Animated Solution for Mathematics - Complex Numbers: Let $z$ be a complex number such that $|\frac{z-2i}{z+i}|=2,z\ne -i$. Then $z$ lies on the circle of radius 2 and centre

Welcome, fellow traveler in the world of complex numbers! Today, we are going to unravel a beautiful problem that sits at the intersection of algebra and geometry.

We are given a complex number $z$ such that:

At first glance, this might look like a simple equation, but it is actually a hidden geometric treasure. Imagine you are standing on the complex plane with two fixed points: $2i$ and $-i$. The equation tells us that the distance from $z$ to $2i$ is exactly twice the distance from $z$ to $-i$.

Squaring for Clarity

Now, how do we turn this geometric intuition into a concrete equation? We start with the property of moduli: $\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$.

This allows us to rewrite our equation as:

By cross-multiplying, we get $|z-2i|=2|z+i|$. To remove the modulus signs, we square both sides:

Note that we squared the $2$ on the right side to get $4$. This is a common trap, so always be vigilant!

From Complex to Cartesian

Now, let us substitute $z=x+iy$. This transforms our complex equation into a familiar Cartesian one:

Using the definition of the modulus squared, $|a+ib{|}^2=a^2+b^2$, we expand this into:

Expanding the squares, we obtain:

Distributing the $4$ gives us:

The Circle Emerges

Now, let us bring everything to one side. The constant $4$ cancels out, which is a delightful surprise!

We are left with:

Dividing by $3$, we get:

To find the center and radius, we complete the square for $y$:

This simplifies to the standard form:

This is the standard equation of a circle $(x-h{)}^2+(y-k{)}^2=R^2$. Comparing our equation, we find the center is $(0,-2)$ and the radius is $2$. You have successfully navigated the path from a complex ratio to a geometric circle. Well done!