Key Concepts and Formulas

• A complex number $z$ is real if and only if $Im(z) = 0$.
• If $z = x + iy$, where $x$ and $y$ are real, then $z^2 = (x+iy)^2 = x^2 - y^2 + 2ixy$.
• The equation of a circle with center $(a, b)$ and radius $r$ is $(x-a)^2 + (y-b)^2 = r^2$.

Step-by-Step Solution

Step 1: Represent the complex number and state the given condition

Let $z = x + iy$, where $x$ and $y$ are real numbers. We are given that $z \ne 1$ and that $\frac{z^2}{z-1}$ is real. This means $Im\left(\frac{z^2}{z-1}\right) = 0$.

Step 2: Express the numerator and denominator in terms of x and y

We have $z^2 = (x + iy)^2 = x^2 + 2ixy - y^2 = (x^2 - y^2) + i(2xy)$ and $z - 1 = (x - 1) + iy$. Therefore,

$\frac{z^2}{z-1} = \frac{(x^2 - y^2) + i(2xy)}{(x-1) + iy}$

Step 3: Rationalize the denominator

To find the imaginary part, we multiply the numerator and denominator by the conjugate of the denominator, which is $(x-1) - iy$:

$\frac{z^2}{z-1} = \frac{(x^2 - y^2) + i(2xy)}{(x-1) + iy} \cdot \frac{(x-1) - iy}{(x-1) - iy}$

Step 4: Simplify the expression

Multiplying the numerator and denominator, we get:

$\frac{z^2}{z-1} = \frac{[(x^2 - y^2)(x-1) + 2xy^2] + i[2xy(x-1) - y(x^2 - y^2)]}{(x-1)^2 + y^2}$

Step 5: Isolate the imaginary part and set it to zero

Since $\frac{z^2}{z-1}$ is real, its imaginary part must be zero. Therefore:

$Im\left(\frac{z^2}{z-1}\right) = \frac{2xy(x-1) - y(x^2 - y^2)}{(x-1)^2 + y^2} = 0$

Since $z \ne 1$, we know that $(x,y) \ne (1,0)$, so $(x-1)^2 + y^2 \ne 0$. Thus, the numerator must be zero:

$2xy(x-1) - y(x^2 - y^2) = 0$

Step 6: Simplify the equation and factor

Factor out $y$:

$y[2x(x-1) - (x^2 - y^2)] = 0$ $y[2x^2 - 2x - x^2 + y^2] = 0$ $y[x^2 - 2x + y^2] = 0$

Step 7: Analyze the two possible cases

This equation gives us two possibilities:

• Case 1: $y = 0$
• Case 2: $x^2 - 2x + y^2 = 0$

Step 8: Interpret the cases geometrically

• Case 1: $y = 0$ means that $z = x$, so $z$ lies on the real axis.
• Case 2: $x^2 - 2x + y^2 = 0$ can be rewritten as $(x-1)^2 + y^2 = 1$. This is the equation of a circle with center $(1, 0)$ and radius $1$. Since the distance from the center $(1,0)$ to the origin $(0,0)$ is 1 (which equals the radius), the circle passes through the origin.

Step 9: Consider the constraint z != 1

The condition $z \ne 1$ means that $(x, y) \ne (1, 0)$.

• If $y=0$ (real axis), then $z=1$ is the point $(1,0)$ on the real axis. So the real axis solution excludes the point $(1,0)$.
• The point $(1,0)$ does NOT satisfy the circle equation: $(1-1)^2 + 0^2 = 0 \ne 1$. So $z=1$ is not a point on the circle.

Therefore, the locus of $z$ is either the real axis (excluding $z=1$) or the circle $(x-1)^2 + y^2 = 1$, which passes through the origin.

Common Mistakes & Tips

• Remember to rationalize the denominator when dealing with complex fractions to isolate the real and imaginary parts.
• Don't forget to consider the given constraints. In this case, $z \ne 1$ is crucial.
• When you have a product equal to zero, remember that it means one or both factors must be zero.

Summary

We found the locus of $z$ given that $\frac{z^2}{z-1}$ is real. By setting the imaginary part of the expression to zero, we derived the condition $y(x^2 + y^2 - 2x) = 0$. This leads to two distinct loci: the real axis ($y=0$) or a circle with center $(1,0)$ and radius $1$ ($x^2 + y^2 - 2x = 0$), which passes through the origin. The constraint $z \ne 1$ means that the point $(1,0)$ is excluded from the real axis solution, but it is not on the circle solution. Therefore, the point represented by the complex number $z$ lies either on the real axis or on a circle passing through the origin. This corresponds to option (A).

Final Answer The final answer is \boxed{either on the real axis or a circle passing through the origin}, which corresponds to option (A).