Question

How do you write \[\dfrac{{ - 2\iota }}{{4 - 2\iota }}\] in the “\[a + b\iota \]” form?

Answer 1

Hint: In the given question, we have been given a fraction with a complex number in its denominator. We have to simplify the value of the fraction to the standard value of a complex number. We are going to do that by rationalizing the denominator. It is done by applying some operations and bringing the denominator to the numerator.

Formula Used:
We are going to use the formula of difference of two squares:
\[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\].

Complete step by step answer:
The given expression is:
\[\dfrac{{ - 2\iota }}{{4 - 2\iota }}\]
To solve this, we are first going to rationalize the denominator,
\[\dfrac{{ - 2\iota }}{{4 - 2\iota }} = \dfrac{{ - 2\iota }}{{4 - 2\iota }} \times \dfrac{{4 + 2\iota }}{{4 + 2\iota }} = \dfrac{{ - 8\iota - 4{\iota ^2}}}{{\left( {4 - 2\iota } \right)\left( {4 + 2\iota } \right)}}\]
Know, we can apply the formula of difference of two squares on the denominator and we know \[{\iota ^2} = - 1\],
\[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
\[\dfrac{{ - 8\iota - 4{\iota ^2}}}{{\left( {4 - 2\iota } \right)\left( {4 + 2\iota } \right)}} = \dfrac{{ - 8\iota - 4\left( { - 1} \right)}}{{{4^2} - {{\left( {2\iota } \right)}^2}}} = \dfrac{{4 - 8\iota }}{{16 - 4{\iota ^2}}} = \dfrac{{4 - 8\iota }}{{16 + 4}} = \dfrac{{4 - 8\iota }}{{20}} = \dfrac{{1 - 2\iota }}{5}\]
Hence, \[\dfrac{{ - 2\iota }}{{4 - 2\iota }}\] in the “\[a + b\iota \]” form is \[\dfrac{{1 - 2\iota }}{5}\].

Thus, \[a = \dfrac{1}{5}\] and \[b = \dfrac{{ - 2}}{5}\]

Additional Information:
The “\[\iota \]” symbol multiplied with the constant is called the complex number. It has a value of \[\sqrt { - 1} \]. It is the imaginary part of the equation, as we know a negative number cannot be square rooted. There are a few properties of the number:
\[{\iota ^2} = - 1\]
\[{\iota ^3} = - \iota \]
\[{\iota ^4} = 1\]

Note:
In the given question, we had to simplify a fraction to be written into the form of a standard complex number with their real and complex parts separated. To do that, we first rationalize the denominator – bringing the complex number from the denominator to the numerator. Then we simplified the expression and solved for the answer.