A real value of x will satisfy the equation l | vedclass

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(C) Given,$\alpha - i\beta = \left(\frac{3-4ix}{3+4ix}\right)$.Taking the modulus on both sides,we get:$|\alpha - i\beta| = \left|\frac{3-4ix}{3+4ix}\

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$\alpha^2 + \beta^2 = 1$

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(C) Given,$\alpha - i\beta = \left(\frac{3-4ix}{3+4ix}ight)$.Taking the modulus on both sides,we get:$|\alpha - i\beta| = \left|\frac{3-4ix}{3+4ix}ight|$.Since the modulus of a quotient is the quotient of the moduli,we have:$|\alpha - i\beta| = \frac{|3-4ix|}{|3+4ix|}$.We know that for any complex number $z = a + ib$,$|z| = \sqrt{a^2 + b^2}$.Thus,$\sqrt{\alpha^2 + (-\beta)^2} = \frac{\sqrt{3^2 + (-4x)^2}}{\sqrt{3^2 + (4x)^2}}$.$\sqrt{\alpha^2 + \beta^2} = \frac{\sqrt{9 + 16x^2}}{\sqrt{9 + 16x^2}}$.Since $x$ is a real value,$9 + 16x^2 
eq 0$,so the ratio is $1$.$\sqrt{\alpha^2 + \beta^2} = 1$.Squaring both sides,we get $\alpha^2 + \beta^2 = 1$.

$A$ real value of $x$ will satisfy the equation $\left(\frac{3-4ix}{3+4ix}\right) = \alpha - i\beta$ (where $\alpha, \beta$ are real),if

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