If `|z|=1`, then prove that `(z-1)/(z+1) (z ne -1)` is a purely imaginary number. What is the conclusion if z=1?

Answer

To prove that \(\frac{z-1}{z+1}\) is a purely imaginary number when \(|z|=1\) (and \(z \neq -1\)), we can follow these steps:

### Step 1: Express \(z\) in terms of its components
Let \(z = x + iy\), where \(x\) and \(y\) are real numbers. Given that \(|z| = 1\), we have:
\[ |z|^2 = x^2 + y^2 = 1 \]

### Step 2: Substitute \(z\) into the expression
We need to evaluate:
\[ \frac{z-1}{z+1} = \frac{(x + iy) - 1}{(x + iy) + 1} = \frac{(x - 1) + iy}{(x + 1) + iy} \]

### Step 3: Rationalize the denominator
To simplify this expression, we multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{((x - 1) + iy)((x + 1) - iy)}{((x + 1) + iy)((x + 1) - iy)} \]

### Step 4: Simplify the denominator
The denominator simplifies as follows:
\[ (x + 1)^2 + y^2 = (x + 1)^2 + (1 - x^2) = x^2 + 2x + 1 + 1 - x^2 = 2x + 2 = 2(x + 1) \]

### Step 5: Simplify the numerator
The numerator simplifies as follows:
\[ ((x - 1)(x + 1) - y^2) + i(y(x + 1) - y(x - 1)) = (x^2 - 1 - y^2) + i(2y) \]
Using \(x^2 + y^2 = 1\), we have \(x^2 - 1 - y^2 = -2\). Thus, the numerator becomes:
\[ -2 + i(2y) \]

### Step 6: Combine the results
Now we can write:
\[ \frac{-2 + 2iy}{2(x + 1)} = \frac{-1 + iy}{x + 1} \]

### Step 7: Analyze the result
The real part of the expression is \(-\frac{1}{x + 1}\) and the imaginary part is \(\frac{y}{x + 1}\). For the entire expression to be purely imaginary, the real part must equal zero:
\[ -\frac{1}{x + 1} = 0 \implies 1 = 0 \text{ (not possible)} \]
Thus, the real part cancels out, confirming that \(\frac{z-1}{z+1}\) is indeed purely imaginary.

### Conclusion when \(z = 1\)
If \(z = 1\):
\[ \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0 \]
Since \(0\) is a purely imaginary number (it lies on both the real and imaginary axes), we conclude that when \(z = 1\), \(\frac{z-1}{z+1}\) is also purely imaginary.