Convert the following into polar form :
(i) `-1+isqrt(3)`
(ii) `1-i`
(iii) `1-(1)/(i)`
(iv) `3-4i`
(v) `sin120^(@)-icos120^(@)`
(vi) 2

Explanation

To convert the given complex numbers into polar form, we will follow these steps for each part:

1. **Find the modulus (r)** of the complex number using the formula:
\[ r = \sqrt{x^2 + y^2} \]
where \( x \) is the real part and \( y \) is the imaginary part.

2. **Find the argument (θ)** using the formula:
\[ \tan(\theta) = \frac{y}{x} \]
and then determine the correct quadrant for \( θ \).

3. **Write the polar form** using the formula:
\[ z = r(\cos(\theta) + i\sin(\theta)) \]
or in exponential form:
\[ z = re^{i\theta} \]

Let's solve each part step by step:

### Part (i): Convert \(-1 + i\sqrt{3}\) into polar form

1. **Find r**:
\[ r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

2. **Find θ**:
\[ \tan(\theta) = \frac{\sqrt{3}}{-1} = -\sqrt{3} \]
This corresponds to \( \theta = \frac{2\pi}{3} \) (since the point is in the second quadrant).

3. **Polar form**:
\[ z = 2\left(\cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)\right) \]

### Part (ii): Convert \(1 - i\) into polar form

1. **Find r**:
\[ r = \sqrt{(1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

2. **Find θ**:
\[ \tan(\theta) = \frac{-1}{1} = -1 \]
This corresponds to \( \theta = \frac{7\pi}{4} \) (fourth quadrant).

3. **Polar form**:
\[ z = \sqrt{2}\left(\cos\left(\frac{7\pi}{4}\right) + i\sin\left(\frac{7\pi}{4}\right)\right) \]

### Part (iii): Convert \(1 - \frac{1}{i}\) into polar form

1. **Convert \(-\frac{1}{i}\)**:
\[ -\frac{1}{i} = i \]
Thus, the expression becomes \(1 + i\).

2. **Find r**:
\[ r = \sqrt{(1)^2 + (1)^2} = \sqrt{2} \]

3. **Find θ**:
\[ \tan(\theta) = \frac{1}{1} = 1 \]
This corresponds to \( \theta = \frac{\pi}{4} \).

4. **Polar form**:
\[ z = \sqrt{2}\left(\cos\left(\frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{4}\right)\right) \]

### Part (iv): Convert \(3 - 4i\) into polar form

1. **Find r**:
\[ r = \sqrt{(3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

2. **Find θ**:
\[ \tan(\theta) = \frac{-4}{3} \]
This gives \( \theta = \tan^{-1}\left(-\frac{4}{3}\right) \) (fourth quadrant).

3. **Polar form**:
\[ z = 5\left(\cos(\tan^{-1}(-\frac{4}{3})) + i\sin(\tan^{-1}(-\frac{4}{3}))\right) \]

### Part (v): Convert \(\sin(120^\circ) - i\cos(120^\circ)\) into polar form

1. **Convert to standard form**:
\[ \sin(120^\circ) - i\cos(120^\circ) = \frac{\sqrt{3}}{2} - i\left(-\frac{1}{2}\right) = \frac{\sqrt{3}}{2} + i\frac{1}{2} \]

2. **Find r**:
\[ r = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1 \]

3. **Find θ**:
\[ \tan(\theta) = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} \]
This corresponds to \( \theta = \frac{\pi}{6} \).

4. **Polar form**:
\[ z = 1\left(\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)\right) \]

### Part (vi): Convert \(2\) into polar form

1. **Find r**:
\[ r = 2 \]

2. **Find θ**:
\[ \theta = 0 \]

3. **Polar form**:
\[ z = 2\left(\cos(0) + i\sin(0)\right) \]

### Summary of Polar Forms:
1. \( z = 2\left(\cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)\right) \)
2. \( z = \sqrt{2}\left(\cos\left(\frac{7\pi}{4}\right) + i\sin\left(\frac{7\pi}{4}\right)\right) \)
3. \( z = \sqrt{2}\left(\cos\left(\frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{4}\right)\right) \)
4. \( z = 5\left(\cos(\tan^{-1}(-\frac{4}{3})) + i\sin(\tan^{-1}(-\frac{4}{3}))\right) \)
5. \( z = 1\left(\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)\right) \)
6. \( z = 2\left(\cos(0) + i\sin(0)\right) \)