Question

If ${{\left( x+iy \right)}^{\dfrac{1}{3}}}=a+ib$, then $\dfrac{x}{a}+\dfrac{y}{b}$ is equal to
(a) $4\left( {{a}^{2}}+{{b}^{2}} \right)$
(b) $4\left( {{a}^{2}}-{{b}^{2}} \right)$
(c) $4\left( {{b}^{2}}-{{a}^{2}} \right)$
(d) None of these

Answer 1

Hint: Take the cube of the given equation on both sides and simplify the expression using the algebraic identity ${{\left( c+d \right)}^{3}}={{c}^{3}}+{{d}^{3}}+3{{c}^{2}}d+3c{{d}^{2}}$. Calculate the higher powers of $i$ using the fact that $i=\sqrt{-1}$. Compare the real part and imaginary part on both sides of the equation and simplify it to calculate the value of the given expression.

Complete step-by-step solution -
We know that ${{\left( x+iy \right)}^{\dfrac{1}{3}}}=a+ib$. We have to calculate the value of $\dfrac{x}{a}+\dfrac{y}{b}$.
We will simplify the above equation ${{\left( x+iy \right)}^{\dfrac{1}{3}}}=a+ib$ by taking the cube on both sides of the given equation.
Thus, we have ${{\left[ {{\left( x+iy \right)}^{\dfrac{1}{3}}} \right]}^{3}}={{\left( a+ib \right)}^{3}}$. So, we have $\left( x+iy \right)={{\left( a+ib \right)}^{3}}.....\left( 1 \right)$.
We know the algebraic identity ${{\left( c+d \right)}^{3}}={{c}^{3}}+{{d}^{3}}+3{{c}^{2}}d+3c{{d}^{2}}$.
Substituting $c=a,d=ib$ in the above identity, we have ${{\left( a+ib \right)}^{3}}={{a}^{3}}+{{\left( ib \right)}^{3}}+3{{a}^{2}}\left( ib \right)+3a{{\left( ib \right)}^{2}}$.
We know that $i=\sqrt{-1}$. So, we have ${{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1,{{i}^{3}}={{i}^{2}}\times i=-i$. Substituting these values in the above equation, we have ${{\left( a+ib \right)}^{3}}={{a}^{3}}+{{\left( ib \right)}^{3}}+3{{a}^{2}}\left( ib \right)+3a{{\left( ib \right)}^{2}}={{a}^{3}}-i{{b}^{3}}+3{{a}^{2}}bi-3a{{b}^{2}}$.
So, we have ${{\left( a+ib \right)}^{3}}={{a}^{3}}-i{{b}^{3}}+3{{a}^{2}}bi-3a{{b}^{2}}={{a}^{3}}-3a{{b}^{2}}+i\left( 3{{a}^{2}}b-{{b}^{3}} \right).....\left( 2 \right)$.
Substituting equation (2) in equation (1), we have $\left( x+iy \right)={{\left( a+ib \right)}^{3}}={{a}^{3}}-3a{{b}^{2}}+i\left( 3{{a}^{2}}b-{{b}^{3}} \right)$.
Thus, we have $\left( x+iy \right)={{a}^{3}}-3a{{b}^{2}}+i\left( 3{{a}^{2}}b-{{b}^{3}} \right)$.
Comparing the real and imaginary parts on both sides of the above equation, we have $x={{a}^{3}}-3a{{b}^{2}}$ and $y=3{{a}^{2}}b-{{b}^{3}}$.
Taking out the common terms in the above equations, we can rewrite them as $x=a\left( {{a}^{2}}-3{{b}^{2}} \right)$ , and $y=b\left( 3{{a}^{2}}-{{b}^{2}} \right)$.
Rearranging the terms of the above equations, we have $\dfrac{x}{a}={{a}^{2}}-3{{b}^{2}}.....\left( 3 \right)$ and $\dfrac{y}{b}=3{{a}^{2}}-{{b}^{2}}.....\left( 4 \right)$.
Substituting equation (3) and (4) in the expression $\dfrac{x}{a}+\dfrac{y}{b}$, we have $\dfrac{x}{a}+\dfrac{y}{b}={{a}^{2}}-3{{b}^{2}}+3{{a}^{2}}-{{b}^{2}}$.
Simplifying the above equation, we have $\dfrac{x}{a}+\dfrac{y}{b}=4{{a}^{2}}-4{{b}^{2}}=4\left( {{a}^{2}}-{{b}^{2}} \right)$.
Hence, the value of the expression $\dfrac{x}{a}+\dfrac{y}{b}$ if ${{\left( x+iy \right)}^{\dfrac{1}{3}}}=a+ib$ is $4\left( {{a}^{2}}-{{b}^{2}} \right)$, which is option (b).

Note: We can’t solve this question without using the algebraic identity. We must also keep in mind that two complex numbers are equal if and only if the real and imaginary parts of both the complex numbers are equal as well.