# The smallest positive integer n for which ${\left( {1 + i} \right)^{2n}} = {\left( {1 - i} \right)^{2n}}$ is:A. $1$B. $2$C. $3$D. $4$

B. $2$

C. $3$

D. $4$

Last updated date: 23rd Mar 2023

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Answer

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**Hint:**In order to evaluate the smallest positive integer, try to bring the given terms in simpler form in order to use the property associated with even power of complex numbers.

**Complete step-by-step solution:**

Given equation is: ${\left( {1 + i} \right)^{2n}} = {\left( {1 - i} \right)^{2n}}$

As we know the rule of product of powers

${\left( x \right)^{a \times b}} = {\left( {{x^a}} \right)^b}$

Applying the same on either side of the equation we obtain:

$ \Rightarrow {\left( {1 + i} \right)^{2n}} = {\left( {1 - i} \right)^{2n}} $

$ \Rightarrow {\left[ {{{\left( {1 + i} \right)}^2}} \right]^n} = {\left[ {{{\left( {1 - i} \right)}^2}} \right]^n} $

Since, we know the algebraic formula for square of sum of numbers i.e.

${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$

${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$

So, using the formula in the above equation, we get

$ \Rightarrow {\left[ {{{\left( {1 + i} \right)}^2}} \right]^n} = {\left[ {{{\left( {1 - i} \right)}^2}} \right]^n} $

$ \Rightarrow {\left[ {{{\left( 1 \right)}^2} + 2 \times \left( 1 \right) \times \left( i \right) + {{\left( i \right)}^2}} \right]^n} = {\left[ {{{\left( 1 \right)}^2} - 2 \times \left( 1 \right) \times \left( i \right) + {{\left( i \right)}^2}} \right]^n}$

$\Rightarrow {\left[ {1 + 2i - 1} \right]^n} = {\left[ {1 - 2i - 1} \right]^n}{\text{ }}\left[ {\because {{\left( i \right)}^2} = - 1} \right] $

$\Rightarrow {\left[ {2i} \right]^n} = {\left[ { - 2i} \right]^n} $

$\Rightarrow {\left( 2 \right)^n}{\left( i \right)^n} = {\left( 2 \right)^n}{\left( { - i} \right)^n}{\text{ }}\left[ {\because {{\left( {ab} \right)}^n} = {{\left( a \right)}^n}{{\left( b \right)}^n}} \right]$

$ \Rightarrow {\left( i \right)^n} = {\left( { - i} \right)^n} $

$\Rightarrow {\left( i \right)^n} = {\left( { - 1} \right)^n}{\left( i \right)^n} $

$\Rightarrow {\left( { - 1} \right)^n} = 1$

Now finally we get the equation

\[{\left( { - 1} \right)^n} = 1\]

This is possible only when the value of n is even or multiple of two. And the smallest multiple of two is two itself out of the given choices.

**Hence, the smallest positive integer n is 2. So, the correct option is B.**

**Note:**In order to solve such problems related to complex numbers, always try to simplify the complex terms by the use of algebraic identities. Never try to directly put the value of $i$ in the question. Try to bring the power of $i$ in even integer as then the solution will be much easier. Also remember the value of power of $i$ for multiples of 2, 3 and 4.

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