Question:

Step 1: Understand the definition of Set A Set A consists of all complex numbers $z$ whose argument is $\frac{\pi}{4}$. Geometrically, this represents a ray starting from the origin $(0,0)$ and extending infinitely, making an angle of $45^\circ$ with the positive real axis. The origin itself is excluded because $\arg(0)$ is undefined. Thus, for $z \in A$, we can write $z = r e^{i\pi/4}$ for some real number $r > 0$. In Cartesian coordinates, if $z = x+iy$, then $\frac{y}{x} = \tan(\frac{\pi}{4}) = 1$, so $y=x$. Since the angle is in the first quadrant, $x > 0$ and $y > 0$. Step 2: Understand the definition of Set B Set B consists of all complex numbers $z$ such that the argument of $z - (3+3i)$ is $\frac{2\pi}{3}$. Geometrically, this represents a ray starting from the point $P(3,3)$ and extending infinitely, making an angle of $\frac{2\pi}{3}$ (or $120^\circ$) with the positive real axis. The starting point $P(3,3)$ itself is excluded because $\arg(0)$ is undefined. Thus, for $z \in B$, we can write $z - (3+3i) = s e^{i2\pi/3}$ for some real number $s > 0$. Rearranging, $z = (3+3i) + s e^{i2\pi/3}$ for $s > 0$. Step 3: Find the intersection $A \cap B$ To find the intersection, we look for complex numbers $z$ that satisfy both conditions. Let $z \in A \cap B$. Then we must have: $$ r e^{i\pi/4} = (3+3i) + s e^{i2\pi/3} $$ where $r > 0$ and $s > 0$. We use Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$ to express the complex exponentials: $e^{i\pi/4} = \cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}$ $e^{i2\pi/3} = \cos(\frac{2\pi}{3}) + i\sin(\frac{2\pi}{3}) = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$ Substitute these into the equation: $$ r \left(\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}\right) = (3+3i) + s \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) $$ $$ \frac{r}{\sqrt{2}} + i\frac{r}{\sqrt{2}} = \left(3 - \frac{s}{2}\right) + i\left(3 + \frac{s\sqrt{3}}{2}\right) $$ Equating the real parts and imaginary parts separately: 1. Real parts: $\frac{r}{\sqrt{2}} = 3 - \frac{s}{2} \quad (*)$ 2. Imaginary parts: $\frac{r}{\sqrt{2}} = 3 + \frac{s\sqrt{3}}{2} \quad (**)$ From equations $(*)$ and $(**)$, since their left-hand sides are equal, their right-hand sides must also be equal: $$ 3 - \frac{s}{2} = 3 + \frac{s\sqrt{3}}{2} $$ Subtract 3 from both sides: $$ -\frac{s}{2} = \frac{s\sqrt{3}}{2} $$ Multiply by 2: $$ -s = s\sqrt{3} $$ Rearrange the terms: $$ s\sqrt{3} + s = 0 $$ Factor out $s$: $$ s(\sqrt{3} + 1) = 0 $$ Since $\sqrt{3} + 1 \neq 0$, it must be that $s = 0$. However, our definition of $z \in B$ requires $s > 0$. A value of $s=0$ would mean $z = 3+3i$, for which $\arg(z-(3+3i)) = \arg(0)$ is undefined. Thus, $3+3i \notin B$. Since we found $s=0$, which contradicts the condition $s>0$, there are no complex numbers $z$ that can satisfy both conditions simultaneously. Therefore, the intersection $A \cap B$ is an empty set, and the number of elements in the intersection, $n(A \cap B)$, is $0$. Step 4: Evaluate the options and state the final answer We found $n(A \cap B) = 0$. Let's check the given options: Option 1: 1 Option 2: $\sum_{r=0}^{99} i^r$ Option 3: 3 Option 4: 0 First, we evaluate Option 2: $\sum_{r=0}^{99} i^r$. The powers of $i$ follow a cycle of four terms: $i^0 = 1$ $i^1 = i$ $i^2 = -1$ $i^3 = -i$ The sum of one complete cycle is $1+i-1-i = 0$. The sum $\sum_{r=0}^{99} i^r$ contains $100$ terms (from $r=0$ to $r=99$). Since $100$ is a multiple of 4 ($100 = 25 \times 4$), the sum consists of 25 complete cycles of powers of $i$. Thus, $\sum_{r=0}^{99} i^r = 25 \times (1+i-1-i) = 25 \times 0 = 0$. Both Option 2 and Option 4 evaluate to 0. Since $n(A \cap B) = 0$, both Option 2 and Option 4 are correct answers. The final answer is $\boxed{0}$. This corresponds to Option 2 and Option 4.