Find the modulus of `(1+i)/(1-i)-(1-i)/(1+i)`.

Explanation

To find the modulus of the complex number \((1+i)/(1-i)-(1-i)/(1+i)\), we will follow these steps:

### Step 1: Define the expression
Let \( z = \frac{1+i}{1-i} - \frac{1-i}{1+i} \).

### Step 2: Find a common denominator
To combine the two fractions, we need a common denominator. The common denominator is \((1-i)(1+i)\).

Thus, we rewrite \( z \) as:
\[ z = \frac{(1+i)(1+i) - (1-i)(1-i)}{(1-i)(1+i)} \]

### Step 3: Simplify the numerator
Now we will simplify the numerator:
- The first term:
\[ (1+i)(1+i) = 1 + 2i + i^2 = 1 + 2i - 1 = 2i \]
- The second term:
\[ (1-i)(1-i) = 1 - 2i + i^2 = 1 - 2i - 1 = -2i \]

Now substituting back into the numerator:
\[ z = \frac{2i - (-2i)}{(1-i)(1+i)} = \frac{2i + 2i}{(1-i)(1+i)} = \frac{4i}{(1-i)(1+i)} \]

### Step 4: Simplify the denominator
Now simplify the denominator:
\[ (1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 2 \]

### Step 5: Combine the results
Now substituting the denominator back into the expression for \( z \):
\[ z = \frac{4i}{2} = 2i \]

### Step 6: Find the modulus of \( z \)
The modulus of a complex number \( a + bi \) is given by:
\[ |z| = \sqrt{a^2 + b^2} \]
In our case, \( z = 0 + 2i \), so:
\[ |z| = \sqrt{0^2 + 2^2} = \sqrt{4} = 2 \]

### Final Answer
The modulus of the given complex number is \( 2 \).

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