If the points z_1, z_2, z_3, z_4 lie on the same circle…
If the points $z_1, z_2, z_3, z_4$ lie on the same circle, show that the following conditions are satisfied: $$ \begin{aligned} \text{Re}(z_1)\text{Im}(z_1) &= \text{Re}(z_2)\text{Im}(z_2)=\text{Re}(z_3)\text{Im}(z_3)=\text{Re}(z_4)\text{Im}(z_4), \\ \text{Re}(z_4) &= \text{Re}(z_1)\text{Re}(z_2)\text{Re}(z_3). \end{aligned} $$
Explanation of Method
Before diving into algebra, it's worth checking whether the claim can possibly be true for any four points on any circle. The first condition says $x_ky_k$ is the same constant for all four points — that only happens if all four points also lie on one specific rectangular hyperbola $xy=\text{const}$, which is not true of a generic quadruple of points on a circle. So the statement can't hold in general; it must implicitly describe a special configuration. This solution first disproves the literal general claim with a concrete counterexample, then identifies and proves the genuine classical result that this problem is almost certainly pointing at.
Step 1 · Counterexample to the literal claim
Take the unit circle $|z|=1$ and four ordinary points on it: $z_1=1$, $z_2=i$, $z_3=-1$, $z_4 = \dfrac{\sqrt3}{2}+\dfrac{i}{2}$ (at angle $30°$). Computing $\text{Re}(z_k)\text{Im}(z_k)$ for each:
$$ 1\cdot 0 = 0, \qquad 0\cdot 1 = 0, \qquad (-1)\cdot 0 = 0, \qquad \frac{\sqrt3}{2}\cdot\frac12 = \frac{\sqrt3}{4} \neq 0 $$
The four products are not all equal, even though all four points genuinely lie on the same circle. So the first condition — and with it the whole claim — is false for four arbitrary points on a circle. This is not a general theorem.
Step 2 · What the conditions actually describe
The condition $x_ky_k=c$ (constant) for all four points says every $z_k$ lies on the rectangular hyperbola $xy=c$. So the statement only makes sense for the special case where $z_1,z_2,z_3,z_4$ are the (up to four) intersection points of a circle and a rectangular hyperbola — not for an arbitrary quadruple on a circle. Restricting further to a circle centered at the origin, $x^2+y^2=r^2$, gives the cleanest version of this classical result, proved below.
Step 3 · Set up the intersection
Let the circle be $x^2+y^2=r^2$ and the hyperbola be $xy=k$. Substituting $y=k/x$ into the circle equation and clearing denominators:
$$ x^2+\frac{k^2}{x^2}=r^2 \quad\Longrightarrow\quad x^4 - r^2x^2 + k^2 = 0 $$
This is a biquadratic (only even powers of $x$) — its four roots $x_1,x_2,x_3,x_4$ automatically come in $\pm$ pairs, and are exactly the real parts of the (up to four) intersection points.
Step 4 · Apply Vieta's formulas
For $x^4+0\cdot x^3-r^2x^2+0\cdot x+k^2=0$, Vieta's formulas give the product of all four roots as the constant term (since the leading coefficient is $1$ and the degree is even):
$$ \boxed{x_1x_2x_3x_4 = k^2} $$
Since $y_k = k/x_k$ for each point, the imaginary parts satisfy the mirror identity:
$$ y_1y_2y_3y_4 = \frac{k^4}{x_1x_2x_3x_4} = \frac{k^4}{k^2} = k^2 $$
So both $\text{Re}(z_1)\text{Re}(z_2)\text{Re}(z_3)\text{Re}(z_4) = k^2$ and $\text{Im}(z_1)\text{Im}(z_2)\text{Im}(z_3)\text{Im}(z_4)=k^2$ — a clean, genuinely general fact about the four intersection points.
Step 5 · Worked numerical example
Take $r^2=2$ and $k=\tfrac12$: the biquadratic is $x^4-2x^2+\tfrac14=0$, giving $x^2 = \dfrac{2\pm\sqrt3}{2}$, so:
$$ x \in \left\{\pm\sqrt{\tfrac{2+\sqrt3}{2}},\ \pm\sqrt{\tfrac{2-\sqrt3}{2}}\right\} \approx \{\pm1.366,\ \pm0.366\} $$
with corresponding $y=k/x$. The four intersection points are approximately $(1.366,0.366)$, $(-1.366,-0.366)$, $(0.366,1.366)$, $(-0.366,-1.366)$, each satisfying $x^2+y^2=2$ and $xy=0.5$. Their product of real parts: $1.366\times(-1.366)\times0.366\times(-0.366) \approx 0.25 = k^2$. ✓
Circle $x^2+y^2=2$ (gray) meeting hyperbola $xy=\tfrac12$ (blue, two branches) at four points; their real parts multiply to $k^2=0.25$, and so do their imaginary parts.
Result
As stated for four arbitrary points on a circle, the claim is false (counterexample in Step 1). The statement only holds for the special case where $z_1,z_2,z_3,z_4$ are the intersection points of a circle centered at the origin with a rectangular hyperbola $xy=k$, in which case the genuine, provable identity is:
obtained directly from Vieta's formulas applied to the biquadratic $x^4-r^2x^2+k^2=0$ satisfied by the four real parts. (The condition "$\text{Re}(z_4)=\text{Re}(z_1)\text{Re}(z_2)\text{Re}(z_3)$" from the original statement is not a valid consequence of this setup — it would additionally require one specific root to equal $1$, which is not generally true.)