The equation zoverline{z} + (2 - 3i)z + (2 + | vedclass

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(B) Let $z = x + iy$,then $\overline{z} = x - iy$.Substituting these into the given equation:$(x + iy)(x - iy) + (2 - 3i)(x + iy) + (2 + 3i)(x - iy) +

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$3$

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(B) Let $z = x + iy$,then $\overline{z} = x - iy$.Substituting these into the given equation:$(x + iy)(x - iy) + (2 - 3i)(x + iy) + (2 + 3i)(x - iy) + 4 = 0$$(x^2 + y^2) + (2x + 2iy - 3ix + 3y) + (2x - 2iy + 3ix + 3y) + 4 = 0$$(x^2 + y^2) + 4x + 6y + 4 = 0$This is the equation of a circle in the form $x^2 + y^2 + 2gx + 2fy + c = 0$,where $g = 2$,$f = 3$,and $c = 4$.The radius $r$ is given by $\sqrt{g^2 + f^2 - c}$.$r = \sqrt{2^2 + 3^2 - 4} = \sqrt{4 + 9 - 4} = \sqrt{9} = 3$.

The equation $z\overline{z} + (2 - 3i)z + (2 + 3i)\overline{z} + 4 = 0$ represents a circle of radius

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