How do you calculate $\dfrac{{1 + i}}{{1 - i}}$?

Hint: Here we are given the fraction in which we are given the numerator as \[1 + i\] and denominator as $1 - i$ and here $i$ refers to iota which is $\sqrt { - 1} $ and therefore we need to proceed by rationalising the denominator and put the value of ${i^2} = - 1$ and simplify the given fraction.

Complete step-by-step answer:
Here we are given that we have the fraction and we need to calculate the value of that fraction which is $\dfrac{{1 + i}}{{1 - i}}$
So we must know that in the above fraction $i$ refers to iota which is $\sqrt { - 1} $ and therefore with this we can calculate the values of ${i^2},{i^3},{i^4}$
So let us calculate that, we will get:
${i^2} = {\left( {\sqrt { - 1} } \right)^2} = - 1$
${i^3} = {(\sqrt { - 1} )^3} = - 1\sqrt { - 1} = - i$
${i^4} = {\left( {{i^2}} \right)^2} = {\left( { - 1} \right)^2} = 1$
In these types of problems where we are given the fraction and there is addition or subtraction in the terms containing $i$ we should know the way by which we can rationalise the denominator. We need to multiply the numerator and denominator with the term same as the denominator but opposite in the sign. Therefore we will get:
$\dfrac{{1 + i}}{{1 - i}} \times \dfrac{{1 + i}}{{1 + i}}$
Now we can use in denominator the formula $(a - b)(a + b) = {a^2} - {b^2}$
Now we can multiply the terms which are in the numerator and denominator and we will get:
$\dfrac{{1 + i}}{{1 - i}} \times \dfrac{{1 + i}}{{1 + i}}$
$\dfrac{{(1 + i)(1 + i)}}{{(1 - i)(1 + i)}}$
$\dfrac{{(1 + i)(1 + i)}}{{({1^2} - {i^2})}}$
In the numerator we can apply the formula of ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
We will get:
$\dfrac{{({1^2} + {i^2} + 2i)}}{{({1^2} - {i^2})}}$
Now in this above term which we have got, we can substitute the value of the term ${i^2} = - 1$ and we can simplify it as:
$\dfrac{{({1^2} + {i^2} + 2i)}}{{({1^2} - {i^2})}} = \dfrac{{(1 - 1 + 2i)}}{{(1 - ( - 1))}}$
Now we can simply solve it after substituting the values and we will get:
$\dfrac{{({1^2} + {i^2} + 2i)}}{{({1^2} - {i^2})}} = \dfrac{{(1 - 1 + 2i)}}{{(1 - ( - 1))}}$
$\dfrac{{2i}}{{1 + 1}} = \dfrac{{2i}}{2} = i$
Hence we get the value of $\dfrac{{1 + i}}{{1 - i}} = i$.

Note: Here the student can also be given the denominator as $\dfrac{{1 - i}}{{1 + i}}$ but then also we need to do the same process but here the value of the multiplication factor will change which will be now $1 - i$ and the formula will be applied accordingly.