
Re \[\dfrac{{{{(1 - i)}^2}}}{{3 - i}}\] is equal to ?
A. \[ - \dfrac{1}{5}\]
B. \[\dfrac{1}{5}\]
C. \[\dfrac{1}{{10}}\]
D. \[ - \dfrac{1}{{10}}\]
Answer
509.4k+ views
Hint: A complex number is of the form a +ib where a and b are real numbers and\[i\]is an imaginary number. Here we only need to find the real part of this given complex number. We will also use the formula\[(a - b)(a + b) = {a^2} - {b^2}\]and then we will multiply by\[3 + i\]. After that we will also use the formula\[{(a - b)^2} = {a^2} + {b^2} - 2ab\]. Also, we know that\[i = \sqrt { - 1} \] and so the square of the imaginary number is\[{i^2} = - 1\].
Complete step by step answer:
Given data is \[\dfrac{{{{(1 - i)}^2}}}{{3 - i}}\].
We will use the formula \[{(a - b)^2} = {a^2} + {b^2} - 2ab\], we will get,
\[ \dfrac{{{{(1 - i)}^2}}}{{3 - i}}= \dfrac{{{1^2} + {{(i)}^2} - 2(1)(i)}}{{3 - i}}\]
We also know that\[{i^2} = - 1\].
Substituting the value and removing the brackets, we get,
\[\dfrac{{{{(1 - i)}^2}}}{{3 - i}} = \dfrac{{1 - 1 - 2i}}{{3 - i}}\]
Simply the given complex number, we will get,
\[ \dfrac{{{{(1 - i)}^2}}}{{3 - i}}= \dfrac{{ - 2i}}{{3 - i}}\]
Multiply by\[3 + i\]in both numerator and denominator, we will get,
\[\dfrac{{{{(1 - i)}^2}}}{{3 - i}} = \dfrac{{ - 2i}}{{3 - i}} \times \dfrac{{3 + i}}{{3 + i}}\]
\[\Rightarrow \dfrac{{{{(1 - i)}^2}}}{{3 - i}}= \dfrac{{ - 2i(3 + i)}}{{(3 - i)(3 + i)}}\]
We will use the formula\[(a - b)(a + b) = {a^2} - {b^2}\], we will get,
\[\dfrac{{{{(1 - i)}^2}}}{{3 - i}} = \dfrac{{ - 6i - 2{i^2}}}{{{3^2} - {i^2}}}\]
Substituting the value\[{i^2} = - 1\], we get,
\[ \dfrac{{{{(1 - i)}^2}}}{{3 - i}}= \dfrac{{ - 6i - 2( - 1)}}{{{3^2} - ( - 1)}}\]
Removing the brackets, we get,
\[\dfrac{{{{(1 - i)}^2}}}{{3 - i}} = \dfrac{{ - 6i + 2}}{{9 + 1}}\]
Taking\[2\]common in the numerator, we get,
\[\dfrac{{{{(1 - i)}^2}}}{{3 - i}} = \dfrac{{2( - 3i + 1)}}{{10}}\]
Simplify the given complex number, we will get,
\[\dfrac{{{{(1 - i)}^2}}}{{3 - i}} = \dfrac{{ - 3i + 1}}{5}\]
Rearranging the above expression, we get,
\[ \dfrac{{{{(1 - i)}^2}}}{{3 - i}}= \dfrac{{1 - 3i}}{5}\]
\[\therefore \dfrac{{{{(1 - i)}^2}}}{{3 - i}}= \dfrac{1}{5} - \dfrac{{3i}}{5}\]
Hence, the real part is \[\dfrac{1}{5}\]. Also, the imaginary part is \[ - \dfrac{3}{5}i\].
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 \[Re\left( z \right) = 0\] and purely real means\[Im\left( z \right) = 0\]. 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 be\[0\], so all Real Numbers and Imaginary Numbers are also Complex Numbers.
Complete step by step answer:
Given data is \[\dfrac{{{{(1 - i)}^2}}}{{3 - i}}\].
We will use the formula \[{(a - b)^2} = {a^2} + {b^2} - 2ab\], we will get,
\[ \dfrac{{{{(1 - i)}^2}}}{{3 - i}}= \dfrac{{{1^2} + {{(i)}^2} - 2(1)(i)}}{{3 - i}}\]
We also know that\[{i^2} = - 1\].
Substituting the value and removing the brackets, we get,
\[\dfrac{{{{(1 - i)}^2}}}{{3 - i}} = \dfrac{{1 - 1 - 2i}}{{3 - i}}\]
Simply the given complex number, we will get,
\[ \dfrac{{{{(1 - i)}^2}}}{{3 - i}}= \dfrac{{ - 2i}}{{3 - i}}\]
Multiply by\[3 + i\]in both numerator and denominator, we will get,
\[\dfrac{{{{(1 - i)}^2}}}{{3 - i}} = \dfrac{{ - 2i}}{{3 - i}} \times \dfrac{{3 + i}}{{3 + i}}\]
\[\Rightarrow \dfrac{{{{(1 - i)}^2}}}{{3 - i}}= \dfrac{{ - 2i(3 + i)}}{{(3 - i)(3 + i)}}\]
We will use the formula\[(a - b)(a + b) = {a^2} - {b^2}\], we will get,
\[\dfrac{{{{(1 - i)}^2}}}{{3 - i}} = \dfrac{{ - 6i - 2{i^2}}}{{{3^2} - {i^2}}}\]
Substituting the value\[{i^2} = - 1\], we get,
\[ \dfrac{{{{(1 - i)}^2}}}{{3 - i}}= \dfrac{{ - 6i - 2( - 1)}}{{{3^2} - ( - 1)}}\]
Removing the brackets, we get,
\[\dfrac{{{{(1 - i)}^2}}}{{3 - i}} = \dfrac{{ - 6i + 2}}{{9 + 1}}\]
Taking\[2\]common in the numerator, we get,
\[\dfrac{{{{(1 - i)}^2}}}{{3 - i}} = \dfrac{{2( - 3i + 1)}}{{10}}\]
Simplify the given complex number, we will get,
\[\dfrac{{{{(1 - i)}^2}}}{{3 - i}} = \dfrac{{ - 3i + 1}}{5}\]
Rearranging the above expression, we get,
\[ \dfrac{{{{(1 - i)}^2}}}{{3 - i}}= \dfrac{{1 - 3i}}{5}\]
\[\therefore \dfrac{{{{(1 - i)}^2}}}{{3 - i}}= \dfrac{1}{5} - \dfrac{{3i}}{5}\]
Hence, the real part is \[\dfrac{1}{5}\]. Also, the imaginary part is \[ - \dfrac{3}{5}i\].
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 \[Re\left( z \right) = 0\] and purely real means\[Im\left( z \right) = 0\]. 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 be\[0\], so all Real Numbers and Imaginary Numbers are also Complex Numbers.
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