Question

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

Answer 1

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.