Let z_1, z_2 and z_3 be three complex numbers | vedclass

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(D) Given $z_1 = e^{-i\pi/4}, z_2 = e^{i0} = 1, z_3 = e^{i\pi/4}$.Since $|z|=1$,$\bar{z} = 1/z$.$z_1 \bar{z}_2 + z_2 \bar{z}_3 + z_3 \bar{z}_1 = e^{-i

Vedclass · Accepted Answer

$29$

Vedclass · Answer

(D) Given $z_1 = e^{-i\pi/4}, z_2 = e^{i0} = 1, z_3 = e^{i\pi/4}$.Since $|z|=1$,$\bar{z} = 1/z$.$z_1 \bar{z}_2 + z_2 \bar{z}_3 + z_3 \bar{z}_1 = e^{-i\pi/4}(1) + 1(e^{-i\pi/4}) + e^{i\pi/4}(e^{i\pi/4}) = 2e^{-i\pi/4} + e^{i\pi/2}$.$2e^{-i\pi/4} + e^{i\pi/2} = 2(\cos(\pi/4) - i\sin(\pi/4)) + i = 2(\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}) + i = \sqrt{2} - i\sqrt{2} + i = \sqrt{2} + i(1 - \sqrt{2})$.The squared modulus is $|\sqrt{2} + i(1 - \sqrt{2})|^2 = (\sqrt{2})^2 + (1 - \sqrt{2})^2 = 2 + (1 - 2\sqrt{2} + 2) = 2 + 3 - 2\sqrt{2} = 5 - 2\sqrt{2}$.Comparing with $\alpha + \beta \sqrt{2}$,we get $\alpha = 5$ and $\beta = -2$.Thus,$\alpha^2 + \beta^2 = 5^2 + (-2)^2 = 25 + 4 = 29$.

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