Convert the following in polar form,
(ii) `((sqrt3 -1)-(sqrt3+1)i)/((2sqrt2))`

Answer

To convert the complex number \(\frac{\sqrt{3} - 1 - (\sqrt{3} + 1)i}{2\sqrt{2}}\) into polar form, we will follow these steps:

### Step 1: Simplify the Expression
First, we simplify the numerator:
\[ \sqrt{3} - 1 - (\sqrt{3} + 1)i = \sqrt{3} - 1 - \sqrt{3}i - i \]
This can be rearranged as:
\[ (\sqrt{3} - 1) - (\sqrt{3} + 1)i \]

### Step 2: Separate Real and Imaginary Parts
Now, we can identify the real part \(a\) and the imaginary part \(b\):
\[ a = \sqrt{3} - 1, \quad b = -(\sqrt{3} + 1) \]

### Step 3: Calculate the Modulus \(r\)
The modulus \(r\) of a complex number \(a + bi\) is given by:
\[ r = \sqrt{a^2 + b^2} \]
Substituting the values of \(a\) and \(b\):
\[ r = \sqrt{(\sqrt{3} - 1)^2 + (-(\sqrt{3} + 1))^2} \]
Calculating \(a^2\):
\[ (\sqrt{3} - 1)^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3} \]
Calculating \(b^2\):
\[ (-(\sqrt{3} + 1))^2 = (\sqrt{3} + 1)^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3} \]
Now, substituting back into the modulus formula:
\[ r = \sqrt{(4 - 2\sqrt{3}) + (4 + 2\sqrt{3})} = \sqrt{8} = 2\sqrt{2} \]

### Step 4: Calculate the Argument \(\theta\)
The argument \(\theta\) is given by:
\[ \theta = \tan^{-1}\left(\frac{b}{a}\right) \]
Substituting \(a\) and \(b\):
\[ \theta = \tan^{-1}\left(\frac{-(\sqrt{3} + 1)}{\sqrt{3} - 1}\right) \]
To simplify this, we can rationalize:
\[ \theta = \tan^{-1}\left(-\frac{\sqrt{3} + 1}{\sqrt{3} - 1}\right) \]
This can be further simplified using known angles. We find that:
\[ \theta = -\frac{5\pi}{12} \]

### Step 5: Write in Polar Form
The polar form of a complex number is given by:
\[ r e^{i\theta} \]
Thus, substituting the values we found:
\[ \frac{\sqrt{3} - 1 - (\sqrt{3} + 1)i}{2\sqrt{2}} = \frac{2\sqrt{2}}{2\sqrt{2}} e^{-i\frac{5\pi}{12}} = e^{-i\frac{5\pi}{12}} \]

### Final Answer
The polar form of the given complex number is:
\[ e^{-i\frac{5\pi}{12}} \]