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Question: Find the smallest positive integer value of ‘n’ for which \(\dfrac{{{\left( 1+i \right)}^{n}}}{{{\le...

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

Explanation

Solution

Hint: Use the fact that we can simplify the expression of the form a+ibc+id\dfrac{a+ib}{c+id} by multiplying and dividing it by c−idc-id. Simplify the given expression and then further use the fact that i=−1i=\sqrt{-1} to calculate higher powers of ii. Then observe the least value of ‘n’ for which the expression (1+i)n(1−i)n−2\dfrac{{{\left( 1+i \right)}^{n}}}{{{\left( 1-i \right)}^{n-2}}} is real.

Complete step-by-step solution -
We have to calculate the least positive integral value of ‘n’ for which (1+i)n(1−i)n−2\dfrac{{{\left( 1+i \right)}^{n}}}{{{\left( 1-i \right)}^{n-2}}} is real.
We can rewrite this expression as (1+i)n(1−i)n(1−i)−2\dfrac{{{\left( 1+i \right)}^{n}}}{{{\left( 1-i \right)}^{n}}{{\left( 1-i \right)}^{-2}}}.
Thus, we have (1+i)n(1−i)n−2=(1+i)n(1−i)2(1−i)n=(1+i1−i)n(1−i)2.....(1)\dfrac{{{\left( 1+i \right)}^{n}}}{{{\left( 1-i \right)}^{n-2}}}=\dfrac{{{\left( 1+i \right)}^{n}}{{\left( 1-i \right)}^{2}}}{{{\left( 1-i \right)}^{n}}}={{\left( \dfrac{1+i}{1-i} \right)}^{n}}{{\left( 1-i \right)}^{2}}.....\left( 1 \right).
We know the algebraic identity (x−y)2=x2+y2−2xy{{\left( x-y \right)}^{2}}={{x}^{2}}+{{y}^{2}}-2xy.
Substituting x=1,y=ix=1,y=i in the above equation, we have (1−i)2=12+i2−2(1)(i){{\left( 1-i \right)}^{2}}={{1}^{2}}+{{i}^{2}}-2\left( 1 \right)\left( i \right).
We know that i=−1i=\sqrt{-1}. Thus, we have i2=(−1)2=−1{{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1.
Thus, we have (1−i)2=12+i2−2(1)(i)=1−1−2i=−2i.....(2){{\left( 1-i \right)}^{2}}={{1}^{2}}+{{i}^{2}}-2\left( 1 \right)\left( i \right)=1-1-2i=-2i.....\left( 2 \right).
Substituting equation (2) in equation (1), we have (1+i)n(1−i)n−2=(1+i1−i)n(1−i)2=(1+i1−i)n(−2i).....(3)\dfrac{{{\left( 1+i \right)}^{n}}}{{{\left( 1-i \right)}^{n-2}}}={{\left( \dfrac{1+i}{1-i} \right)}^{n}}{{\left( 1-i \right)}^{2}}={{\left( \dfrac{1+i}{1-i} \right)}^{n}}\left( -2i \right).....\left( 3 \right).
We will now simplify the expression 1+i1−i\dfrac{1+i}{1-i}.
We know that we can simplify the expression of the form a+ibc+id\dfrac{a+ib}{c+id} by multiplying and dividing it by c−idc-id.
Substituting a=1,b=1,c=1,d=−1a=1,b=1,c=1,d=-1 in the above expression, we can rewrite 1+i1−i\dfrac{1+i}{1-i} as 1+i1−i=1+i1−i×1+i1+i=(1+i)2(1+i)(1−i)\dfrac{1+i}{1-i}=\dfrac{1+i}{1-i}\times \dfrac{1+i}{1+i}=\dfrac{{{\left( 1+i \right)}^{2}}}{\left( 1+i \right)\left( 1-i \right)}.
We know the algebraic identities (x+y)2=x2+y2+2xy{{\left( x+y \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy and (x+y)(x−y)=x2−y2\left( x+y \right)\left( x-y \right)={{x}^{2}}-{{y}^{2}}.
