If frac{z-1}{2z+1} is a purely imaginary numb | vedclass

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(B) Let $w = \frac{z-1}{2z+1}$. Since $w$ is purely imaginary,$w + \overline{w} = 0$. Substituting $w$,we get $\frac{z-1}{2z+1} + \frac{\overline{z}-1

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$\frac{3}{4}$ units

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(B) Let $w = \frac{z-1}{2z+1}$. Since $w$ is purely imaginary,$w + \overline{w} = 0$. Substituting $w$,we get $\frac{z-1}{2z+1} + \frac{\overline{z}-1}{2\overline{z}+1} = 0$. $(z-1)(2\overline{z}+1) + (\overline{z}-1)(2z+1) = 0$. $2z\overline{z} + z - 2\overline{z} - 1 + 2z\overline{z} + \overline{z} - 2z - 1 = 0$. $4z\overline{z} - z - \overline{z} - 2 = 0$. Dividing by $4$,we get $z\overline{z} - \frac{1}{4}z - \frac{1}{4}\overline{z} - \frac{1}{2} = 0$. This is the equation of a circle in the form $z\overline{z} - \overline{\alpha}z - \alpha\overline{z} + k = 0$,where the center is $\alpha = \frac{1}{8}$ and the radius $r = \sqrt{|\alpha|^2 - k}$. Here,$\alpha = \frac{1}{8}$ and $k = -\frac{1}{2}$. $r = \sqrt{(\frac{1}{8})^2 - (-\frac{1}{2})} = \sqrt{\frac{1}{64} + \frac{1}{2}} = \sqrt{\frac{1+32}{64}} = \sqrt{\frac{33}{64}} = \frac{\sqrt{33}}{8}$. Wait,re-evaluating the standard form: $z\overline{z} - \frac{1}{4}z - \frac{1}{4}\overline{z} = \frac{1}{2}$. $(z - \frac{1}{4})(\overline{z} - \frac{1}{4}) = \frac{1}{2} + \frac{1}{16} = \frac{9}{16}$. $|z - \frac{1}{4}|^2 = (\frac{3}{4})^2$. Thus,the radius is $\frac{3}{4}$ units.

If $\frac{z-1}{2z+1}$ is a purely imaginary number,then the locus of $z$ represents a circle. Find its radius.

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