` sin^(-1) {1/i ( z-1)}`, where z is non real and `i = sqrt(-1)`, can be the angle of a triangle If:

Options

A. Re(z)=1, Im(z)=2
B. Re(z)=1,`-1 le "Im"(z) le 1`
C. Re(z)+Im(z)=0
D. None of these

Explanation

To solve the problem, we need to analyze the expression \( \sin^{-1} \left( \frac{1}{i(z-1)} \right) \) where \( z \) is a non-real complex number. Let's break down the solution step by step.

### Step 1: Rewrite the expression
We start with the expression:
\[ \sin^{-1} \left( \frac{1}{i(z-1)} \right) \]
To simplify this, we multiply both the numerator and the denominator by \( i \):
\[ \sin^{-1} \left( \frac{i}{i^2(z-1)} \right) \]
Since \( i^2 = -1 \), we can rewrite this as:
\[ \sin^{-1} \left( \frac{i}{-1(z-1)} \right) = \sin^{-1} \left( -\frac{i}{z-1} \right) \]

### Step 2: Express \( z \) in terms of its real and imaginary parts
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. Then:
\[ z - 1 = (x - 1) + iy \]
Substituting this into our expression gives:
\[ \sin^{-1} \left( -\frac{i}{(x-1) + iy} \right) \]

### Step 3: Simplify the expression further
To simplify \( -\frac{i}{(x-1) + iy} \), we multiply the numerator and denominator by the conjugate of the denominator:
\[ -\frac{i((x-1) - iy)}{(x-1)^2 + y^2} \]
This results in:
\[ -\frac{(x-1)i + y}{(x-1)^2 + y^2} \]
Thus, we can write:
\[ \sin^{-1} \left( \frac{y - (x-1)i}{(x-1)^2 + y^2} \right) \]

### Step 4: Analyze the imaginary and real parts
For \( \sin^{-1}(w) \) to be defined, the argument \( w \) must lie within the range of the sine function, which is between -1 and 1. Therefore, we need to ensure that:
1. The imaginary part \( y \) must be between -1 and 1.
2. The real part \( x - 1 \) must be such that the overall expression remains valid.

### Step 5: Conditions for \( z \) to be an angle of a triangle
Since we are looking for conditions under which \( \sin^{-1} \left( \frac{1}{i(z-1)} \right) \) can represent an angle of a triangle, we note that angles must be non-negative. This implies:
- The imaginary part \( y \) must be non-negative (i.e., \( y \geq 0 \)).
- The real part \( x \) must be equal to 1 (i.e., \( x = 1 \)).

### Conclusion
Thus, the conditions for \( z \) are:
- The real part \( x = 1 \)
- The imaginary part \( y \) lies in the interval \( [0, 1] \)

### Final Answer
The conditions for \( z \) are:
- \( \text{Re}(z) = 1 \)
- \( 0 \leq \text{Im}(z) \leq 1 \)