
On multiplying, we get:
$\Rightarrow \dfrac{2+i}{7+5i}\times \dfrac{7-5i}{7-5i}$
On multiplying the terms, we get:
$\Rightarrow \dfrac{2\times 7-2\times 5i+i\times 7-i\times 5i}{7\times 7-7\times 5i+5i\times 7-5i\times 5i}$
On simplifying the terms, we get:
$\Rightarrow \dfrac{14-10i+7i-5{{i}^{2}}}{49-30i+30i-5{{i}^{2}}}$
On simplifying the terms, we get:
\[\Rightarrow \dfrac{14-3i-5{{i}^{2}}}{49-5{{i}^{2}}}\]
Now we know that ${{i}^{2}}=-1$ , therefore on substituting it in the above expression, we get:
\[\Rightarrow \dfrac{14-3i-5(-1)}{49-5(-1)}\]
On simplifying the brackets, we get:
\[\Rightarrow \dfrac{14-3i+5}{49+5}\]
On adding the terms, we get:
\[\Rightarrow \dfrac{19-3i}{54}\]
On splitting the denominator, we get:
\[\Rightarrow \dfrac{19}{54}-\dfrac{3i}{54}\], which is the required solution in the form of $a+bi$.
Note: It is to be remembered that when a number is multiplied with both the numerator and denominator of the fraction, the value of the number does not decrease which is the reason we multiplied both the numerator and denominator of the expression with the complex conjugate of the denominator.
The value of $i=\sqrt{-1}$ , which should be remembered when doing questions on complex numbers.
Complete step by step solution:
We have the expression as $\dfrac{2+i}{7+5i}$
Now to simplify the expression, we will multiply both the numerator and denominator of the fraction by the complex conjugate of the denominator. The complex conjugate is a complex number which on multiplying with the original complex number, removes the imaginary part from the number.
The complex conjugate of a complex number $a+bi$ is $a-bi$ therefore, the complex conjugate of $7+5i$ is $7-5i$ which we will multiply with the numerator and the denominator.
On multiplying, we get:
$\Rightarrow \dfrac{2+i}{7+5i}\times \dfrac{7-5i}{7-5i}$
On multiplying the terms, we get:
$\Rightarrow \dfrac{2\times 7-2\times 5i+i\times 7-i\times 5i}{7\times 7-7\times 5i+5i\times 7-5i\times 5i}$
On simplifying the terms, we get:
$\Rightarrow \dfrac{14-10i+7i-5{{i}^{2}}}{49-30i+30i-5{{i}^{2}}}$
On simplifying the terms, we get:
\[\Rightarrow \dfrac{14-3i-5{{i}^{2}}}{49-5{{i}^{2}}}\]
Now we know that ${{i}^{2}}=-1$ , therefore on substituting it in the above expression, we get:
\[\Rightarrow \dfrac{14-3i-5(-1)}{49-5(-1)}\]
On simplifying the brackets, we get:
\[\Rightarrow \dfrac{14-3i+5}{49+5}\]
On adding the terms, we get:
\[\Rightarrow \dfrac{19-3i}{54}\]
On splitting the denominator, we get:
\[\Rightarrow \dfrac{19}{54}-\dfrac{3i}{54}\], which is the required solution in the form of $a+bi$.
Note: It is to be remembered that when a number is multiplied with both the numerator and denominator of the fraction, the value of the number does not decrease which is the reason we multiplied both the numerator and denominator of the expression with the complex conjugate of the denominator.
The value of $i=\sqrt{-1}$ , which should be remembered when doing questions on complex numbers.
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