Find the modulus and argument of the following :
(i) `-sqrt (3)+i`
(ii) `-1-isqrt(3)`
(iii) `5+12i`
(iv) `3(cos300^(@)-isin30^(@))`

Explanation

Let's solve the question step by step.

### Question: Find the modulus and argument of the following:
(i) \(-\sqrt{3} + i\)
(ii) \(-1 - i\sqrt{3}\)
(iii) \(5 + 12i\)
(iv) \(3(\cos 300^\circ - i \sin 30^\circ)\)

---

### (i) For \(-\sqrt{3} + i\):

1. **Find the modulus**:
\[ |z| = \sqrt{x^2 + y^2} = \sqrt{(-\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \]

2. **Find the argument**:
\[ \tan \theta = \frac{y}{x} = \frac{1}{-\sqrt{3}} = -\frac{1}{\sqrt{3}} \]
This corresponds to an angle in the second quadrant:
\[ \theta = 150^\circ \]

**Final answer for (i)**: Modulus = 2, Argument = \(150^\circ\)

---

### (ii) For \(-1 - i\sqrt{3}\):

1. **Find the modulus**:
\[ |z| = \sqrt{x^2 + y^2} = \sqrt{(-1)^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

2. **Find the argument**:
\[ \tan \theta = \frac{y}{x} = \frac{-\sqrt{3}}{-1} = \sqrt{3} \]
This corresponds to an angle in the third quadrant:
\[ \theta = 240^\circ \]

**Final answer for (ii)**: Modulus = 2, Argument = \(240^\circ\)

---

### (iii) For \(5 + 12i\):

1. **Find the modulus**:
\[ |z| = \sqrt{x^2 + y^2} = \sqrt{(5)^2 + (12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \]

2. **Find the argument**:
\[ \tan \theta = \frac{y}{x} = \frac{12}{5} \]
Thus, \(\theta = \tan^{-1}\left(\frac{12}{5}\right)\).

**Final answer for (iii)**: Modulus = 13, Argument = \(\tan^{-1}\left(\frac{12}{5}\right)\)

---

### (iv) For \(3(\cos 300^\circ - i \sin 30^\circ)\):

1. **Convert to standard form**:
\[ z = 3(\cos 300^\circ - i \sin 30^\circ) = 3(\cos 300^\circ) - 3(i \sin 30^\circ) \]
Knowing that \(\cos 300^\circ = \frac{1}{2}\) and \(\sin 30^\circ = \frac{1}{2}\):
\[ z = 3 \left(\frac{1}{2}\right) - 3\left(i \cdot \frac{1}{2}\right) = \frac{3}{2} - \frac{3}{2}i \]

2. **Find the modulus**:
\[ |z| = \sqrt{\left(\frac{3}{2}\right)^2 + \left(-\frac{3}{2}\right)^2} = \sqrt{\frac{9}{4} + \frac{9}{4}} = \sqrt{\frac{18}{4}} = \frac{3\sqrt{2}}{2} \]

3. **Find the argument**:
\[ \tan \theta = \frac{-\frac{3}{2}}{\frac{3}{2}} = -1 \]
This corresponds to an angle in the fourth quadrant:
\[ \theta = 315^\circ \]

**Final answer for (iv)**: Modulus = \(\frac{3\sqrt{2}}{2}\), Argument = \(315^\circ\)

---

### Summary of Answers:
1. (i) Modulus = 2, Argument = \(150^\circ\)
2. (ii) Modulus = 2, Argument = \(240^\circ\)
3. (iii) Modulus = 13, Argument = \(\tan^{-1}\left(\frac{12}{5}\right)\)
4. (iv) Modulus = \(\frac{3\sqrt{2}}{2}\), Argument = \(315^\circ\)

---