If `sum_(k=0)^(200)i^k+prod_(p=1)^(50)i^p =x+iy` then `(x,y)` is

Options

A. `(0,1)`
B. `(1,-1)`
C. `(2,3)`
D. `(4,8)`

Answer

To solve the problem, we need to evaluate the expression given in the question:

\[ \sum_{k=0}^{200} i^k + \prod_{p=1}^{50} i^p = x + iy \]

### Step 1: Evaluate the summation \(\sum_{k=0}^{200} i^k\)

The powers of \(i\) cycle every 4 terms:
- \(i^0 = 1\)
- \(i^1 = i\)
- \(i^2 = -1\)
- \(i^3 = -i\)
- \(i^4 = 1\), and so on.

Since the powers repeat every 4, we can find how many complete cycles fit into 200:
\[ 200 \div 4 = 50 \quad \text{(with a remainder of 0)} \]
This means we have 50 complete cycles of \(1 + i - 1 - i\).

The sum of one complete cycle is:
\[ 1 + i - 1 - i = 0 \]
Thus, the total sum for 50 cycles is:
\[ 50 \times 0 = 0 \]

### Step 2: Evaluate the product \(\prod_{p=1}^{50} i^p\)

The product can be simplified as:
\[ \prod_{p=1}^{50} i^p = i^{1 + 2 + 3 + \ldots + 50} \]
The sum of the first 50 natural numbers is:
\[ \frac{50 \times 51}{2} = 1275 \]
Thus, we have:
\[ \prod_{p=1}^{50} i^p = i^{1275} \]

### Step 3: Simplify \(i^{1275}\)

To simplify \(i^{1275}\), we find the remainder when 1275 is divided by 4:
\[ 1275 \div 4 = 318 \quad \text{(remainder 3)} \]
Thus,
\[ i^{1275} = i^3 = -i \]

### Step 4: Combine the results

Now we combine the results from Steps 1 and 2:
\[ \sum_{k=0}^{200} i^k + \prod_{p=1}^{50} i^p = 0 + (-i) = -i \]
This can be expressed as:
\[ 0 + (-1)i \]
Thus, we have:
\[ x + iy = 0 - i \]
From this, we can identify:
- \(x = 0\)
- \(y = -1\)

### Final Answer

The coordinates \((x, y)\) are:
\[ (0, -1) \]