If \({(x + iy)^{\frac{1}{3}}} = a + ib\), where \(x,y,a,b \in R\), then \(\frac{x}{a} - \frac{y}{b} = \)
Practice this Complex Number question from Engineering Entrance Exam.
Question Details
Mathematics
Complex Number
Engineering Entrance Exam
Question
If \({(x + iy)^{\frac{1}{3}}} = a + ib\), where \(x,y,a,b \in R\), then \(\frac{x}{a} - \frac{y}{b} = \)
Solution
\({(x + iy)^{\frac{1}{3}}} = a + ib\)
\( \Rightarrow x + iy = {(a + ib)^3}\)
\( \Rightarrow x + iy = {a^3} - i{b^3} + i3{a^2}b - 3a{b^2}\)
\( = {a^3} - 3a{b^2} + i(3{a^2}b - {b^3})\)
\( \Rightarrow x = {a^3} - 3a{b^2}\)
and \(y = 3{a^2}b - {b^3}\)
So, \(\frac{x}{a} - \frac{y}{b} = {a^2} - 3{b^2} - 3{a^2} + {b^2}\)
\( = - 2{a^2} - 2{b^2} = - 2({a^2} + {b^2})\)