NCERT Solutions · Class 11 Mathematics Complex Numbers and Quadratic Equations

Use \(\displaystyle \beta\bar\beta = |\beta|^2 \) to factor the denominator.Since \(\displaystyle |\beta| = 1 \), we have \(\displaystyle \beta\bar{\beta} = |\beta|^2 = 1 \). Replace the \(\displaystyle 1 \) in the denominator by \(\displaystyle \beta\bar{\beta} \) and factor: \[1 - \bar{\alpha}\beta = \beta\bar{\beta} - \bar{\alpha}\beta = \beta\left(\bar{\beta} - \bar{\alpha}\right) = \beta\,\overline{\left(\beta - \alpha\right)} \]using that conjugation is additive, \(\displaystyle \bar\beta - \bar\alpha = \overline{\beta - \alpha} \).Now take moduli, using \(\displaystyle |zw| = |z||w| \) and \(\displaystyle |\bar{z}| = |z| \): \[\left|1 - \bar{\alpha}\beta\right| = |\beta|\left|\overline{\beta - \alpha}\right| = 1 \cdot |\beta - \alpha| = |\beta - \alpha| \]Because \(\displaystyle \alpha \neq \beta \), the quantity \(\displaystyle |\beta - \alpha| \) is non-zero, so the denominator is non-zero and we may divide: \[\left|\frac{\beta-\alpha}{1-\bar{\alpha}\beta}\right| = \frac{|\beta-\alpha|}{|1-\bar{\alpha}\beta|} = \frac{|\beta-\alpha|}{|\beta-\alpha|} = 1 \]The value of \(\displaystyle \left|\dfrac{\beta-\alpha}{1-\bar{\alpha}\beta}\right| \) is \(\displaystyle 1 \).

Evaluate the modulus, then compare exponents.First, \[|1 - i| = \sqrt{1^2 + (-1)^2} = \sqrt{2} = 2^{1/2} \]So the equation becomes \[\left(2^{1/2}\right)^{x} = 2^{x} \quad\Longrightarrow\quad 2^{x/2} = 2^{x} \]Since the base \(\displaystyle 2 \) is positive and not equal to \(\displaystyle 1 \), the exponential function is one-one, so the exponents must be equal: \[\frac{x}{2} = x \;\Longrightarrow\; x = 2x \;\Longrightarrow\; x = 0 \]Thus \(\displaystyle x = 0 \) is the only solution, and it is excluded because the question asks for non-zero integral solutions.There are \(\displaystyle 0 \) non-zero integral solutions.

Take the modulus of both sides.The key rule is that the modulus is multiplicative: for complex numbers \(\displaystyle z_1, z_2 \), \[|z_1 z_2| = |z_1|\,|z_2| \] and this extends to any finite product. Also \(\displaystyle |p + iq| = \sqrt{p^2 + q^2} \), so \(\displaystyle |p+iq|^2 = p^2 + q^2 \).Taking the modulus of the given equation: \[|a+ib|\,|c+id|\,|e+if|\,|g+ih| = |\mathrm{A} + i\mathrm{B}| \] \[\sqrt{a^2+b^2}\;\sqrt{c^2+d^2}\;\sqrt{e^2+f^2}\;\sqrt{g^2+h^2} = \sqrt{\mathrm{A}^2 + \mathrm{B}^2} \]Both sides are non-negative real numbers, so squaring preserves the equality: \[\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)\left(e^{2}+f^{2}\right)\left(g^{2}+h^{2}\right) = \mathrm{A}^{2}+\mathrm{B}^{2} \]Hence the required identity is proved.

Simplify the base first.Multiply numerator and denominator by the conjugate of the denominator: \[\frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1 + 2i + i^2}{1 - i^2} = \frac{2i}{2} = i \]So the equation becomes \(\displaystyle i^m = 1 \).The powers of \(\displaystyle i \) repeat with period \(\displaystyle 4 \): \[i^1 = i, \qquad i^2 = -1, \qquad i^3 = -i, \qquad i^4 = 1 \]So among \(\displaystyle m = 1, 2, 3, 4 \) the first value giving \(\displaystyle 1 \) is \(\displaystyle m = 4 \) (and in general \(\displaystyle i^m = 1 \) exactly when \(\displaystyle m \) is a multiple of \(\displaystyle 4 \)).The least positive integral value is \(\displaystyle m = 4 \).