Express the following in the polar form:
(i) `1+i`
(ii) `-1-i`
(iii) `1+sqrt(3)i`
(iv) `(-16)/(1+sqrt(3)i)`.

Answer

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

1. **Calculate the modulus (r)** using the formula \( r = \sqrt{a^2 + b^2} \), where \( a \) is the real part and \( b \) is the imaginary part of the complex number.
2. **Determine the argument (θ)** using the formula \( \theta = \tan^{-1}(\frac{b}{a}) \). We also need to consider the quadrant in which the complex number lies to find the correct angle.
3. **Express the complex number in polar form** using the formula \( z = r(\cos \theta + i \sin \theta) \).

Now, let's solve each part step by step.

### (i) For the complex number \( 1 + i \):

**Step 1: Calculate the modulus (r)**
\[ r = \sqrt{1^2 + 1^2} = \sqrt{2} \]

**Step 2: Determine the argument (θ)**
\[ \theta = \tan^{-1}\left(\frac{1}{1}\right) = \tan^{-1}(1) = \frac{\pi}{4} \]
(The angle is in the first quadrant.)

**Step 3: Express in polar form**
\[ z = r(\cos \theta + i \sin \theta) = \sqrt{2}\left(\cos\frac{\pi}{4} + i \sin\frac{\pi}{4}\right) \]

### (ii) For the complex number \( -1 - i \):

**Step 1: Calculate the modulus (r)**
\[ r = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2} \]

**Step 2: Determine the argument (θ)**
\[ \theta = \tan^{-1}\left(\frac{-1}{-1}\right) = \tan^{-1}(1) = \frac{\pi}{4} \]
(The angle is in the third quadrant, so we add \( \pi \)):
\[ \theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \]

**Step 3: Express in polar form**
\[ z = \sqrt{2}\left(\cos\frac{5\pi}{4} + i \sin\frac{5\pi}{4}\right) \]

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

**Step 1: Calculate the modulus (r)**
\[ r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

**Step 2: Determine the argument (θ)**
\[ \theta = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \]
(The angle is in the first quadrant.)

**Step 3: Express in polar form**
\[ z = 2\left(\cos\frac{\pi}{3} + i \sin\frac{\pi}{3}\right) \]

### (iv) For the complex number \( \frac{-16}{1 + \sqrt{3}i} \):

**Step 1: Rationalize the denominator**
Multiply numerator and denominator by the conjugate:
\[ \frac{-16(1 - \sqrt{3}i)}{(1 + \sqrt{3}i)(1 - \sqrt{3}i)} = \frac{-16(1 - \sqrt{3}i)}{1^2 - (\sqrt{3})^2} = \frac{-16(1 - \sqrt{3}i)}{1 - 3} = \frac{-16(1 - \sqrt{3}i)}{-2} = 8(1 - \sqrt{3}i) \]

**Step 2: Calculate the modulus (r)**
\[ r = \sqrt{(8)^2 + (-8\sqrt{3})^2} = \sqrt{64 + 192} = \sqrt{256} = 16 \]

**Step 3: Determine the argument (θ)**
\[ \theta = \tan^{-1}\left(\frac{-8\sqrt{3}}{8}\right) = \tan^{-1}(-\sqrt{3}) \]
(The angle is in the fourth quadrant):
\[ \theta = -\frac{\pi}{3} \]

**Step 4: Express in polar form**
\[ z = 16\left(\cos\left(-\frac{\pi}{3}\right) + i \sin\left(-\frac{\pi}{3}\right)\right) \]

### Final Answers:
1. \( z = \sqrt{2}\left(\cos\frac{\pi}{4} + i \sin\frac{\pi}{4}\right) \)
2. \( z = \sqrt{2}\left(\cos\frac{5\pi}{4} + i \sin\frac{5\pi}{4}\right) \)
3. \( z = 2\left(\cos\frac{\pi}{3} + i \sin\frac{\pi}{3}\right) \)
4. \( z = 16\left(\cos\left(-\frac{\pi}{3}\right) + i \sin\left(-\frac{\pi}{3}\right)\right) \)