Step 1: Understanding the Concept:
A complex number $w$ is purely imaginary if its real part is zero, which is algebraically expressed as $w + \bar{w} = 0$.
We apply this condition to the given expression to derive the Cartesian equation of the locus, which is stated to be a circle.
Step 2: Key Formula or Approach:
Let $w = \frac{z-1}{2z+1}$. Since $w$ is purely imaginary, $w + \bar{w} = 0$.
Substitute $z = x + iy$ into the resulting equation to get the form $x^2 + y^2 + 2gx + 2fy + c = 0$.
The radius of such a circle is given by $R = \sqrt{g^2 + f^2 - c}$.
Step 3: Detailed Explanation:
Given $w = \frac{z-1}{2z+1}$ is purely imaginary.
\[ \frac{z-1}{2z+1} + \overline{\left(\frac{z-1}{2z+1}\right)} = 0 \]
\[ \frac{z-1}{2z+1} + \frac{\bar{z}-1}{2\bar{z}+1} = 0 \]
Take a common denominator and set the numerator to zero:
\[ (z-1)(2\bar{z}+1) + (\bar{z}-1)(2z+1) = 0 \]
Expand the terms:
\[ (2z\bar{z} + z - 2\bar{z} - 1) + (2\bar{z}z + \bar{z} - 2z - 1) = 0 \]
Combine like terms. Note that $z\bar{z} = \bar{z}z = |z|^2$:
\[ 4|z|^2 - z - \bar{z} - 2 = 0 \]
Now, let $z = x + iy$. Then $|z|^2 = x^2 + y^2$ and $z + \bar{z} = 2x$.
Substitute these into the equation:
\[ 4(x^2 + y^2) - (2x) - 2 = 0 \]
Divide the entire equation by 4 to get the standard form of a circle:
\[ x^2 + y^2 - \frac{1}{2}x - \frac{1}{2} = 0 \]
Comparing this with the general circle equation $x^2 + y^2 + 2gx + 2fy + c = 0$:
$2g = -\frac{1}{2} \implies g = -\frac{1}{4}$
$2f = 0 \implies f = 0$
$c = -\frac{1}{2}$
The radius $R$ is calculated as:
\[ R = \sqrt{g^2 + f^2 - c} \]
\[ R = \sqrt{\left(-\frac{1}{4}\right)^2 + 0^2 - \left(-\frac{1}{2}\right)} \]
\[ R = \sqrt{\frac{1}{16} + \frac{1}{2}} = \sqrt{\frac{1}{16} + \frac{8}{16}} = \sqrt{\frac{9}{16}} \]
\[ R = \frac{3}{4} \]
Step 4: Final Answer:
The radius of the circle is $\frac{3}{4}$ units.