Question:

Step 1: Understand the geometric implication of the given condition. The given condition states that $\frac{z_3 - z_2}{z_1 - z_2}$ is a purely imaginary number. This means that the argument of this complex number is $\pm \frac{\pi}{2}$. Geometrically, this implies that the vector from $z_2$ to $z_3$ ($\vec{Z_2Z_3}$) is perpendicular to the vector from $z_2$ to $z_1$ ($\vec{Z_2Z_1}$). Therefore, the angle at vertex $B(z_2)$ in $\triangle ABC$ is $90^\circ$. $$ \arg\left(\frac{z_3 - z_2}{z_1 - z_2}\right) = \pm \frac{\pi}{2} $$ Step 2: Express the center of the square $z_4$ in terms of $z_1, z_2, z_3$. A square is drawn outwardly on side $AC$, with its center at $P(z_4)$. For a square with adjacent vertices $z_A$ and $z_C$, its center $z_P$ is given by the formula: $$ z_P = \frac{z_A+z_C}{2} + \frac{i}{2}(z_C-z_A) $$ Applying this for vertices $A(z_1)$ and $C(z_3)$, the center of the square $z_4$ is: $$ z_4 = \frac{z_1+z_3}{2} + \frac{i}{2}(z_3-z_1) $$ To relate $z_4$ to $z_2$, we subtract $z_2$ from both sides: $$ z_4 - z_2 = \frac{z_1+z_3}{2} + \frac{i}{2}(z_3-z_1) - z_2 $$ Rearranging the terms to group with $z_2$: $$ z_4 - z_2 = \frac{(z_1-z_2) + (z_3-z_2)}{2} + \frac{i}{2}((z_3-z_2) - (z_1-z_2)) $$ Let $A' = z_1-z_2$, $C' = z_3-z_2$, and $P' = z_4-z_2$. The equation can be written as: $$ P' = \frac{A'+C'}{2} + \frac{i}{2}(C'-A') $$ Step 3: Evaluate Option 3. Option 3 states $\arg\left(\frac{z_1 - z_2}{z_4 - z_2}\right) + \arg\left(\frac{z_3 - z_2}{z_4 - z_2}\right) = 0$. Using the notation from Step 2, this is equivalent to $\arg\left(\frac{A'}{P'}\right) + \arg\left(\frac{C'}{P'}\right) = 0$. By the properties of arguments, this can be combined as $\arg\left(\frac{A'C'}{(P')^2}\right) = 0$. For this argument to be $0$, the complex number $\frac{A'C'}{(P')^2}$ must be a positive real number. From Step 1, $\frac{C'}{A'}$ is purely imaginary. Let $\frac{C'}{A'} = k i$ for some real number $k \ne 0$. So, $C' = k i A'$. Substitute $C' = k i A'$ into the expression for $P'$: $$ P' = \frac{A' + k i A'}{2} + \frac{i}{2}(k i A' - A') $$ $$ P' = \frac{A'(1+ki)}{2} + \frac{A'(i^2k-i)}{2} = \frac{A'}{2}(1+ki-k-i) = \frac{A'}{2}((1-k) + i(k-1)) $$ $$ P' = \frac{A'}{2}(1-k)(1-i) $$ Now, let's compute the expression $\frac{A'C'}{(P')^2}$: $$ \frac{A'C'}{(P')^2} = \frac{A'(k i A')}{\left(\frac{A'}{2}(1-k)(1-i)\right)^2} = \frac{k i (A')^2}{\frac{(A')^2}{4}(1-k)^2(1-i)^2} $$ $$ = \frac{4 k i}{(1-k)^2 (1 - 2i + i^2)} = \frac{4 k i}{(1-k)^2 (-2i)} = \frac{-2k}{(1-k)^2} $$ For $\arg\left(\frac{A'C'}{(P')^2}\right) = 0$, the expression $\frac{-2k}{(1-k)^2}$ must be a positive real number. Since $(1-k)^2$ is always positive (for $k \ne 1$, and $k=1$ implies $P'=0$ which is not possible for a square center if $A' \ne 0$), we must have $-2k > 0$, which implies $k < 0$. This means that option 3 is correct if and only if $\frac{z_3 - z_2}{z_1 - z_2}$ is purely imaginary with a negative imaginary component (i.e., $\arg\left(\frac{z_3 - z_2}{z_1 - z_2}\right) = -\frac{\pi}{2}$). In this specific case, the arguments $\arg\left(\frac{z_1 - z_2}{z_4 - z_2}\right)$ and $\arg\left(\frac{z_3 - z_2}{z_4 - z_2}\right)$ have equal magnitudes and opposite signs, summing to 0. Step 4: Evaluate Option 4. Option 4 states that $z_1, z_2, z_3$ and $z_4$ lie on a circle. From Step 1, we know that the angle at $z_2$ is $\angle z_1 z_2 z_3 = 90^\circ$. Now consider the angle at $z_4$, which is $\angle z_1 z_4 z_3$. This is the angle $\angle CPA$. Let's find the complex number ratio $\frac{z_1-z_4}{z_3-z_4}$. Using the relations from Step 2: $$ z_1-z_4 = (z_1-z_2) - (z_4-z_2) = A' - P' $$ $$ z_3-z_4 = (z_3-z_2) - (z_4-z_2) = C' - P' $$ Substitute $P' = \frac{A'+C'}{2} + \frac{i}{2}(C'-A')$: $$ A' - P' = A' - \left( \frac{A'+C'}{2} + \frac{i}{2}(C'-A') \right) = \frac{A'-C'}{2} - \frac{i}{2}(C'-A') = \frac{1+i}{2}(A'-C') $$ $$ C' - P' = C' - \left( \frac{A'+C'}{2} + \frac{i}{2}(C'-A') \right) = \frac{C'-A'}{2} - \frac{i}{2}(C'-A') = \frac{1-i}{2}(C'-A') $$ Now, form the ratio: $$ \frac{A'-P'}{C'-P'} = \frac{\frac{1+i}{2}(A'-C')}{\frac{1-i}{2}(C'-A')} = \frac{-(1+i)}{1-i} = \frac{-(1+i)^2}{(1-i)(1+i)} = \frac{-(1+2i+i^2)}{1-i^2} = \frac{-2i}{2} = -i $$ Since $\frac{z_1-z_4}{z_3-z_4} = -i$, this implies that the vector $\vec{Z_4Z_1}$ is obtained by rotating the vector $\vec{Z_4Z_3}$ by $-\frac{\pi}{2}$ (clockwise $90^\circ$). Therefore, the angle at $z_4$ is $\angle z_1 z_4 z_3 = 90^\circ$. Consider the quadrilateral with vertices $z_1, z_2, z_3, z_4$. The sum of opposite angles is $\angle z_1 z_2 z_3 + \angle z_1 z_4 z_3 = 90^\circ + 90^\circ = 180^\circ$. A quadrilateral whose opposite angles sum to $180^\circ$ is cyclic, meaning all four vertices lie on a single circle. Thus, Option 4 is correct. The final answer is $\boxed{\text{3,4}}$.