If the equation a|z|^2 + overline{bar{alpha}z | vedclass

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Question

(B) The given equation is $a|z|^2 + \overline{\bar{\alpha}z + \alpha\bar{z}} + d = 0$.Since $\overline{\bar{\alpha}z + \alpha\bar{z}} = \alpha\bar{z}

Vedclass · Accepted Answer

$|\alpha|^2 - ad > 0$ and $a \in \mathbb{R} - \{0\}$

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(B) The given equation is $a|z|^2 + \overline{\bar{\alpha}z + \alpha\bar{z}} + d = 0$.Since $\overline{\bar{\alpha}z + \alpha\bar{z}} = \alpha\bar{z} + \bar{\alpha}z$,the equation becomes $az\bar{z} + \bar{\alpha}z + \alpha\bar{z} + d = 0$.Dividing by $a$ (assuming $a 
eq 0$),we get $z\bar{z} + \frac{\bar{\alpha}}{a}z + \frac{\alpha}{a}\bar{z} + \frac{d}{a} = 0$.This is the standard form of a circle $z\bar{z} + \bar{\beta}z + \beta\bar{z} + c = 0$ where $\beta = \frac{\alpha}{a}$.The radius $r$ is given by $r = \sqrt{|\beta|^2 - c} = \sqrt{\left|\frac{\alpha}{a}ight|^2 - \frac{d}{a}} = \sqrt{\frac{|\alpha|^2 - ad}{a^2}}$.For the equation to represent a circle,the radius must be real and positive,so $|\alpha|^2 - ad > 0$ and $a 
eq 0$.

If the equation $a|z|^2 + \overline{\bar{\alpha}z + \alpha\bar{z}} + d = 0$ represents a circle where $a, d$ are real constants,then which of the following conditions is correct?

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