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Question: Re \[\dfrac{{{{(1 - i)}^2}}}{{3 - i}}\] is equal to ? A. \[ - \dfrac{1}{5}\] B. \[\dfrac{1}{5}\...

Re (1−i)23−i\dfrac{{{{(1 - i)}^2}}}{{3 - i}} is equal to ?
A. −15 - \dfrac{1}{5}
B. 15\dfrac{1}{5}
C. 110\dfrac{1}{{10}}
D. −110 - \dfrac{1}{{10}}

Explanation

Solution

A complex number is of the form a +ib where a and b are real numbers andiiis an imaginary number. Here we only need to find the real part of this given complex number. We will also use the formula(a−b)(a+b)=a2−b2(a - b)(a + b) = {a^2} - {b^2}and then we will multiply by3+i3 + i. After that we will also use the formula(a−b)2=a2+b2−2ab{(a - b)^2} = {a^2} + {b^2} - 2ab. Also, we know thati=−1i = \sqrt { - 1} and so the square of the imaginary number isi2=−1{i^2} = - 1.

Complete step by step answer:
Given data is (1−i)23−i\dfrac{{{{(1 - i)}^2}}}{{3 - i}}.
We will use the formula (a−b)2=a2+b2−2ab{(a - b)^2} = {a^2} + {b^2} - 2ab, we will get,
(1−i)23−i=12+(i)2−2(1)(i)3−i\dfrac{{{{(1 - i)}^2}}}{{3 - i}}= \dfrac{{{1^2} + {{(i)}^2} - 2(1)(i)}}{{3 - i}}
We also know thati2=−1{i^2} = - 1.
Substituting the value and removing the brackets, we get,
(1−i)23−i=1−1−2i3−i\dfrac{{{{(1 - i)}^2}}}{{3 - i}} = \dfrac{{1 - 1 - 2i}}{{3 - i}}
Simply the given complex number, we will get,
(1−i)23−i=−2i3−i\dfrac{{{{(1 - i)}^2}}}{{3 - i}}= \dfrac{{ - 2i}}{{3 - i}}
Multiply by3+i3 + iin both numerator and denominator, we will get,
(1−i)23−i=−2i3−i×3+i3+i\dfrac{{{{(1 - i)}^2}}}{{3 - i}} = \dfrac{{ - 2i}}{{3 - i}} \times \dfrac{{3 + i}}{{3 + i}}
⇒(1−i)23−i=−2i(3+i)(3−i)(3+i)\Rightarrow \dfrac{{{{(1 - i)}^2}}}{{3 - i}}= \dfrac{{ - 2i(3 + i)}}{{(3 - i)(3 + i)}}
We will use the formula(a−b)(a+b)=a2−b2(a - b)(a + b) = {a^2} - {b^2}, we will get,
(1−i)23−i=−6i−2i232−i2\dfrac{{{{(1 - i)}^2}}}{{3 - i}} = \dfrac{{ - 6i - 2{i^2}}}{{{3^2} - {i^2}}}

Substituting the valuei2=−1{i^2} = - 1, we get,
(1−i)23−i=−6i−2(−1)32−(−1)\dfrac{{{{(1 - i)}^2}}}{{3 - i}}= \dfrac{{ - 6i - 2( - 1)}}{{{3^2} - ( - 1)}}
Removing the brackets, we get,
(1−i)23−i=−6i+29+1\dfrac{{{{(1 - i)}^2}}}{{3 - i}} = \dfrac{{ - 6i + 2}}{{9 + 1}}
Taking22common in the numerator, we get,
(1−i)23−i=2(−3i+1)10\dfrac{{{{(1 - i)}^2}}}{{3 - i}} = \dfrac{{2( - 3i + 1)}}{{10}}
Simplify the given complex number, we will get,
(1−i)23−i=−3i+15\dfrac{{{{(1 - i)}^2}}}{{3 - i}} = \dfrac{{ - 3i + 1}}{5}
Rearranging the above expression, we get,
(1−i)23−i=1−3i5\dfrac{{{{(1 - i)}^2}}}{{3 - i}}= \dfrac{{1 - 3i}}{5}
∴(1−i)23−i=15−3i5\therefore \dfrac{{{{(1 - i)}^2}}}{{3 - i}}= \dfrac{1}{5} - \dfrac{{3i}}{5}

Hence, the real part is 15\dfrac{1}{5}. Also, the imaginary part is −35i - \dfrac{3}{5}i.

Note: A complex number is a sum of real numbers and imaginary numbers. It is represented by ‘z’. A complex number is said to be purely imaginary means Re(z)=0Re\left( z \right) = 0 and purely real meansIm(z)=0Im\left( z \right) = 0. The complex number is of the form a+ib, where real number a is called the real part and the real number b is called the imaginary part (also called “iota”). But either part can be00, so all Real Numbers and Imaginary Numbers are also Complex Numbers.