Animated Solution for Mathematics - Complex Numbers: If $z\ne 1$ and $\frac{z^2}{z-1}$ is real, then the point represented by the complex number $z$ lies

Imagine you are standing on the complex plane, a vast, two-dimensional landscape where every point is a number. We are exploring a condition where a complex number $z$ satisfies the requirement that the expression

is purely real. This is a geometric quest to find the 'locus'—the set of all possible points $z$ that satisfy this condition.

Our first step is to translate the word 'real' into the language of complex numbers. A complex number $w$ is real if and only if it has no imaginary part, which is equivalent to the condition $w=\bar{w}$.

We set our expression equal to its conjugate:

This equation serves as the foundation for our entire derivation.

Using the properties of conjugates, we know that the conjugate of a quotient is the quotient of the conjugates, and the conjugate of $z^2$ is $(\bar{z}{)}^2$. Since $1$ is a real number, its conjugate is simply $1$.

This yields the following equality:

To clear the fractions, we cross-multiply:

Expanding this expression, we obtain:

Bringing all terms to one side, we have:

We can factor this expression by grouping. From the first part, we factor out $z\bar{z}$ to get $z\bar{z}(z-\bar{z})$. The second part is a difference of squares: $(z-\bar{z})(z+\bar{z})$.

Factoring out the common term $(z-\bar{z})$, we arrive at the simplified equation:

This product being zero provides two distinct, elegant cases.

Case 1: $z-\bar{z}=0$, which implies $z=\bar{z}$. This is the definition of the real axis. Thus, any point on the real axis (excluding $z=1$ to avoid division by zero) is a valid solution.

Case 2: $z\bar{z}-(z+\bar{z})=0$. To identify this shape, we use Cartesian coordinates where $z=x+iy$. We know that $z\bar{z}=x^2+y^2$ and $z+\bar{z}=2x$.

Substituting these into the equation, we get:

Completing the square, we find:

This is the equation of a circle with center $(1,0)$ and radius $1$. Note that this circle passes through the origin $(0,0)$.

The locus of $z$ is the union of the real axis and this circle. We must strictly adhere to the constraint $z eq 1$.

Since the point $(1,0)$ lies on the real axis, we must exclude it from the final set. Thus, the point $z$ lies either on the real axis (excluding $z=1$) or on the circle defined by $(x-1{)}^2+y^2=1$.