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

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

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

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(C) Given the equation: $\alpha - i\beta = \frac{3 - 4ix}{3 + 4ix}$.Taking the modulus on both sides,we get: $|\alpha - i\beta| = \left| \frac{3 - 4ix}{3 + 4ix} \right|$.Since the modulus of a quotient is the quotient of the moduli,we have: $\sqrt{\alpha^2 + \beta^2} = \frac{|3 - 4ix|}{|3 + 4ix|}$.Since $x$ is real,the modulus of $3 - 4ix$ is $\sqrt{3^2 + (-4x)^2} = \sqrt{9 + 16x^2}$ and the modulus of $3 + 4ix$ is $\sqrt{3^2 + (4x)^2} = \sqrt{9 + 16x^2}$.Thus,$\sqrt{\alpha^2 + \beta^2} = \frac{\sqrt{9 + 16x^2}}{\sqrt{9 + 16x^2}} = 1$.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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