Question

The value of the complex number \[\dfrac{{1 + 7i}}{{{{\left( {2 - i} \right)}^2}}}\] is equal to
A. \[\sqrt 2 \left( {\cos \dfrac{{3\pi }}{4} + i\sin \dfrac{{3\pi }}{4}} \right)\]
B. \[\sqrt 2 \left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)\]
C. \[\cos \dfrac{{3\pi }}{4} + i\sin \dfrac{{3\pi }}{4}\]
D. None of these

Answer 1

Hint: In this question, we have given a complex number and we need to find the value of this complex number. For that, we first apply an algebraic identity in the denominator and simplify it then rationalize the number and after simplifying it we will get the desired result.

Formula used:
We have been using the following formulas:
1. \[{\left( {a - b} \right)^2} = \left( {{a^2} - 2ab + {b^2}} \right)\]
2. \[{\left( i \right)^2} = - 1\]
3. \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]

Complete step-by-step solution:
Given complex number is \[\dfrac{{1 + 7i}}{{{{\left( {2 - i} \right)}^2}}}\]
Now we apply algebraic identity in the denominator \[{\left( {a - b} \right)^2} = \left( {{a^2} - 2ab + {b^2}} \right)\], we have
\[\dfrac{{1 + 7i}}{{{{\left( {2 - i} \right)}^2}}} = \dfrac{{1 + 7i}}{{4 + {i^2} - 4i}}\]
Now we know that \[{\left( i \right)^2} = - 1\], From the above equation we get
\[
  \dfrac{{1 + 7i}}{{{{\left( {2 - i} \right)}^2}}} = \dfrac{{1 + 7i}}{{4 - 1 - 4i}} \\
   = \dfrac{{1 + 7i}}{{3 - 4i}} \\
 \]
Now by rationalizing the resultant number, we obtain
\[
  \dfrac{{1 + 7i}}{{3 - 4i}} = \dfrac{{1 + 7i}}{{3 - 4i}} \times \dfrac{{3 + 4i}}{{3 + 4i}} \\
   = \dfrac{{\left( {1 + 7i} \right)\left( {3 + 4i} \right)}}{{\left( {3 - 4i} \right)\left( {3 + 4i} \right)}} \\
 \]
Now by applying algebraic identity \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\], we obtain
\[
  \dfrac{{1 + 7i}}{{3 - 4i}} = \dfrac{{\left( {1 + 7i} \right)\left( {3 + 4i} \right)}}{{{3^2} - {{\left( {4i} \right)}^2}}} \\
   = \dfrac{{\left( {1 + 7i} \right)\left( {3 + 4i} \right)}}{{9 - 16{i^2}}} \\
 \]
Now by simplifying and substitute \[{\left( i \right)^2} = - 1\], we get
\[
  \dfrac{{1 + 7i}}{{3 - 4i}} = \dfrac{{1\left( {3 + 4i} \right) + 7i\left( {3 + 4i} \right)}}{{9 + 16}} \\
   = \dfrac{{3 + 4i + 21i + 28{{(i)}^2}}}{{25}} \\
   = \dfrac{{3 + 25i - 28}}{{25}}\left( {\because {{\left( i \right)}^2} = - 1} \right) \\
   = \dfrac{{25i - 25}}{{25}} \\
 \]
Further solving, we get
\[
  \dfrac{{1 + 7i}}{{3 - 4i}} = \dfrac{{25\left( {i - 1} \right)}}{{25}} \\
   = i - 1 \\
 \]
Hence, option(D) is correct answer

Note: Students learn that there are two types of numbers: complex numbers and real numbers. A real number is one that can be shown on a number line, whereas a complex number is one that cannot be shown on a number line. We cannot find the square root of negative numbers, so we assumed \[\sqrt { - 1} \] to be iota \[i\] which represents the complex number \[a + ib\]. Converting complex numbers into polar form is used for graph paper expression, so we can convert any complex number into the polar form using a similar approach.