Question

If \[{{(x+iy)}^{\dfrac{1}{3}}}=a+ib\] , then \[\left( \dfrac{x}{a} \right)+\left( \dfrac{y}{b} \right)\] is equal to
A. \[4({{a}^{2}}+{{b}^{2}})\]
B. \[4({{a}^{2}}-{{b}^{2}})\]
C. \[4({{b}^{2}}-{{a}^{2}})\]
D.None of these

Answer 1

Hint: \[i\] stands for iota. It is also called the imaginary part. A complex number has the real part and the imaginary part. The complex numbers are the field of numbers of the form \[x+iy\] , where $x$ and \[y\] are real numbers and \[i\] is the imaginary unit equal to the square root of \[-1\] , \[\sqrt{-1}\] . When a single letter \[z=x+iy\] is used to denote a complex number, it is also called an "affix." In component notation, \[z=x+iy\] can be written \[(x,y)\] . The field of complex numbers has the field of real numbers as a subfield.

Formula used:
$i^1=1$
$i^2=-1$
$i^3=-i$
$i^4=1$

Complete step-by-step answer:
Algebra is a very important component of Elementary Mathematics. The standard algebraic expressions and identities are of equal conditions that specifically hold for all the values of the variables. The Binomial Theorem gives out a standard way of expanding the powers of binomials or other terms. Algebraic identity used in this question is \[{{(a+b)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}\] .
Given as \[{{(x+iy)}^{\dfrac{1}{3}}}=a+ib\]
Taking cube on both sides we get
 \[(x+iy)={{(a+ib)}^{3}}\]
Using the algebraic identity \[{{(a+b)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}\] we get
 \[x+iy={{a}^{3}}-i{{b}^{3}}+3{{a}^{2}}bi-3a{{b}^{2}}\]
Further simplifying we get
 \[x+iy={{a}^{3}}-3a{{b}^{2}}+i(-{{b}^{3}}+3{{a}^{2}}b)\]
Comparing both sides we get
 \[x={{a}^{3}}-3a{{b}^{2}}\]
And \[y=-{{b}^{3}}+3{{a}^{2}}b\]
Further rearranging we get
 \[\dfrac{x}{a}={{a}^{2}}-3{{b}^{2}}\]
And \[\dfrac{y}{b}=-{{b}^{2}}+3{{a}^{2}}\]
Now adding the two equations we get
 \[\left( \dfrac{x}{a} \right)+\left( \dfrac{y}{b} \right)={{a}^{2}}-3{{b}^{2}}+(-{{b}^{2}}+3{{a}^{2}})\]
Further solving we get
$\therefore$ \[\left( \dfrac{x}{a} \right)+\left( \dfrac{y}{b} \right)=4({{a}^{2}}-{{b}^{2}})\]
Therefore, the correct answer is “Option B”.

Note: To solve this type of questions all the properties of matrices and the algebraic identities have to be known properly. Two complex numbers are equal if and only if their real and imaginary numbers are equal. A matrix whose elements contain complex numbers called a complex matrix. The sum of two conjugate complex numbers is always real.