Question

Find the square root of $\frac{x^2}{y^2}+\frac{y^2}{x^2}-\frac{1}{i}(\frac{x}{y}-\frac{y}{x})-\frac{9}{4}$

• A. $\pm [(\frac{x}{y}-\frac{y}{x})+\frac{i}{2}]$
• B. $\pm [(\frac{x}{y}+\frac{y}{x})-\frac{i}{2}]$
• C. $\pm [(\frac{x}{y}+\frac{y}{x})-\frac{i}{2}]$
• D. $\pm [(\frac{x}{y}-\frac{y}{x})-\frac{i}{2}]$

Answer 1

The correct option is A $\pm [(\frac{x}{y}-\frac{y}{x})+\frac{i}{2}]$

Multiply and divide by i in the imaginary part of $\frac{x^2}{y^2}+\frac{y^2}{x^2}-\frac{1}{i}(\frac{x}{y}-\frac{y}{x})-\frac{9}{4}$

$\frac{x^2}{y^2}+\frac{y^2}{x^2}+i(\frac{x}{y}-\frac{y}{x})-\frac{9}{4}$

Let $\frac{x}{y}-\frac{y}{x}=u$
$\frac{x^2}{y^2}+\frac{y^2}{x^2}=u^2+2$
Given complex number: $u^2+2-\frac{9}{4}+i\left(u\right)=(u^2-\frac{1}{4})+iu$
Let $\sqrt{(u^2-\frac{1}{4})+iu}=A+iB$
Square on both sides
$(u^2-\frac{1}{4})+iu=(A^2-B^2)+2ABi$
$A^2-B^2=u^2-\frac{1}{4}....\left(1\right)$