Thus, we can simplify the expression 1+i1−i=1+i1−i×1+i1+i=(1+i)2(1+i)(1−i)\dfrac{1+i}{1-i}=\dfrac{1+i}{1-i}\times \dfrac{1+i}{1+i}=\dfrac{{{\left( 1+i \right)}^{2}}}{\left( 1+i \right)\left( 1-i \right)} as 1+i1−i=(1+i)2(1+i)(1−i)=12+i2+2(1)(i)12−(i)2\dfrac{1+i}{1-i}=\dfrac{{{\left( 1+i \right)}^{2}}}{\left( 1+i \right)\left( 1-i \right)}=\dfrac{{{1}^{2}}+{{i}^{2}}+2\left( 1 \right)\left( i \right)}{{{1}^{2}}-{{\left( i \right)}^{2}}}.
We know that i=−1i=\sqrt{-1}. Thus, we have i2=(−1)2=−1{{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1.
So, we can rewrite the expression 1+i1−i=(1+i)2(1+i)(1−i)=12+i2+2(1)(i)12−(i)2\dfrac{1+i}{1-i}=\dfrac{{{\left( 1+i \right)}^{2}}}{\left( 1+i \right)\left( 1-i \right)}=\dfrac{{{1}^{2}}+{{i}^{2}}+2\left( 1 \right)\left( i \right)}{{{1}^{2}}-{{\left( i \right)}^{2}}} as 1+i1−i=12+i2+2(1)(i)12−(i)2=1−1+2i1−(−1)=2i1+1=2i2=i\dfrac{1+i}{1-i}=\dfrac{{{1}^{2}}+{{i}^{2}}+2\left( 1 \right)\left( i \right)}{{{1}^{2}}-{{\left( i \right)}^{2}}}=\dfrac{1-1+2i}{1-\left( -1 \right)}=\dfrac{2i}{1+1}=\dfrac{2i}{2}=i.
Thus, we can rewrite the expression (1+i1−i)n{{\left( \dfrac{1+i}{1-i} \right)}^{n}} as (1+i1−i)n=in.....(4){{\left( \dfrac{1+i}{1-i} \right)}^{n}}={{i}^{n}}.....\left( 4 \right).
Substituting equation (4) in equation (3), we have (1+i)n(1−i)n−2=(1+i1−i)n(−2i)=in(−2i)\dfrac{{{\left( 1+i \right)}^{n}}}{{{\left( 1-i \right)}^{n-2}}}={{\left( \dfrac{1+i}{1-i} \right)}^{n}}\left( -2i \right)={{i}^{n}}\left( -2i \right).
Thus, we have (1+i)n(1−i)n−2=(1+i1−i)n(−2i)=in(−2i)=−2in+1\dfrac{{{\left( 1+i \right)}^{n}}}{{{\left( 1-i \right)}^{n-2}}}={{\left( \dfrac{1+i}{1-i} \right)}^{n}}\left( -2i \right)={{i}^{n}}\left( -2i \right)=-2{{i}^{n+1}}
We will now calculate the smallest integral value of ‘n’ for which the expression in+1{{i}^{n+1}} is real.
We know that i=−1i=\sqrt{-1}. Thus, we have i2=(−1)2=−1{{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1.
So, we have in+1=i2{{i}^{n+1}}={{i}^{2}}. Thus, we have n+1=2n+1=2.
Rearranging the terms of the above equation, we have n=2−1=1n=2-1=1.
So, the expression (1+i)n(1−i)n−2=(1+i1−i)n(−2i)=in(−2i)=−2in+1\dfrac{{{\left( 1+i \right)}^{n}}}{{{\left( 1-i \right)}^{n-2}}}={{\left( \dfrac{1+i}{1-i} \right)}^{n}}\left( -2i \right)={{i}^{n}}\left( -2i \right)=-2{{i}^{n+1}} takes the value 2 for n=1n=1
Hence, the smallest positive value of n for which (1+i)n(1−i)n−2\dfrac{{{\left( 1+i \right)}^{n}}}{{{\left( 1-i \right)}^{n-2}}} has real value isn=1n=1.

Note: We can’t solve this question without simplifying the expression 1+i1−i\dfrac{1+i}{1-i} and using the algebraic identities. We can write any complex number in the form a+iba+ib, where ibib is the imaginary part and aa is the real part.