The equation zoverline{z} + aoverline{z} + ov | vedclass

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(B) Given the equation $z\overline{z} + a\overline{z} + \overline{a}z + b = 0$.Adding $|a|^2$ to both sides,we get:$z\overline{z} + a\overline{z} + \o

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$|a|^2 > b$

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(B) Given the equation $z\overline{z} + a\overline{z} + \overline{a}z + b = 0$.Adding $|a|^2$ to both sides,we get:$z\overline{z} + a\overline{z} + \overline{a}z + |a|^2 = |a|^2 - b$This can be factored as:$(z + a)(\overline{z} + \overline{a}) = |a|^2 - b$Since $(z + a)(\overline{z} + \overline{a}) = |z + a|^2$,the equation becomes:$|z + a|^2 = |a|^2 - b$This represents a circle with center $-a$ and radius $\sqrt{|a|^2 - b}$ if the radius squared is positive,i.e.,$|a|^2 - b > 0$.Therefore,$|a|^2 > b$.

The equation $z\overline{z} + a\overline{z} + \overline{a}z + b = 0$,where $b \in \mathbb{R}$,represents a circle if

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