Question

Find the smallest positive integer value of n for which \[\dfrac{{{{\left( {1 + i} \right)}^n}}}{{{{\left( {1 - i} \right)}^{n - 2}}}}\] is a real number.

Answer 1

Hint: Complex numbers are the numbers which are expressed in the form of \[a + ib\] where ‘i’ is an imaginary number called iota and has the value of \[\left( {\surd - 1} \right).\] For example, \[2 + 3i\] is a complex number, where \[2\] is a real number and \[3i\] is an imaginary number. Therefore, the combination of both the real number and imaginary number is a complex number.
The main application of these numbers is to represent periodic motions such as water waves, alternating current, light waves, etc., which relies on sine or cosine waves etc. There are certain formulas which are used to solve the problems based on complex numbers.
Rationalization is the property, when denominator and numerator is multiplied by opposite sign of denominator with same power such that the rationalized term gets cancelled.
\[{\left( {1 - i} \right)^{n - 2}}\] is rationalized with \[{\left( {1 + i} \right)^{n - 2}}\] in numerator and denominator.

Complete step-by-step answer:
\[ \Rightarrow \dfrac{{{{\left( {1 + i} \right)}^n}}}{{{{\left( {1 - i} \right)}^{n - 2}}}}\]
Rationalizing
\[ \Rightarrow \dfrac{{{{\left( {1 + i} \right)}^n}}}{{{{\left( {1 - i} \right)}^{n - 2}}}} \times \dfrac{{{{\left( {1 + i} \right)}^{n - 2}}}}{{{{\left( {1 + i} \right)}^{n - 2}}}}\] [After rationalization]
\[ \Rightarrow \dfrac{{{{\left( {1 + i} \right)}^{n + n - 2}}}}{{{{\left[ {\left( {1 + i} \right)\left( {1 - i} \right)} \right]}^{n - 2}}}}\]
Using concept of $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$
We get,
\[ \Rightarrow \dfrac{{{{\left( {1 + i} \right)}^{2n - 2}}}}{{{{\left( {1 - {i^2}} \right)}^{n - 2}}}}\]
\[ \Rightarrow \dfrac{{{{\left( {1 + i} \right)}^{2\left( {n - 1} \right)}}}}{{{{\left( {1 - \left( { - 1} \right)} \right)}^{n - 2}}}}\], As we know \[\left[ \begin{gathered}
  i = \sqrt { - 1} ,\,\,{i^2} = - 1 \\
  {i^3} = - i,\,\,\,{i^4}\, = \,\,1 \\
\end{gathered} \right]\]
\[ \Rightarrow \dfrac{{{{\left( {1 + i} \right)}^{2\left( {n - 1} \right)}}}}{{{{\left( 2 \right)}^{n - 2}}}}\]
Simplify
\[ \Rightarrow \dfrac{{{{\left( {{{\left( {1 + i} \right)}^2}} \right)}^{\left( {n - 1} \right)}}}}{{{{\left( 2 \right)}^{n - 2}}}}\]
\[ \Rightarrow \dfrac{{{{\left( {1 + {i^2} + 2i} \right)}^{n - 1}}}}{{{2^{n - 2}}}}\]
Solve square of whole term
\[ \Rightarrow \dfrac{{{{\left( {1 + \left( { - 1} \right) + 2i} \right)}^{n - 1}}}}{{{2^{n - 2}}}}\]
\[ \Rightarrow \dfrac{{{{\left( {2i} \right)}^{n - 1}}}}{{{2^{n - 2}}}}\]
\[ \Rightarrow \dfrac{{{{\left( 2 \right)}^{n - 1}}{i^{n - 1}}}}{{{2^{n - 2}}}}\]
Solve
\[ \Rightarrow {2^{n + ( - n) + 2}}{i^{n - 1}}\]
\[ \Rightarrow {2^1}{i^{n - 1}}\]
To make identity real
\[n - 1 = 0\]
\[n = 1\]

So, the smallest positive integer for n is \[1\].

Note: In cases where you have a fraction with a radical in the denominator, you can use a technique called rationalizing a denominator to eliminate the radical. The point of rationalizing the denominator is to make it easier to understand what the quantity really is, by removing radicals from the denominator.