Question

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

Answer 1

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

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

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