Courses
Courses for Kids
Free study material
Offline Centres
More Last updated date: 02nd Dec 2023
Total views: 380.4k
Views today: 11.80k

# 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. $4$ Verified
380.4k+ views
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.