Question

If ${{\left( \dfrac{1+i}{1-i} \right)}^{x}}=1$ then
(a) x = 2n+1, where n is any positive integer.
(b) x = 4n, where n is any positive integer.
(c) x = 2n, where n is any positive integer.
(d) x = 4n+1, where n is any positive integer.

Answer 1

Hint: In this question, we can use the concept of power of Iota. Iota is square root of minus 1 means $\sqrt{-1}$. For example, what is the value of Iota's power 3? Iota $i=\sqrt{-1}$ so ${{i}^{2}}$ is $\sqrt{-1}\times \sqrt{-1}=-1$ and hence ${{i}^{3}}={{i}^{2}}\times i=-i$.

Complete step-by-step answer:

Let us consider the given expression,
${{\left( \dfrac{1+i}{1-i} \right)}^{x}}=1................(1)$
To simplify the term $\left( \dfrac{1+i}{1-i} \right)$ by multiply numerator and denominator by the conjugate of the denominator, the process by which a fraction is rewritten so that the denominator contains only rational numbers is known as rationalization.
$\dfrac{1+i}{1-i}=\dfrac{1+i}{1-i}\times \dfrac{1+i}{1+i}=\dfrac{{{(1+i)}^{2}}}{1-{{i}^{2}}}=\dfrac{1+2i+{{i}^{2}}}{1-{{i}^{2}}}$
We have ${{i}^{2}}=-1$
$\dfrac{1+i}{1-i}=\dfrac{1+2i-1}{1+1}=\dfrac{2i}{2}=i$
Now put this value in the equation (1), we get
${{\left( \dfrac{1+i}{1-i} \right)}^{x}}={{(i)}^{x}}=1$
${{(i)}^{x}}={{(i)}^{4}}\text{ or }{{(i)}^{8}}\text{ or }{{(i)}^{12}}\text{ or }{{(i)}^{16}}$
Hence x = 4n for any positive value of n
Therefore, the correct option for the given question is option (b).

Note: We might get confused with the question, are imaginary numbers positive or negative?. The answer is No. An imaginary number is not positive or negative. A positive number is greater than zero, and a negative number is less than zero; but “greater than” and “less than” don't exist for Complex numbers, only for Real Numbers, and Imaginary numbers are always Complex numbers.