Question

The smallest positive integer \[n\] for which \[{\left( {1 + i} \right)^{2n}} = {\left( {1 - i} \right)^{2n}}\] is ?

Answer 1

Hint: Here in this question, we have to calculate the smallest positive integer \[n\] in such a way that satisfies the given complex equation. To find the possible smallest positive integer \[n\] by multiplying and dividing the complex equation by its conjugate and further simplifying using algebraic identities to get the required solution.

Complete answer:
A positive number, also known as a positive integer, is a number that is bigger than zero. The smallest positive integer is 1.
Consider the given complex equation
\[{\left( {1 + i} \right)^{2n}} = {\left( {1 - i} \right)^{2n}}\]
Divide both side by \[{\left( {1 - i} \right)^{2n}}\], then we have
\[ \Rightarrow \,\,\,\dfrac{{{{\left( {1 + i} \right)}^{2n}}}}{{{{\left( {1 - i} \right)}^{2n}}}} = 1\] ------(1)
We know the power of a quotient property of exponent is \[\dfrac{{{a^m}}}{{{b^m}}} = {\left( {\dfrac{a}{b}} \right)^m}\], then equation (1) becomes
\[ \Rightarrow \,\,\,{\left( {\dfrac{{1 + i}}{{1 - i}}} \right)^{2n}} = 1\] ------(2)
The conjugate of the denominator of complex equations \[\dfrac{{1 + i}}{{1 - i}}\] is \[1 + i\].
Multiply and divide the complex equation by \[\left( {1 + i} \right)\], then equation (2) becomes
\[ \Rightarrow \,\,\,{\left( {\dfrac{{1 + i}}{{1 - i}} \times \dfrac{{1 + i}}{{1 + i}}} \right)^{2n}} = 1\]
\[ \Rightarrow \,\,\,{\left( {\dfrac{{{{\left( {1 + i} \right)}^2}}}{{\left( {1 - i} \right)\left( {1 + i} \right)}}} \right)^{2n}} = 1\]
Apply the algebraic identities \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\] in numerator and \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\] in denominator, then
\[ \Rightarrow \,\,\,{\left( {\dfrac{{{1^2} + {i^2} + 2i}}{{{1^2} - {i^2}}}} \right)^{2n}} = 1\]
\[ \Rightarrow \,\,\,{\left( {\dfrac{{1 + {i^2} + 2i}}{{1 - {i^2}}}} \right)^{2n}} = 1\]
As we know the value of the \[i\] in complex numbers is \[i = \sqrt { - 1} \], then \[{i^2} = - 1\].
On simplifying the value of \[{i^2} = - 1\], then we have
\[ \Rightarrow \,\,\,{\left( {\dfrac{{1 + \left( { - 1} \right) + 2i}}{{1 - \left( { - 1} \right)}}} \right)^{2n}} = 1\]
On using a sign convention, we can written as
\[ \Rightarrow \,\,\,{\left( {\dfrac{{1 - 1 + 2i}}{{1 + 1}}} \right)^{2n}} = 1\]
\[ \Rightarrow \,\,\,{\left( {\dfrac{{2i}}{2}} \right)^{2n}} = 1\]
On simplification we get
\[ \Rightarrow \,\,\,{\left( i \right)^{2n}} = 1\]
Now to find the positive integer of \[n\] by giving a value as \[n = 1,2,3,4.....\].
At \[n = 1\]
\[ \Rightarrow \,\,\,{\left( i \right)^{2\left( 1 \right)}} = {i^2} = - 1\]
At \[n = 2\]
\[ \Rightarrow \,\,\,{\left( i \right)^{2\left( 2 \right)}} = {i^4}\]
\[ \Rightarrow \,\,\,{i^2} \cdot {i^2}\]
\[ \Rightarrow \,\,\,\left( { - 1} \right) \cdot \,\left( { - 1} \right)\]
\[\therefore \,\,{i^4} = 1\]
Hence for the smallest integer \[n = 4\] the given complex equation \[{\left( {1 + i} \right)^{2n}} = {\left( {1 - i} \right)^{2n}}\] satisfy.

Note:
Any number of the form \[a + ib\] is known as a complex number. If the given complex equation is in the form of fraction it can be simplified by applying a process of rationalization by multiplying both numerator and denominator by its conjugate. Conjugate obtained by changing the sign of the imaginary part of a given complex number.