
The smallest positive integer n for which \[{{\left( \dfrac{1+i}{1-i} \right)}^{n}}=1\] is
(a) n = 8
(b) n = 16
(c) n = 12
(d) none of these
Answer
612.3k+ views
Hint: In the question, take \[z=\left( \dfrac{1+i}{1-i} \right)\]. Rationalize the expression and get the value of z. Substitute in the question and determine for what value of n in the expression will produce 1. Use basic rules of complex identities like \[{{i}^{2}}=-1,{{i}^{3}}=i\] etc.
Complete step-by-step answer:
We have to find the smallest positive integer n for the given expression, \[{{\left( \dfrac{1+i}{1-i} \right)}^{n}}=1\]
Now let us first take, \[z=\left( \dfrac{1+i}{1-i} \right)\].
Now let us rationalize the given expression by multiplying \[\left( 1+i \right)\] in the numerator and denominator.
\[z=\dfrac{1+i}{1-i}\times \dfrac{\left( 1+i \right)}{\left( 1+i \right)}=\dfrac{\left( 1+i \right)\left( 1+i \right)}{\left( 1-i \right)\left( 1+i \right)}=\dfrac{{{\left( 1+i \right)}^{2}}}{\left( 1-i \right)\left( 1+i \right)}\]
We know the basic identities,
\[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]
\[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\], let us apply them in the above expression.
\[\begin{align}
& \therefore z=\dfrac{{{\left( 1+i \right)}^{2}}}{\left( 1-i \right)\left( 1+i \right)} \\
& z=\dfrac{1+2i+{{\left( i \right)}^{2}}}{1-{{\left( i \right)}^{2}}}\left\{ \because {{i}^{2}}=-1 \right\} \\
& z=\dfrac{1+2i-1}{1-\left( -1 \right)}=\dfrac{2i}{1+1}=\dfrac{2i}{2}=i \\
\end{align}\]
Thus we got the value of z = i.
Hence we took, \[z=\left( \dfrac{1+i}{1-i} \right)=i\]
i.e. \[{{\left( \dfrac{1+i}{1-i} \right)}^{n}}=1\], which means that, z = i.
\[{{i}^{n}}=1\]
So, \[{{i}^{4}}=1\], when the power of i is 4, the value is 1.
Thus we can say that n = 4, which is the smallest positive integer.
Thus none of the option matches, n = 4.
\[\therefore \] Option (d) is the correct answer, which is none of these.
Note: It is important that you take the expression, \[\left( \dfrac{1+i}{1-i} \right)=z\]. That is one of the important steps in this question. Whenever a complex expression comes in the denominator, you need to rationalize. Here \[\left( 1-i \right)\] is a complex number, so you need to rationalize z to get the value.
Complete step-by-step answer:
We have to find the smallest positive integer n for the given expression, \[{{\left( \dfrac{1+i}{1-i} \right)}^{n}}=1\]
Now let us first take, \[z=\left( \dfrac{1+i}{1-i} \right)\].
Now let us rationalize the given expression by multiplying \[\left( 1+i \right)\] in the numerator and denominator.
\[z=\dfrac{1+i}{1-i}\times \dfrac{\left( 1+i \right)}{\left( 1+i \right)}=\dfrac{\left( 1+i \right)\left( 1+i \right)}{\left( 1-i \right)\left( 1+i \right)}=\dfrac{{{\left( 1+i \right)}^{2}}}{\left( 1-i \right)\left( 1+i \right)}\]
We know the basic identities,
\[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]
\[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\], let us apply them in the above expression.
\[\begin{align}
& \therefore z=\dfrac{{{\left( 1+i \right)}^{2}}}{\left( 1-i \right)\left( 1+i \right)} \\
& z=\dfrac{1+2i+{{\left( i \right)}^{2}}}{1-{{\left( i \right)}^{2}}}\left\{ \because {{i}^{2}}=-1 \right\} \\
& z=\dfrac{1+2i-1}{1-\left( -1 \right)}=\dfrac{2i}{1+1}=\dfrac{2i}{2}=i \\
\end{align}\]
Thus we got the value of z = i.
Hence we took, \[z=\left( \dfrac{1+i}{1-i} \right)=i\]
i.e. \[{{\left( \dfrac{1+i}{1-i} \right)}^{n}}=1\], which means that, z = i.
\[{{i}^{n}}=1\]
So, \[{{i}^{4}}=1\], when the power of i is 4, the value is 1.
Thus we can say that n = 4, which is the smallest positive integer.
Thus none of the option matches, n = 4.
\[\therefore \] Option (d) is the correct answer, which is none of these.
Note: It is important that you take the expression, \[\left( \dfrac{1+i}{1-i} \right)=z\]. That is one of the important steps in this question. Whenever a complex expression comes in the denominator, you need to rationalize. Here \[\left( 1-i \right)\] is a complex number, so you need to rationalize z to get the value.
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