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$\sin^{- 1} [\frac{1}{i} (z - 1)]$ , where z is non-real, can be the angle of a triangle if

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Re (z) = 1, Im (z) = 2

Re (z) = Im (z)

Re (z) = 1, –1 ≤ Im (z) ≤ 1

Re (z) + Im (z) = 0

answer is B.

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Detailed Solution

$\frac{z - 1}{i}$ should be real

$i . e ., \frac{x + iy - 1}{i} = y - i (x - 1)$

$\begin{array}{l} \Rightarrow x - 1 = 0 i.e., x = 1 \\ ∴ \sin^{- 1} \frac{z - 1}{i} = \sin^{- 1} y \\ ∴ - 1 \leq y \leq 1 \end{array}$

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