Question

Given that $x,yϵR$, Solve the complex equation: $4x^2+3xy+(2xy-3x^2)i=4y^2-\left(\frac{x^2}{2}\right)+(3xy-2y^2)i$

• A. $x=\frac{3K}{2},y=KKϵR$
• B. $x=K,y=\frac{3K}{2}KϵR$
• C. $x=K,y=\frac{K}{2}KϵR$
• D. $x=\frac{K}{2},y=KKϵR$

Answer 1

The correct option is B $x=K,y=\frac{3K}{2}KϵR$

Given that $4x^2+3xy+(2xy-3x^2)i=4y^2-\left(x^2/2\right)+(3xy-2y^2)i$
We proceed by equating the real and imaginary parts.
Thus, $4x^2+3xy=4y^2-x^2/2$ and
$2xy-3x^2=3xy-2y^2$
$\Rightarrow 9x^2+6xy-8y^2=0$ and $3x^2+xy-2y^2=0$
Let us take $\frac{x}{y}$ to be $a$.
The first equation becomes $9a^2+6a-8=0$, solving which gives $a=\frac{-6\pm \sqrt{36+288}}{18}=\frac{2}{3},\frac{-4}{3}$
The second equation similarly becomes $3a^2+a-2=0$, solving which gives $a=\frac{-1\pm \sqrt{1+24}}{6}=\frac{2}{3},-1$
Thus, we have a common root in both the equations and therefore $\frac{x}{y}=\frac{2}{3}$
Hence if $x=K,y=\frac{3K}{2}$