Question

If ${{\left( x+iy \right)}^{3}}=u+iv$ then show that $\dfrac{u}{x}+\dfrac{v}{y}=4\left( {{x}^{2}}-{{y}^{2}} \right)$.

Answer 1

Hint: In the above question we will expand the cubic expression and by comparing both sides of equality, we will get the value of u and v. The formula to expand a cubic expression is as follows:
${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab(a+b)$

Complete step-by-step answer:
Also, we will use the property of iota $(i)$ which is as follows:

& {{i}^{4n+2}}=-1 \\\ & {{i}^{4n+3}}=-i \\\ \end{aligned}$$ We have been given $${{\left( x+iy \right)}^{3}}=u+iv$$ and we have to prove that $$\dfrac{u}{x}+\dfrac{v}{y}=4\left( {{x}^{2}}-{{y}^{2}} \right)$$. $${{\left( x+iy \right)}^{3}}=u+iv$$ On applying the formula $${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab(a+b)$$ to the above expression, we get, $$\begin{aligned} & {{\left( x+iy \right)}^{3}}=u+iv \\\ & {{x}^{3}}+{{\left( iy \right)}^{3}}+3xyi\left( x+iy \right)=u+iv \\\ & {{x}^{3}}+{{i}^{3}}{{y}^{3}}+3{{x}^{2}}yi+3x{{y}^{2}}{{i}^{2}}=u+iv \\\ \end{aligned}$$ Now $${{i}^{3}}$$ can be written in the form of $${{i}^{\left( 4\times 0+3 \right)}}$$. We know that, $$\begin{aligned} & {{i}^{\left( 4n+3 \right)}}=-i \\\ & {{i}^{\left( 4\times 0+3 \right)}}={{i}^{3}}=-i \\\ \end{aligned}$$ Also, $${{i}^{2}}$$ can be written in the form of $${{i}^{\left( 4\times 0+3 \right)}}$$. We know that, $$\begin{aligned} & {{i}^{\left( 4n+2 \right)}}=-1 \\\ & {{i}^{\left( 4\times 0+2 \right)}}={{i}^{2}}=-1 \\\ \end{aligned}$$ On substituting the values of $${{i}^{2}}$$ and $${{i}^{3}}$$ in the above expressions, we get, $$\begin{aligned} & {{x}^{3}}-{{y}^{3}}i+3{{x}^{2}}yi-3x{{y}^{2}}=u+iv \\\ & {{x}^{3}}-3x{{y}^{2}}+\left( 3{{x}^{2}}y-{{y}^{3}} \right)i=u+iv \\\ \end{aligned}$$ On comparing both the sides of equality we get, $$u={{x}^{3}}-3x{{y}^{2}}$$ and $$v=3{{x}^{2}}y-{{y}^{3}}$$. Now we have to prove $$\dfrac{u}{x}+\dfrac{v}{y}=4\left( {{x}^{2}}-{{y}^{2}} \right)$$. Taking left hand side of the expression, we get, $$\dfrac{u}{x}+\dfrac{v}{y}$$ On substituting the values of u and v, we get, $$\begin{aligned} & \dfrac{u}{x}+\dfrac{v}{y}=\dfrac{{{x}^{3}}-3x{{y}^{2}}}{x}+\dfrac{3{{x}^{2}}y-{{y}^{3}}}{y}=\dfrac{x\left( {{x}^{2}}-3{{y}^{2}} \right)}{x}+\dfrac{y\left( 3{{x}^{2}}-{{y}^{2}} \right)}{y} \\\ & ={{x}^{2}}-3{{y}^{2}}+3{{x}^{2}}-{{y}^{2}}=4{{x}^{2}}-4{{y}^{2}}=4\left( {{x}^{2}}-{{y}^{2}} \right) \\\ \end{aligned}$$ Hence, it is proved that $$\dfrac{u}{x}+\dfrac{v}{y}=4\left( {{x}^{2}}-{{y}^{2}} \right)$$. Note: Be careful while expanding the expression $${{\left( x+iy \right)}^{3}}$$, using the cubic formula, don’t miss the term ‘i' while expanding it. Also remember that two imaginary numbers are equal if and only if the real part of another number and similarly for the imaginary part also.