If Z = alpha + i beta satisfies the equation | vedclass

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(B) Given $Z = \alpha + i \beta$. Since $|Z| - Z = 1 + 2i$,we have $\sqrt{\alpha^2 + \beta^2} - (\alpha + i \beta) = 1 + 2i$. Equating real and imagin

Vedclass · Accepted Answer

$\frac{25}{4}$

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(B) Given $Z = \alpha + i \beta$. Since $|Z| - Z = 1 + 2i$,we have $\sqrt{\alpha^2 + \beta^2} - (\alpha + i \beta) = 1 + 2i$. Equating real and imaginary parts: $\sqrt{\alpha^2 + \beta^2} - \alpha = 1$ and $-\beta = 2 \Rightarrow \beta = -2$. Substitute $\beta = -2$ into the real part equation: $\sqrt{\alpha^2 + (-2)^2} - \alpha = 1 \Rightarrow \sqrt{\alpha^2 + 4} = \alpha + 1$. Squaring both sides: $\alpha^2 + 4 = (\alpha + 1)^2 $ $\Rightarrow \alpha^2 + 4 = \alpha^2 + 2\alpha + 1 $ $\Rightarrow 2\alpha = 3 $ $\Rightarrow \alpha = \frac{3}{2}$. We need to find $Z \bar{Z} = |Z|^2 = \alpha^2 + \beta^2$. $Z \bar{Z} = (\frac{3}{2})^2 + (-2)^2 = \frac{9}{4} + 4 = \frac{9 + 16}{4} = \frac{25}{4}$.

If $Z = \alpha + i \beta$ satisfies the equation $|Z| - Z = 1 + 2i$ and $|Z| = \sqrt{\alpha^2 + \beta^2}$,then $Z \bar{Z} = $

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