QUESTION

# Write the following in the form of A + iB:$\dfrac{{{{\left( {a + ib} \right)}^2}}}{{\left( {a - ib} \right)}} - \dfrac{{{{\left( {a - ib} \right)}^2}}}{{\left( {a + ib} \right)}}$

Hint: To convert the given equation in the standard form (i.e. A + iB), we will be applying rationalization.

Given, $\dfrac{{{{\left( {a + ib} \right)}^2}}}{{\left( {a - ib} \right)}} - \dfrac{{{{\left( {a - ib} \right)}^2}}}{{\left( {a + ib} \right)}}$
To rationalize, let's multiply and divide each term with (a + ib) and (a – ib) respectively, we get,
$\dfrac{{{{\left( {a + ib} \right)}^2}}}{{\left( {a - ib} \right)}} \times \dfrac{{\left( {a + ib} \right)}}{{\left( {a + ib} \right)}} - \dfrac{{{{\left( {a - ib} \right)}^2}}}{{\left( {a + ib} \right)}} \times \dfrac{{\left( {a - ib} \right)}}{{\left( {a - ib} \right)}}$
We know that, $(a+b)(a-b) = (a^2-b^2)$
$\Rightarrow \dfrac{{{{\left( {a + ib} \right)}^3}}}{{{a^2} - {i^2}{b^2}}} - \dfrac{{{{\left( {a - ib} \right)}^3}}}{{{a^2} - {i^2}{b^2}}}$
Now you know the value of ${i^2} = - 1$
$\Rightarrow \dfrac{{{{\left( {a + ib} \right)}^3}}}{{{a^2} + {b^2}}} - \dfrac{{{{\left( {a - ib} \right)}^3}}}{{{a^2} + {b^2}}}$
We know that $(a+b)^3 = (a^3+b^3+3ab(a+b))$ and $(a-b)^3 = (a^3-b^3-3ab(a-b))$
Applying it, we get
$\Rightarrow \dfrac{{{a^3} + {i^3}{b^3} + 3i{a^2}b + 3{i^2}a{b^2}}}{{{a^2} + {b^2}}} - \dfrac{{{a^3} - {i^3}{b^3} - 3i{a^2}b + 3{i^2}a{b^2}}}{{{a^2} + {b^2}}}$
$\Rightarrow \dfrac{{{a^3} + {i^3}{b^3} + 3i{a^2}b + 3{i^2}a{b^2} - {a^3} + {i^3}{b^3} + 3i{a^2}b - 3{i^2}a{b^2}}}{{{a^2} + {b^2}}}$
$\Rightarrow \dfrac{{2{i^3}{b^3} + 6i{a^2}b}}{{{a^2} + {b^2}}} = \dfrac{{ - 2i{b^3} + 6i{a^2}b}}{{{a^2} + {b^2}}} = i\left( {\dfrac{{ - 2{b^3} + 6{a^2}b}}{{{a^2} + {b^2}}}} \right)$
$\Rightarrow 0 + i\left( {\dfrac{{ - 2{b^3} + 6{a^2}b}}{{{a^2} + {b^2}}}} \right)$
So the above equation is in standard form.

Note: To solve such problems, apply the concept of rationalization and use the necessary algebraic identities to arrive at the solution. Remember the values of higher powers of i.