Question

$\display \text{Find the square root of}\\\\\displaystyle{\frac{x^2}{y^2}+\frac{y^2}{x^2}+\frac{1}{2i}\left[\frac{x}{y}+\frac{y}{x}\right]+\frac{31}{16}}$

Answer 1

Answer:

We want the square root of

$\frac{ {x}^{2} }{ {y}^{2} } + \frac{ {y}^{2} }{ {x}^{2} } + \frac{1}{2i}( \frac{x}{y} + \frac{y}{x} ) + \frac{31}{16}$

$let \: a = \sqrt{ \frac{ {x}^{2} }{ {y}^{2} } + \frac{ {y}^{2} }{ {x}^{2} } + \frac{1}{2i}( \frac{x}{y} + \frac{y}{x} ) + \frac{31}{16} }$

$Let \: \frac{x}{y} \: be \: \: u = >$

$then \sqrt{ {u}^{2} + \frac{1}{ {u}^{2} } + \frac{1}{2i} (u + \frac{1}{u}) + \frac{31}{16} } \\ = \sqrt{ {u}^{2} + 2 + \frac{1}{ {u}^{2} } + \frac{1}{2i} (u + \frac{1}{2} ) + \frac{31}{16} - 2} \\ = \sqrt{ {(u + \frac{1}{u}) }^{2} + \frac{1}{2i} (u + \frac{1}{u} ) - \frac{1}{16} }$

Let

$u + \frac{1}{u} = v \\ a = \sqrt{ {v}^{2} + \frac{1}{2i}v - \frac{1}{16} } \\ = \frac{1}{4} \sqrt{16 {v}^{2} - 8iv - 1} \\ = \frac{1}{4} \sqrt{16 {v}^{2} - 8iv - {i}^{2} } \\ = \frac{1}{4} \sqrt{ {(4v - i)}^{2} } \\ a = + or - \frac{1}{4}(4v - i) \\ = + or - \frac{1}{4}(4( \frac{x}{y} + \frac{y}{x} ) - i) \\ = + or - (\frac{x}{y} - \frac{y}{x} - \frac{i}{4}) \\ we \: therefore \: get \\ \sqrt{ \frac{ {x}^{2} }{ {y}^{2} } + \frac{ {y}^{2} }{ {x}^{2} } + \frac{1}{2i}( \frac{x}{y} + \frac{y}{x}) + \frac{31}{16} } \\ = + or - ( \frac{x}{y} - \frac{y}{x} - \frac{i}{4} )$

Answer 2

$\huge \sf \green{hello \: mate}$

As, we know that we have to find square root or given equation

$\tt \boxed{\frac{x^2}{y^2}+\frac{y^2}{x^2}+\frac{1}{2i}\left[\frac{x}{y}+\frac{y}{x}\right]+\frac{31}{16}}</p><p>$

________________________

so , simply we can take the root of whole equation and then solve it

$let \: \frac{x}{y} \: be \: \: u \implies$

$\tt\begin{lgathered}then \sqrt{ {u}^{2} + \frac{1}{ {u}^{2} } + \frac{1}{2i} (u + \frac{1}{u}) + \frac{31}{16} } \\ = \sqrt{ {u}^{2} + 2 + \frac{1}{ {u}^{2} } + \frac{1}{2i} (u + \frac{1}{2} ) + \frac{31}{16} - 2} \\ = \sqrt{ {(u + \frac{1}{u}) }^{2} + \frac{1}{2i} (u + \frac{1}{u} ) - \frac{1}{16} }\end{lgathered}$

_______________________

let,

$\tt\begin{lgathered}u + \frac{1}{u} = v \\ a = \sqrt{ {v}^{2} + \frac{1}{2i}v - \frac{1}{16} } \\ = \frac{1}{4} \sqrt{16 {v}^{2} - 8iv - 1} \\ = \frac{1}{4} \sqrt{16 {v}^{2} - 8iv - {i}^{2} } \\ = \frac{1}{4} \sqrt{ {(4v - i)}^{2} } \\ a = + or - \frac{1}{4}(4v - i) \\ = + or - \frac{1}{4}(4( \frac{x}{y} + \frac{y}{x} ) - i) \\ = + or - (\frac{x}{y} - \frac{y}{x} - \frac{i}{4}) \\ we \: therefore \: get \\ \sqrt{ \frac{ {x}^{2} }{ {y}^{2} } + \frac{ {y}^{2} }{ {x}^{2} } + \frac{1}{2i}( \frac{x}{y} + \frac{y}{x}) + \frac{31}{16} } \\ = + or - ( \frac{x}{y} - \frac{y}{x} - \frac{i}{4} )\end{lgathered}$