Question

Find the modulus of $\dfrac{{1 + i}}{{1 - i}} - \dfrac{{1 - i}}{{1 + i}}$.

Answer 1

Hint: To solve such questions we need to transform it in the form of $z = a + ib$. Here, we have to rationalize it or take LCM to transform it to $a + ib$ form and then calculate the modulus.

Complete step-by-step answer:
We need to find the modulus of $\dfrac{{1 + i}}{{1 - i}} - \dfrac{{1 - i}}{{1 + i}}$.
Consider $z = \dfrac{{1 + i}}{{1 - i}} - \dfrac{{1 - i}}{{1 + i}}$
Taking LCM, we get $\dfrac{{{{(1 + i)}^2} - {{(1 - i)}^2}}}{{(1 - i)(1 + i)}}$
 $\Rightarrow z = \dfrac{{1 + {i^2} + 2i - 1 - {i^2} + 2i}}{{{1^2} - {i^2}}}$
 $\Rightarrow z = \dfrac{{4i}}{1 + 1}$ (as we know ${i^2} = - 1$)
$\Rightarrow z = 2i$
We obtain $z = 2i$ which is the form $z = a + ib$
$\therefore \left| z \right| = \left| {2i} \right| = \sqrt {{2^2}} = 2$ (if $z = a + ib$ then $\left| z \right| = \sqrt {{a^2} + {b^2}} $)

Note: Modulus of any complex number $z$ where $z = a + ib$, then $\left| z \right| = \sqrt {{a^2} + {b^2}} $ where $a$ is real part and $b$ is imaginary part of the complex number $z$. Also ${i^2} = - 1$.