Key Concepts and Formulas

• A complex number $w$ is purely imaginary if its real part is zero, i.e., $Re(w) = 0$.
• The modulus $|z_1 - z_2|$ represents the distance between two complex numbers $z_1$ and $z_2$ in the Argand plane.
• The equation of a circle with center $(h, k)$ and radius $r$ is given by $(x - h)^2 + (y - k)^2 = r^2$. The minimum distance from a point $P$ outside the circle to the circle is $PC - r$, where $C$ is the center of the circle.

Step-by-Step Solution

1. Define the given purely imaginary complex number We are given that $w = \frac{z - i}{z - 1}$ is purely imaginary. Our goal is to find the locus of $z$. For $w$ to be purely imaginary, $Re(w) = 0$.

2. Express $z$ in Cartesian form and substitute into the expression Let $z = x + iy$, where $x$ and $y$ are real numbers. We substitute this into the expression for $w$ to separate the real and imaginary parts. $w = \frac{x + iy - i}{x + iy - 1} = \frac{x + i(y - 1)}{(x - 1) + iy}$

3. Separate the real and imaginary parts of $w$ We multiply the numerator and denominator by the conjugate of the denominator to rationalize the expression and separate the real and imaginary parts. The conjugate of $(x - 1) + iy$ is $(x - 1) - iy$. $w = \frac{x + i(y - 1)}{(x - 1) + iy} \cdot \frac{(x - 1) - iy}{(x - 1) - iy}$ $w = \frac{[x + i(y - 1)][(x - 1) - iy]}{[(x - 1) + iy][(x - 1) - iy]}$ Expanding the numerator: $[x + i(y - 1)][(x - 1) - iy] = x(x - 1) - ixy + i(y - 1)(x - 1) - i^2y(y - 1)$ $= x^2 - x - ixy + i(xy - x - y + 1) + y^2 - y$ $= (x^2 - x + y^2 - y) + i(-xy + xy - x - y + 1)$ $= (x^2 - x + y^2 - y) + i(1 - x - y)$ Expanding the denominator: $[(x - 1) + iy][(x - 1) - iy] = (x - 1)^2 - (iy)^2 = (x - 1)^2 + y^2$ Therefore, $w = \frac{x^2 - x + y^2 - y}{(x - 1)^2 + y^2} + i\frac{1 - x - y}{(x - 1)^2 + y^2}$

4. Apply the purely imaginary condition to find the locus of $z$ Since $w$ is purely imaginary, its real part must be zero: $Re(w) = \frac{x^2 - x + y^2 - y}{(x - 1)^2 + y^2} = 0$ This implies that the numerator must be zero, provided the denominator is non-zero (i.e., $z \neq 1$): $x^2 - x + y^2 - y = 0$ Completing the square for both $x$ and $y$: $(x^2 - x + \frac{1}{4}) + (y^2 - y + \frac{1}{4}) = \frac{1}{4} + \frac{1}{4}$ $(x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = \frac{1}{2}$ This is a circle with center $C(\frac{1}{2}, \frac{1}{2})$ and radius $R = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.

5. Calculate the minimum value of $|z - (3 + 3i)|$ We want to find the minimum value of $|z - (3 + 3i)|$, which represents the distance between a point $z$ on the circle and the point $P(3, 3)$. The minimum distance is $PC - R$, where $PC$ is the distance between the center of the circle and the point $P$, and $R$ is the radius of the circle. First, we calculate $PC$: $PC = \sqrt{(3 - \frac{1}{2})^2 + (3 - \frac{1}{2})^2} = \sqrt{(\frac{5}{2})^2 + (\frac{5}{2})^2} = \sqrt{\frac{25}{4} + \frac{25}{4}} = \sqrt{\frac{50}{4}} = \frac{5\sqrt{2}}{2} = \frac{5}{\sqrt{2}}$ Now, we calculate the minimum distance: Minimum value of $|z - (3 + 3i)| = PC - R = \frac{5}{\sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$

Common Mistakes & Tips:

• Be careful when rationalizing the denominator to avoid algebraic errors.
• Remember to exclude points where the original expression is undefined (in this case, $z = 1$).
• Visualizing the problem geometrically can simplify the solution.

Summary

The problem involves finding the locus of a complex number $z$ given that $\frac{z - i}{z - 1}$ is purely imaginary. This leads to the equation of a circle. The minimum distance from a fixed point to the circle is then calculated using the distance formula and the radius of the circle. The final answer is $2\sqrt{2}$.

Final Answer The final answer is $\boxed{2\sqrt{2}}$, which corresponds to option (D).