Convert the complex number `z=(i-1)/(cospi/3+isinpi/3)`in the polar form.

Explanation

To convert the complex number \( z = \frac{i - 1}{\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}} \) into polar form, we will follow these steps:

### Step 1: Simplify the denominator
The denominator is given as \( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \). We know that:
\[ \cos \frac{\pi}{3} = \frac{1}{2}, \quad \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \]
Thus, the denominator becomes:
\[ \frac{1}{2} + i \frac{\sqrt{3}}{2} \]

### Step 2: Rewrite \( z \)
Now we can rewrite \( z \):
\[ z = \frac{i - 1}{\frac{1}{2} + i \frac{\sqrt{3}}{2}} \]

### Step 3: Multiply numerator and denominator by the conjugate of the denominator
To eliminate the complex number in the denominator, we multiply the numerator and denominator by the conjugate of the denominator:
\[ z = \frac{(i - 1)(\frac{1}{2} - i \frac{\sqrt{3}}{2})}{(\frac{1}{2} + i \frac{\sqrt{3}}{2})(\frac{1}{2} - i \frac{\sqrt{3}}{2})} \]

### Step 4: Calculate the denominator
Using the formula \( (a + bi)(a - bi) = a^2 + b^2 \):
\[ \text{Denominator} = \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1 \]

### Step 5: Calculate the numerator
Now, let's calculate the numerator:
\[ (i - 1)(\frac{1}{2} - i \frac{\sqrt{3}}{2}) = i \cdot \frac{1}{2} - i \cdot i \frac{\sqrt{3}}{2} - 1 \cdot \frac{1}{2} + 1 \cdot i \frac{\sqrt{3}}{2} \]
\[ = \frac{i}{2} + \frac{\sqrt{3}}{2} - \frac{1}{2} + i \frac{\sqrt{3}}{2} \]
Combining real and imaginary parts:
\[ = \left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right) + i \left(\frac{1}{2} + \frac{\sqrt{3}}{2}\right) \]
\[ = \frac{\sqrt{3} - 1}{2} + i \frac{1 + \sqrt{3}}{2} \]

### Step 6: Write \( z \)
Thus, we have:
\[ z = \frac{\sqrt{3} - 1}{2} + i \frac{1 + \sqrt{3}}{2} \]

### Step 7: Find \( r \) and \( \theta \)
In polar form, \( z = r(\cos \theta + i \sin \theta) \).
To find \( r \):
\[ r = \sqrt{\left(\frac{\sqrt{3} - 1}{2}\right)^2 + \left(\frac{1 + \sqrt{3}}{2}\right)^2} \]
Calculating:
\[ = \sqrt{\frac{(\sqrt{3} - 1)^2 + (1 + \sqrt{3})^2}{4}} = \frac{1}{2} \sqrt{(\sqrt{3} - 1)^2 + (1 + \sqrt{3})^2} \]
Expanding:
\[ (\sqrt{3} - 1)^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3} \]
\[ (1 + \sqrt{3})^2 = 1 + 2\sqrt{3} + 3 = 4 + 2\sqrt{3} \]
Adding:
\[ (4 - 2\sqrt{3}) + (4 + 2\sqrt{3}) = 8 \]
Thus:
\[ r = \frac{1}{2} \sqrt{8} = \sqrt{2} \]

### Step 8: Find \( \theta \)
To find \( \theta \):
\[ \tan \theta = \frac{\text{Imaginary part}}{\text{Real part}} = \frac{\frac{1 + \sqrt{3}}{2}}{\frac{\sqrt{3} - 1}{2}} = \frac{1 + \sqrt{3}}{\sqrt{3} - 1} \]
This can be simplified using the tangent addition formula:
\[ \theta = 75^\circ \text{ or } \frac{5\pi}{12} \]

### Final Polar Form
Thus, the polar form of \( z \) is:
\[ z = \sqrt{2} \left( \cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12} \right) \]