Question

How do you divide $\dfrac{{4 - 2i}}{{3 + 7i}}$ ?

Answer 1

Hint: In order to solve this sum, we multiply both the numerator and denominator with the conjugate term of the denominator. After that we express the denominator in ${a^2} - {b^2}$ form, and expand the numerator. We replace the imaginary number ${i^2}$ with $ - 1$ , and simplify further to get our required answer.

Complete step by step solution:
In the given question, we need to divide $\dfrac{{4 - 2i}}{{3 + 7i}}$, $i$ here represents an imaginary number: $i = \sqrt { - 1} $
A complex number is represented in the form of ‘\[a + bi\]’, where ‘a’ is the real number and ‘bi’ is the imaginary number.
In order to divide the given number, we take the conjugate of the denominator. Conjugate means to take a similar complex number but opposite sign before the imaginary number.
Thus, conjugate of the denominator: $3 - 7i$
Let us multiply both the denominator and numerator with the conjugate number:
$ \Rightarrow \dfrac{{\left( {4 - 2i} \right)\left( {3 - 7i} \right)}}{{\left( {3 + 7i} \right)\left( {3 - 7i} \right)}}$
Now we know that $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$
Thus, we have:
$ \Rightarrow \dfrac{{\left( {4 - 2i} \right)\left( {3 - 7i} \right)}}{{\left( {{3^2} - {{\left( {7i} \right)}^2}} \right)}}$
Simplifying the numerator and the denominator further, we get:
$ \Rightarrow \dfrac{{12 - 28i - 6i + 14{i^2}}}{{9 - 49{i^2}}}$
Adding the like terms:
$ \Rightarrow \dfrac{{12 - 34i + 14{i^2}}}{{9 - 49{i^2}}}$
Now we know that $i = \sqrt { - 1} $
Therefore, ${i^2} = - 1$
Placing this value in our required expression, we get:
$ \Rightarrow \dfrac{{12 - 34i - 14}}{{9 + 49}}$
On simplifying further, we get:
$ \Rightarrow \dfrac{{ - 2 - 34i}}{{58}}$
We can also write the above term as:
$ \Rightarrow - \dfrac{2}{{58}} - \dfrac{{34}}{{58}}i$
On reducing the numbers to the simplest form, we get:
$ \Rightarrow - \dfrac{1}{{29}} - \dfrac{{17}}{{29}}i$
Thus, we have our required answer.

Note: A complex number is a number that can be represented as $a + bi$, where a and b are real numbers, and $i$ represents the imaginary unit, satisfying the equation $i = - 1$. Because no real number satisfies the equation, therefore $i$ is called an imaginary number. Some properties of complex numbers are:
When a, b, c and d are real numbers and \[a + b = c + d\], then \[a = c\] and \[b = d\]
The sum of two conjugate complex numbers is real. For example, if we have a number as $z = a + ib$ , where $a$ and $b$ are real numbers, and the conjugate number $\overline z = a - ib$ , then the sum of $z + \overline z $ is a real number.
The product of two conjugate complex numbers is real.