Respuesta :
Let [tex]\bar{z}[/tex]denote the complex conjugate of a complex number z.
z is a non - zero complex number for which both real and imaginary parts are integers.
[tex]\displaystyle (\bar{z} )^{2} + \frac{1}{z^{2} } = x + \iota y[/tex]
Let, we assume that,
[tex]\displaystyle z = r.e^{\iota \theta}z[/tex]
[tex]\sf \displaystyle \bar{z} = r.e^{- \iota \theta}[/tex]
[tex]\sf \displaystyle \frac{1}{z} = \frac{1}{r} e^{- \iota \theta} [/tex]
Put the values in the expression, we get.
[tex]\sf \displaystyle (r.e^{\iota \theta} )^{2} + \bigg(\frac{1}{r} e^{- \iota \theta} \bigg)^{2} = x + \iota y[/tex]
[tex]\displaystyle r^{2} e^{- \iota 2 \theta} + \frac{1}{r^{2} } e^{- \iota 2 \theta} = x + \iota y[/tex]
Taking modulus on both sides of the equation, we get.
[tex]\sf \displaystyle \bigg|r^{2} e^{- \iota 2 \theta} + \frac{1}{r^{2} } e^{- \iota 2 \theta} \bigg| = \bigg| x + \iota y \bigg| [/tex]
[tex]\sf \displaystyle \bigg| \bigg(r^{2} + \frac{1}{r^{2} }\bigg) e^{- \iota 2 \theta} \bigg| = \bigg| x + \iota y \bigg| [/tex]
[tex]\sf \displaystyle r^{2} + \frac{1}{r^{2} } =\sqrt{x^{2} + y^{2} }[/tex]
[tex]\sf \displaystyle r^{4} + \frac{1}{r^{4} } + 2 = (x^{2} + y^{2} )[/tex]
[tex]\sf \displaystyle r^{4} + \frac{1}{r^{4} } = (x^{2} + y^{2} - 2 )[/tex]
[tex]\sf \displaystyle (r^{4} )^{2} - (x^{2} + y^{2} - 2)(r^{4} ) + 1 = 0[/tex]
[tex]\sf \displaystyle r^{4}= \frac{(x^{2} + y^{2}- 2) \pm \sqrt{(x^{2} + y^{2}- 2)^{2} - 4 } }{2}[/tex]
Taking power 1/4 on both sides of the equation, we get.
[tex]\sf \displaystyle r= \bigg[\frac{(x^{2} + y^{2}- 2) \pm \sqrt{(x^{2} + y^{2}- 2)^{2} - 4 } }{2} \bigg]^{1/4}[/tex]
Now, we can check options one by one.
First we see all options have 1/4 in power it means our equation is correct.
Option = 1 :
[tex] \sf \displaystyle \bigg(\frac{43 + 3\sqrt{205} }{2} \bigg)^{1/4}[/tex]
We can write expression as,
[tex]\sf \displaystyle x^{2} + y^{2} - 2 = 43[/tex]
[tex]\begin{gathered}\sf \displaystyle x^{2} + y^{2} = 45 \\ \\ x = 6 \ and \: y = 3\end{gathered}[/tex]
It means this options satisfy the equation.
Option = 2 :
[tex]\sf \displaystyle \bigg(\frac{7+ \sqrt{33} }{4} \bigg)^{1/4}[/tex]
As we can see that,
In this option denominator is 4 but in our equation denominator is 2,
Multiply numerator and denominator of the equations by 2, we get.
[tex]\sf \displaystyle r= \bigg[\frac{2(x^{2} + y^{2}- 2) \pm \sqrt{(x^{2} + y^{2}- 2)^{2} - 4 } }{4} \bigg]^{1/4}[/tex]
[tex]\displaystyle 2(x^{2} + y^{2} - 2) = 7[/tex]
[tex]\sf \displaystyle 2x^{2} + 2y^{2} - 4 = 7[/tex]
[tex]\sf \displaystyle 2x^{2} + 2y^{2} = 11[/tex]
This is not possible it means option = 2 is incorrect.
Option = 3 :
[tex]\sf \displaystyle \bigg(\frac{9+ \sqrt{65} }{4} \bigg)^{1/4}[/tex]
As we can see that,
In this option denominator is 4 but in our equation denominator is 2,
Multiply numerator and denominator of the equations by 2, we get.
[tex]\sf \displaystyle r= \bigg[\frac{2(x^{2} + y^{2}- 2) \pm \sqrt{(x^{2} + y^{2}- 2)^{2} - 4 } }{4} \bigg]^{1/4}[/tex]
[tex]\sf \displaystyle 2(x^{2} + y^{2} - 2 ) = 9[/tex]
[tex]\sf \displaystyle 2x^{2} + 2y^{2} - 4 = 9[/tex]
[tex]\sf \displaystyle 2x^{2} + 2y^{2} = 13[/tex]
This is not possible it means option = 3 is incorrect.
Option = 4 :
[tex]\sf \displaystyle \bigg(\frac{7+ \sqrt{13} }{6} \bigg)^{1/4}[/tex]
As we can see that,
In this option denominator is 6 but in our equation denominator is 2,
Multiply numerator and denominator of the equations by 3, we get.
[tex]\sf \displaystyle r= \bigg[\frac{3(x^{2} + y^{2}- 2) \pm \sqrt{(x^{2} + y^{2}- 2)^{2} - 4 } }{6} \bigg]^{1/4}[/tex]
[tex]\sf \displaystyle 3(x^{2} + y^{2} - 2 ) = 7[/tex]
[tex]\sf \displaystyle 3x^{2} + 3y^{2} - 6= 7[/tex]
[tex]\sf \displaystyle 3x^{2} + 3y^{2} = 13[/tex]
This is not possible it means option = 4 is incorrect.
∴ Option [1] is only correct answer.
