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Question: How do you simplify \(\dfrac{{3 - 2i}}{{1 - i}}\)?...

How do you simplify 3−2i1−i\dfrac{{3 - 2i}}{{1 - i}}?

Explanation

Solution

As we are given an expression which contains the imaginary number iota i.e., ii, we can say that this whole expression is an imaginary number. In order to simplify the expression, we will multiply the numerator and denominator with the conjugate of the denominator to keep the expression balanced. Multiplying the conjugate with the denominator will give us a real number in the denominator and then we will simplify it further.

Complete step by step solution:
(i)
As we are given:
3−2i1−i\dfrac{{3 - 2i}}{{1 - i}}

In order to simplify the given expression, we will multiply the numerator and the denominator of the expression with the conjugate of the denominator (1−i)(1 - i) i.e., (1+i)(1 + i)

Therefore, we will get:
3−2i1−i(1+i1+i)\dfrac{{3 - 2i}}{{1 - i}}\left( {\dfrac{{1 + i}}{{1 + i}}} \right)

Now, we will multiply the numerators among themselves and denominators among themselves.
(3−2i)(1+i)(1−i)(1+i)\dfrac{{(3 - 2i)(1 + i)}}{{(1 - i)(1 + i)}}

(ii)
As we can see, we have (1−i)(1+i)(1 - i)(1 + i) in the denominator. To simplify this term, we can apply the
identity:
(a+b)(a−b)=a2−b2(a + b)(a - b) = {a^2} - {b^2}

Therefore, applying the above-mentioned identity in the denominator we get:
(3−2i)(1+i)12−i2\dfrac{{(3 - 2i)(1 + i)}}{{{1^2} - {i^2}}}

(iii)
As we know that the value of iota is −1\sqrt { - 1} i.e.,
i=−1i = \sqrt { - 1}

Squaring both the sides, we will get:
i2=−1{i^2} = - 1

Therefore, we can substitute i2{i^2} as −1 - 1 in the expression, and we will get:
(3−2i)(1+i)12−(−1)\dfrac{{(3 - 2i)(1 + i)}}{{{1^2} - ( - 1)}}

On simplifying, it will become:
(3−2i)(1+i)1+1\dfrac{{(3 - 2i)(1 + i)}}{{1 + 1}}

On further simplifying,
(3−2i)(1+i)2\dfrac{{(3 - 2i)(1 + i)}}{2}

(iv)
Now, multiplying the terms in numerator by opening the parentheses we will get:
3(1+i)−2i(1+i)2\dfrac{{3(1 + i) - 2i(1 + i)}}{2}

On multiplying, we will get:
3+3i−2i−2i22\dfrac{{3 + 3i - 2i - 2{i^2}}}{2}

Adding the imaginary terms and substituting i2{i^2} as −1 - 1 in the expression, and we will get:
3+i−2(−1)2 3+i+22 5+i2  \dfrac{{3 + i - 2( - 1)}}{2} \\\ \dfrac{{3 + i + 2}}{2} \\\ \dfrac{{5 + i}}{2} \\\

Separating the terms and writing the expression in the form of a+bia + bi, we will get:
52+12i\dfrac{5}{2} + \dfrac{1}{2}i

Hence, the simplification of (3−2i1−i)\left( {\dfrac{{3 - 2i}}{{1 - i}}} \right) is (52+12i)\left( {\dfrac{5}{2} + \dfrac{1}{2}i} \right).

Note: Always remember when the simplification of an imaginary expression is asked which is in fraction, it is suitable to multiply the numerator and the denominator with the conjugate of the denominator as it rationalizes the denominator and after simplification try to leave the term in the standard form of complex numbers i.e., a+bia + bi.