Convert the complex number ` (1+i)/(1-i)` in the polar form

Answer

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

### Step 1: Rationalize the complex number
Multiply the numerator and the denominator by the conjugate of the denominator.

\[ \frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} \]

### Step 2: Simplify the numerator
Using the formula \( (a+b)^2 = a^2 + 2ab + b^2 \):

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

### Step 3: Simplify the denominator
Using the difference of squares formula \( (a-b)(a+b) = a^2 - b^2 \):

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

### Step 4: Combine the results
Now we can combine the simplified numerator and denominator:

\[ \frac{2i}{2} = i \]

### Step 5: Convert to polar form
The polar form of a complex number \( z = r(\cos \theta + i \sin \theta) \) requires finding the modulus \( r \) and the argument \( \theta \).

1. **Modulus \( r \)**:
\[ r = |i| = 1 \]

2. **Argument \( \theta \)**:
Since \( i \) lies on the positive imaginary axis, the angle \( \theta \) is:
\[ \theta = \frac{\pi}{2} \]

### Step 6: Write in polar form
Now substituting \( r \) and \( \theta \) into the polar form:

\[ i = 1 \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right) \]

Thus, the polar form of the complex number \( \frac{1+i}{1-i} \) is:

\[ 1 \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right) \]

### Final Answer:
The polar form of the complex number \( \frac{1+i}{1-i} \) is:

\[ 1 \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right) \]

---