Let a = text{Im}left( frac{1 + z^2}{2iz} righ | vedclass

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(A) Let $z = x + iy$. Since $|z| = 1$,we have $x^2 + y^2 = 1$.Consider the expression $\frac{1 + z^2}{2iz} = \frac{1 + (x + iy)^2}{2i(x + iy)} = \frac

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$(-1, 1)$

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(A) Let $z = x + iy$. Since $|z| = 1$,we have $x^2 + y^2 = 1$.Consider the expression $\frac{1 + z^2}{2iz} = \frac{1 + (x + iy)^2}{2i(x + iy)} = \frac{1 + x^2 - y^2 + 2ixy}{2ix - 2y} = \frac{(1 + x^2 - y^2) + 2ixy}{-2y + 2ix}$.Since $x^2 + y^2 = 1$,we have $1 - y^2 = x^2$. Substituting this into the numerator:$\frac{(x^2 + x^2) + 2ixy}{-2y + 2ix} = \frac{2x^2 + 2ixy}{2i(x + iy)} = \frac{2x(x + iy)}{2i(x + iy)} = \frac{x}{i} = -ix$.Thus,$a = 	ext{Im}(-ix) = -x$.Since $|z| = 1$,$x$ ranges from $-1$ to $1$. Given $z 
e \pm 1$,$x 
e \pm 1$.Therefore,$x \in (-1, 1)$,which implies $a = -x \in (-1, 1)$.Hence,the set $A = (-1, 1)$.

Let $a = \text{Im}\left( \frac{1 + z^2}{2iz} \right)$,where $z$ is any non-zero complex number. The set $A = \{ a : |z| = 1 \text{ and } z \ne \pm 1 \}$ is equal to

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