Question

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 1

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.

& \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{aligned}$$ Thus we got the value of z = i. Hence we took, $$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.