Solveeit Logo
Login

Question

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

How do you write −2ι4−2ι\dfrac{{ - 2\iota }}{{4 - 2\iota }} in the “a+bιa + b\iota ” form?

Explanation

Solution

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:
a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right).

Complete step by step answer:
The given expression is:
−2ι4−2ι\dfrac{{ - 2\iota }}{{4 - 2\iota }}
To solve this, we are first going to rationalize the denominator,
−2ι4−2ι=−2ι4−2ι×4+2ι4+2ι=−8ι−4ι2(4−2ι)(4+2ι)\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 ι2=−1{\iota ^2} = - 1,
a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)
−8ι−4ι2(4−2ι)(4+2ι)=−8ι−4(−1)42−(2ι)2=4−8ι16−4ι2=4−8ι16+4=4−8ι20=1−2ι5\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, −2ι4−2ι\dfrac{{ - 2\iota }}{{4 - 2\iota }} in the “a+bιa + b\iota ” form is 1−2ι5\dfrac{{1 - 2\iota }}{5}.

Thus, a=15a = \dfrac{1}{5} and b=−25b = \dfrac{{ - 2}}{5}

Additional Information:
The “ι\iota ” symbol multiplied with the constant is called the complex number. It has a value of −1\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:
ι2=−1{\iota ^2} = - 1
ι3=−ι{\iota ^3} = - \iota
ι4=1{\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.