Animated Solution for Mathematics - Complex Numbers: If $z$ is a complex number such that $\frac{z-i}{z-1}$ is purely imaginary, then the minimum value of $|z-(3+3i)|$ is:

A Masterclass

Welcome, students. Today, we are going to peel back the layers of a classic JEE Advanced problem. It is not just about crunching numbers; it is about visualizing the hidden geometry of the complex plane.

We are given a complex number $z$ and a condition: $\frac{z-i}{z-1}$ is purely imaginary. Our mission is to find the minimum value of $|z-(3+3i)|$. Let us embark on this journey.

Decoding the Locus

When you see a condition like "purely imaginary," your first instinct should be to translate it into the language of real and imaginary parts. Let $z=x+iy$, where $x$ and $y$ are real numbers.

The expression becomes:

This looks intimidating, but we have a powerful tool: rationalization. We multiply the numerator and the denominator by the conjugate of the denominator, which is $(x-1)-iy$.

This clears the imaginary component from the denominator, leaving us with a real value: $(x-1{)}^2+y^2$. Now, the numerator becomes $(x+i(y-1))((x-1)-iy)$.

Expanding this, we focus only on the real part because the problem dictates the entire fraction is purely imaginary. The real part is $x(x-1)+y(y-1)$. Setting this to zero, we get:

The Circle Revealed

Look at that equation: $x^2-x+y^2-y=0$. Does it look familiar? It is the equation of a circle!

To see it clearly, we complete the square. We add $\frac{1}{4}$ to both sides for the $x$ terms and $\frac{1}{4}$ for the $y$ terms. This transforms our equation into:

We have successfully identified the locus of $z$. It is a circle centered at $C(\frac{1}{2},\frac{1}{2})$ with a radius $r=\frac{1}{\sqrt{2}}$. Our point $z$ is not just wandering aimlessly; it is constrained to dance along the circumference of this circle.

The Final Minimization

Now, we pivot to the second part of the problem. We need to minimize $|z-(3+3i)|$. Geometrically, this is the distance between our moving point $z$ and the fixed point $P(3,3)$.

In the world of coordinate geometry, the shortest distance from a point to a circle is found along the normal—the line connecting the external point $P$ to the center $C$. The minimum distance is simply the distance $PC$ minus the radius $r$.

First, we calculate $PC$ using the distance formula:

Finally, we subtract the radius:

By rationalizing $\frac{1}{\sqrt{2}}$ as $\frac{\sqrt{2}}{2}$, we get:

And there you have it! Through the power of algebraic manipulation and geometric intuition, we have navigated the complex plane to find our answer: $2\sqrt{2}$. Never fear the complexity of these problems; break them down, visualize the path, and the math will always guide you home.