Express `z=(-1+isqrt(3))/(1+i)` in polar form and then find the modulus and argument of z. Hence deduce the value of cos `(5pi)/(12)`

Answer

To express \( z = \frac{-1 + i\sqrt{3}}{1 + i} \) in polar form and find its modulus and argument, we will follow these steps:

### Step 1: Multiply by the Conjugate
To simplify the expression, we multiply the numerator and the denominator by the conjugate of the denominator, which is \( 1 - i \).

\[ z = \frac{(-1 + i\sqrt{3})(1 - i)}{(1 + i)(1 - i)} \]

### Step 2: Simplify the Denominator
Calculating the denominator:

\[ (1 + i)(1 - i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2 \]

### Step 3: Simplify the Numerator
Now, for the numerator:

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

### Step 4: Combine the Results
Putting it all together, we have:

\[ z = \frac{(-1 + \sqrt{3}) + (1 + \sqrt{3})i}{2} \]

This can be rewritten as:

\[ z = \frac{-1 + \sqrt{3}}{2} + i \cdot \frac{1 + \sqrt{3}}{2} \]

### Step 5: Find Modulus
The modulus \( |z| \) is given by:

\[ |z| = \sqrt{\left(\frac{-1 + \sqrt{3}}{2}\right)^2 + \left(\frac{1 + \sqrt{3}}{2}\right)^2} \]

Calculating each term:

1. \( \left(\frac{-1 + \sqrt{3}}{2}\right)^2 = \frac{(-1 + \sqrt{3})^2}{4} = \frac{1 - 2\sqrt{3} + 3}{4} = \frac{4 - 2\sqrt{3}}{4} = 1 - \frac{\sqrt{3}}{2} \)

2. \( \left(\frac{1 + \sqrt{3}}{2}\right)^2 = \frac{(1 + \sqrt{3})^2}{4} = \frac{1 + 2\sqrt{3} + 3}{4} = \frac{4 + 2\sqrt{3}}{4} = 1 + \frac{\sqrt{3}}{2} \)

Adding these results:

\[ |z|^2 = \left(1 - \frac{\sqrt{3}}{2}\right) + \left(1 + \frac{\sqrt{3}}{2}\right) = 2 \]

Thus,

\[ |z| = \sqrt{2} \]

### Step 6: Find Argument
The argument \( \theta \) can be found using:

\[ \tan \theta = \frac{y}{x} = \frac{\frac{1 + \sqrt{3}}{2}}{\frac{-1 + \sqrt{3}}{2}} = \frac{1 + \sqrt{3}}{-1 + \sqrt{3}} \]

This simplifies to:

\[ \tan \theta = \frac{1 + \sqrt{3}}{-1 + \sqrt{3}} = \frac{(1 + \sqrt{3})(-1 - \sqrt{3})}{(-1 + \sqrt{3})(-1 - \sqrt{3})} = \frac{-1 - \sqrt{3} + \sqrt{3} + 3}{1 - 3} = \frac{2}{-2} = -1 \]

Thus, \( \theta = \tan^{-1}(-1) = -\frac{\pi}{4} \).

### Step 7: Polar Form
The polar form of \( z \) is:

\[ z = |z| \left( \cos \theta + i \sin \theta \right) = \sqrt{2} \left( \cos\left(-\frac{\pi}{4}\right) + i \sin\left(-\frac{\pi}{4}\right) \right) \]

### Step 8: Deduce \( \cos\left(\frac{5\pi}{12}\right) \)
From the argument, we can deduce that:

\[ \cos\left(\frac{5\pi}{12}\right) = \frac{\sqrt{3} - 1}{2\sqrt{2}} \]

### Summary of Results
- Modulus: \( |z| = \sqrt{2} \)
- Argument: \( \theta = -\frac{\pi}{4} \)
- Polar Form: \( z = \sqrt{2} \left( \cos\left(-\frac{\pi}{4}\right) + i \sin\left(-\frac{\pi}{4}\right) \right) \)
- Value of \( \cos\left(\frac{5\pi}{12}\right) = \frac{\sqrt{3} - 1}{2\sqrt{2}} \)