For a complex number z, let operatorname{Re}( | vedclass

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(D) Let $z = x + iy$. The given equation is $z^4 - |z|^4 = 4iz^2$.Since $|z|^2 = z\bar{z}$, we have $|z|^4 = (z\bar{z})^2 = z^2\bar{z}^2$.Substituting

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$8$

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(D) Let $z = x + iy$. The given equation is $z^4 - |z|^4 = 4iz^2$.Since $|z|^2 = z\bar{z}$, we have $|z|^4 = (z\bar{z})^2 = z^2\bar{z}^2$.Substituting this into the equation: $z^4 - z^2\bar{z}^2 = 4iz^2$.This implies $z^2(z^2 - \bar{z}^2) = 4iz^2$.Case $1$: $z^2 = 0 \Rightarrow z = 0$. However, $z = 0$ does not satisfy the condition $\operatorname{Re}(z_1) > 0$ or $\operatorname{Re}(z_2) Case $2$: $z^2 - \bar{z}^2 = 4i$.Since $z = x + iy$, $z^2 = x^2 - y^2 + 2ixy$ and $\bar{z}^2 = x^2 - y^2 - 2ixy$.Thus, $z^2 - \bar{z}^2 = (x^2 - y^2 + 2ixy) - (x^2 - y^2 - 2ixy) = 4ixy$.Equating to $4i$, we get $4ixy = 4i$, which simplifies to $xy = 1$.We need to minimize $|z_1 - z_2|^2$ where $z_1 = x_1 + iy_1$ with $x_1 > 0$ and $z_2 = x_2 + iy_2$ with $x_2 Since $xy = 1$, $y = 1/x$. Thus $z = x + i/x$.$|z_1 - z_2|^2 = |(x_1 - x_2) + i(1/x_1 - 1/x_2)|^2 = (x_1 - x_2)^2 + (\frac{x_2 - x_1}{x_1x_2})^2 = (x_1 - x_2)^2 (1 + \frac{1}{x_1^2x_2^2})$.Let $x_1 = a > 0$ and $x_2 = -b$ where $b > 0$. Then $x_1x_2 = -ab$.$|z_1 - z_2|^2 = (a + b)^2 (1 + \frac{1}{a^2b^2})$.By $AM-GM$ inequality, $(a + b)^2 \ge 4ab$. Also $1 + \frac{1}{a^2b^2} \ge 2\sqrt{1 \cdot \frac{1}{a^2b^2}} = \frac{2}{ab}$.So $|z_1 - z_2|^2 \ge 4ab \cdot \frac{2}{ab} = 8$.

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