Key Concepts and Formulas

• Complex Number Representation: A complex number $z$ can be represented as $z = x + iy$, where $x$ and $y$ are real numbers, and $i = \sqrt{-1}$.
• Modulus of a Complex Number: The modulus (or magnitude) of a complex number $z = x + iy$ is given by $|z| = \sqrt{x^2 + y^2}$, and thus $|z|^2 = x^2 + y^2$.
• Equality of Complex Numbers: Two complex numbers $a + ib$ and $c + id$ are equal if and only if their real parts are equal ($a = c$) and their imaginary parts are equal ($b = d$).

Step-by-Step Solution

Step 1: Substitute the algebraic form of $z$ and its modulus into the given equation. We are given $z = x + iy$ and $z^2 = i|z|^2$. We substitute these expressions into the given equation to express it in terms of real and imaginary components. $(x + iy)^2 = i(x^2 + y^2)$

Step 2: Expand and simplify the equation. Expand the left side and express both sides in the form $a + ib$. This prepares the equation for separating real and imaginary parts. $(x + iy)^2 = x^2 + 2ixy + (iy)^2 = x^2 + 2ixy - y^2 = (x^2 - y^2) + i(2xy)$ The right side is: $i(x^2 + y^2) = 0 + i(x^2 + y^2)$ Now, we have: $(x^2 - y^2) + i(2xy) = 0 + i(x^2 + y^2)$

Step 3: Equate the real and imaginary parts. Using the principle of equality of complex numbers, we equate the real parts and the imaginary parts of the equation separately. This transforms the complex equation into a system of two real equations. Equating the real parts: $x^2 - y^2 = 0 \quad \text{(Equation 1)}$ Equating the imaginary parts: $2xy = x^2 + y^2 \quad \text{(Equation 2)}$

Step 4: Solve the system of real equations. Solve the two equations to find the relationship between $x$ and $y$. From Equation 1: $x^2 = y^2$ This implies $y = x$ or $y = -x$.

• Case 1: $y = x$ Substitute $y = x$ into Equation 2: $2x(x) = x^2 + x^2$ $2x^2 = 2x^2$ This equation holds true for all $x$. Since $z$ is a non-zero complex number, $x$ cannot be zero. Thus, $y = x$ is a valid solution when $x \neq 0$.

• Case 2: $y = -x$ Substitute $y = -x$ into Equation 2: $2x(-x) = x^2 + (-x)^2$ $-2x^2 = x^2 + x^2$ $-2x^2 = 2x^2$ $4x^2 = 0$ $x = 0$ If $x = 0$, then $y = -x = 0$. This implies $z = 0$, which contradicts the condition that $z$ is a non-zero complex number. Therefore, this case is invalid.

Step 5: Determine the locus of $z$. The only valid solution is $y = x$ with $x \neq 0$. This represents a straight line passing through the origin with a slope of 1, excluding the origin itself.

Common Mistakes & Tips

• Forgetting the non-zero condition: Always remember to check if your solution satisfies all the given conditions, especially the "non-zero" condition.
• Incorrect algebraic manipulation: Double-check your algebraic manipulations, especially when expanding and simplifying complex expressions.
• Not considering all cases: Make sure to consider all possible cases when solving equations. For example, when $x^2 = y^2$, consider both $y = x$ and $y = -x$.

Summary

By substituting $z = x + iy$ into the given equation $z^2 = i|z|^2$ and equating the real and imaginary parts, we obtained a system of two real equations. Solving this system, we found that the only valid solution for a non-zero complex number $z$ is $y = x$. Therefore, $z$ lies on the line $y = x$.

Final Answer The final answer is \boxed{y = x}, which corresponds to option (C).