The sum of principal arguments of complex numbers `1+i,-1+isqrt3,-sqrt3-i,sqrt3-i,i,-3i,2,-1`is

Options

A. `(11pi)/12`
B. `(13pi)/12`
C. `(12pi)/13`
D. `pi/15`

Answer

To find the sum of the principal arguments of the complex numbers \(1+i\), \(-1+\sqrt{3}i\), \(-\sqrt{3}-i\), \(\sqrt{3}-i\), \(i\), \(-3i\), \(2\), and \(-1\), we will calculate the principal argument for each complex number step by step.

### Step 1: Calculate the principal argument of \(1+i\)

The complex number \(1+i\) can be expressed in polar form.

- \(x = 1\), \(y = 1\)
- The modulus \(r = \sqrt{x^2 + y^2} = \sqrt{1^2 + 1^2} = \sqrt{2}\)
- The principal argument \(\theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}(1) = \frac{\pi}{4}\)

**Hint:** Use the formula \(\theta = \tan^{-1}\left(\frac{y}{x}\right)\) to find the argument.

### Step 2: Calculate the principal argument of \(-1+\sqrt{3}i\)

For the complex number \(-1+\sqrt{3}i\):

- \(x = -1\), \(y = \sqrt{3}\)
- The modulus \(r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2\)
- The principal argument \(\theta = \tan^{-1}\left(\frac{\sqrt{3}}{-1}\right)\). This is in the second quadrant, so \(\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\)

**Hint:** Remember to adjust the angle based on the quadrant.

### Step 3: Calculate the principal argument of \(-\sqrt{3}-i\)

For the complex number \(-\sqrt{3}-i\):

- \(x = -\sqrt{3}\), \(y = -1\)
- The modulus \(r = \sqrt{(-\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2\)
- The principal argument \(\theta = \tan^{-1}\left(\frac{-1}{-\sqrt{3}}\right)\). This is in the third quadrant, so \(\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}\)

**Hint:** For angles in the third quadrant, add \(\pi\) to the reference angle.

### Step 4: Calculate the principal argument of \(\sqrt{3}-i\)

For the complex number \(\sqrt{3}-i\):

- \(x = \sqrt{3}\), \(y = -1\)
- The modulus \(r = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2\)
- The principal argument \(\theta = \tan^{-1}\left(\frac{-1}{\sqrt{3}}\right)\). This is in the fourth quadrant, so \(\theta = -\frac{\pi}{6}\)

**Hint:** For angles in the fourth quadrant, use the negative of the reference angle.

### Step 5: Calculate the principal argument of \(i\)

For the complex number \(i\):

- \(x = 0\), \(y = 1\)
- The modulus \(r = \sqrt{0^2 + 1^2} = 1\)
- The principal argument \(\theta = \frac{\pi}{2}\)

**Hint:** The argument of \(i\) is a standard angle.

### Step 6: Calculate the principal argument of \(-3i\)

For the complex number \(-3i\):

- \(x = 0\), \(y = -3\)
- The modulus \(r = \sqrt{0^2 + (-3)^2} = 3\)
- The principal argument \(\theta = -\frac{\pi}{2}\)

**Hint:** The argument of \(-3i\) is also a standard angle.

### Step 7: Calculate the principal argument of \(2\)

For the complex number \(2\):

- \(x = 2\), \(y = 0\)
- The modulus \(r = \sqrt{2^2 + 0^2} = 2\)
- The principal argument \(\theta = 0\)

**Hint:** The argument of a positive real number is \(0\).

### Step 8: Calculate the principal argument of \(-1\)

For the complex number \(-1\):

- \(x = -1\), \(y = 0\)
- The modulus \(r = \sqrt{(-1)^2 + 0^2} = 1\)
- The principal argument \(\theta = \pi\)

**Hint:** The argument of a negative real number is \(\pi\).

### Step 9: Sum the principal arguments

Now we sum all the principal arguments calculated:

\[ \text{Sum} = \frac{\pi}{4} + \frac{2\pi}{3} + \frac{7\pi}{6} - \frac{\pi}{6} + \frac{\pi}{2} - \frac{\pi}{2} + 0 + \pi \]

To simplify, convert all terms to a common denominator (which is 12):

\[ \text{Sum} = \frac{3\pi}{12} + \frac{8\pi}{12} + \frac{14\pi}{12} - \frac{2\pi}{12} + \frac{6\pi}{12} - \frac{6\pi}{12} + 0 + \frac{12\pi}{12} \]

Combining these gives:

\[ \text{Sum} = \frac{3 + 8 + 14 - 2 + 6 - 6 + 0 + 12}{12}\pi = \frac{35}{12}\pi \]

### Final Answer

The sum of the principal arguments is \(\frac{35\pi}{12}\).