The principal argument `(1+isqrt(3))^(2)` is

Options

A. `(pi)/(3)`
B. `-(pi)/(3)`
C. `(2pi)/(3)`
D. `-(2pi)/(3)`

Answer

To find the principal argument of the complex number \( (1 + i\sqrt{3})^2 \), we will follow these steps:

### Step 1: Expand the expression
We start with the expression \( (1 + i\sqrt{3})^2 \). We can use the formula for the square of a binomial, which is \( (a + b)^2 = a^2 + 2ab + b^2 \).

Here, \( a = 1 \) and \( b = i\sqrt{3} \).

\[ (1 + i\sqrt{3})^2 = 1^2 + 2(1)(i\sqrt{3}) + (i\sqrt{3})^2 \]

### Step 2: Calculate each term
Calculating each term:
- \( 1^2 = 1 \)
- \( 2(1)(i\sqrt{3}) = 2i\sqrt{3} \)
- \( (i\sqrt{3})^2 = i^2 \cdot 3 = -1 \cdot 3 = -3 \)

Putting it all together:
\[ (1 + i\sqrt{3})^2 = 1 + 2i\sqrt{3} - 3 \]

### Step 3: Combine like terms
Now combine the real parts:
\[ 1 - 3 + 2i\sqrt{3} = -2 + 2i\sqrt{3} \]

### Step 4: Identify the real and imaginary parts
From the expression \( -2 + 2i\sqrt{3} \):
- The real part \( a = -2 \)
- The imaginary part \( b = 2\sqrt{3} \)

### Step 5: Determine the quadrant
Since \( a < 0 \) and \( b > 0 \), the complex number lies in the second quadrant.

### Step 6: Calculate the argument
The argument \( \theta \) can be calculated using the tangent function:
\[ \tan(\theta) = \frac{b}{a} = \frac{2\sqrt{3}}{-2} = -\sqrt{3} \]

The angle whose tangent is \( -\sqrt{3} \) corresponds to \( \frac{\pi}{3} \) in the first quadrant. However, since we are in the second quadrant, we need to adjust:
\[ \theta = \pi - \frac{\pi}{3} = \frac{3\pi}{3} - \frac{\pi}{3} = \frac{2\pi}{3} \]

### Conclusion
Thus, the principal argument of \( (1 + i\sqrt{3})^2 \) is:
\[ \text{Principal Argument} = \frac{2\pi}{3} \]