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

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

Explanation

Solution

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

Complete step-by-step answer:
Given, (a+ib)2(a−ib)−(a−ib)2(a+ib)\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,
(a+ib)2(a−ib)×(a+ib)(a+ib)−(a−ib)2(a+ib)×(a−ib)(a−ib)\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)=(a2−b2)(a+b)(a-b) = (a^2-b^2)
⇒(a+ib)3a2−i2b2−(a−ib)3a2−i2b2\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 i2=−1{i^2} = - 1
⇒(a+ib)3a2+b2−(a−ib)3a2+b2\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=(a3+b3+3ab(a+b))(a+b)^3 = (a^3+b^3+3ab(a+b)) and (a−b)3=(a3−b3−3ab(a−b))(a-b)^3 = (a^3-b^3-3ab(a-b))
Applying it, we get
⇒a3+i3b3+3ia2b+3i2ab2a2+b2−a3−i3b3−3ia2b+3i2ab2a2+b2\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}}}
⇒a3+i3b3+3ia2b+3i2ab2−a3+i3b3+3ia2b−3i2ab2a2+b2\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}}}
⇒2i3b3+6ia2ba2+b2=−2ib3+6ia2ba2+b2=i(−2b3+6a2ba2+b2)\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)
⇒0+i(−2b3+6a2ba2+b2)\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.