Question

Write the least positive value of n for which ${{\left( \dfrac{1+i}{1-i} \right)}^{n}}$ is real.

Answer 1

We have to rationalize the given equation in the question and then we compare the following to the general term of a complex number which is $a\text{ }+\text{ }ib$ and then we find the value of n such that b = 0.

Complete step-by-step answer:
Complex numbers are numbers which are represented on the imaginary plane. They are represented in the following number:$a\text{ }+\text{ }ib$, where a denotes the real part of the complex number and b denotes the imaginary part.

Some of the basic identities we need to remember before we proceed into the question are

1. ${{i}^{2}}=-1$
2. ${{i}^{3}}=-i$
3. ${{i}^{4}}=1$

With these in mind, let us proceed with the question.

First, we rationalize the denominator of the term inside the bracket which is $\dfrac{1+i}{1-i}$. Rationalizing means multiplying the number with its conjugate. For example, if we have to rationalize $a\text{ }+\text{ }ib$ , we multiply the term with$a\text{-}\text{ }ib$ . However, to ensure that the value of the term remains the same, we should multiply both numerator and denominator by $1+i$ to rationalise the denominator without changing the value of the fraction. Therefore, rationalising the fraction, we obtain

$\dfrac{1+i}{1-i}=\left( \dfrac{1+i}{1-\text{i}}\times \dfrac{1+\text{i}}{1+\text{i}} \right)$

Using the identity (a + b) (a – b) = ${a}^{2}-{b}^{2}$, in the above expression, we get
= $\left( \dfrac{{{\left( 1+\text{i} \right)}^{2}}}{{{1}^{2}}-{{\left( \text{i} \right)}^{2}}} \right)$,
= $\left( \dfrac{1-1+2\text{i}}{1-\left( -1 \right)} \right)$,
=$\left( \dfrac{2i}{2} \right)$,
= $i$ .

Putting the rationalized value back into the parent term we obtain the expression, we have to find the least value of n such that ${{i}^{n}}$ is real.

Now we are asked the least positive of n for which the term is real. Using our identities about i in (1), we can say that ${{i}^{2}}$ = -1 which is a real value.

So, the value of n =2.

Note: When we multiply with the conjugate it gives us a simplified solution. So, remember that we have to take the conjugate carefully as it leads to elimination of the imaginary part in the denominator which makes the question more approachable.