Convert `z=(i-1)/("cos"(pi)/(3)+"isin"(pi)/(3))` in polar form.

Answer

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

### Step 1: Identify the Denominator
The denominator can be expressed using Euler's formula:
\[ \cos\left(\frac{\pi}{3}\right) + i \sin\left(\frac{\pi}{3}\right) = e^{i\frac{\pi}{3}} \]
Calculating the values:
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, \quad \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
Thus, the denominator is:
\[ \frac{1}{2} + i \frac{\sqrt{3}}{2} = e^{i\frac{\pi}{3}} \]

### Step 2: Rationalize the Numerator
The numerator is \( i - 1 \). We can express this in the form \( a + bi \):
\[ i - 1 = -1 + i \]
Now, we can find the modulus and argument of this complex number.

### Step 3: Calculate the Modulus of the Numerator
The modulus \( r \) of \( -1 + i \) is given by:
\[ r = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Step 4: Calculate the Argument of the Numerator
The argument \( \theta \) can be found using:
\[ \tan(\theta) = \frac{b}{a} = \frac{1}{-1} = -1 \]
This corresponds to an angle in the second quadrant. Therefore, the angle is:
\[ \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \]

### Step 5: Express the Numerator in Polar Form
Thus, we can express the numerator in polar form:
\[ -1 + i = \sqrt{2} e^{i\frac{3\pi}{4}} \]

### Step 6: Combine the Results
Now substituting back into the expression for \( z \):
\[ z = \frac{\sqrt{2} e^{i\frac{3\pi}{4}}}{e^{i\frac{\pi}{3}}} \]
Using the property of exponents:
\[ z = \sqrt{2} e^{i\left(\frac{3\pi}{4} - \frac{\pi}{3}\right)} \]

### Step 7: Simplify the Argument
To simplify the argument:
\[ \frac{3\pi}{4} - \frac{\pi}{3} = \frac{9\pi}{12} - \frac{4\pi}{12} = \frac{5\pi}{12} \]
Thus, we have:
\[ z = \sqrt{2} e^{i\frac{5\pi}{12}} \]

### Step 8: Final Polar Form
The polar form of \( z \) is:
\[ z = \sqrt{2} \left( \cos\left(\frac{5\pi}{12}\right) + i \sin\left(\frac{5\pi}{12}\right) \right) \]