The point z in the Argand plane moves such th | vedclass

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(B) Let $z = x + iy$. Then $iz + 1 = i(x + iy) + 1 = (1 - y) + ix$ and $iz - 1 = i(x + iy) - 1 = -(1 + y) + ix$. Consider the expression $\frac{iz + 1

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a circle

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(B) Let $z = x + iy$. Then $iz + 1 = i(x + iy) + 1 = (1 - y) + ix$ and $iz - 1 = i(x + iy) - 1 = -(1 + y) + ix$. Consider the expression $\frac{iz + 1}{iz - 1} = \frac{(1 - y) + ix}{-(1 + y) + ix}$. Multiply the numerator and denominator by the conjugate of the denominator: $-(1 + y) - ix$. The denominator becomes $(-(1 + y))^2 + x^2 = (1 + y)^2 + x^2$. The real part of the numerator is $(1 - y)(-(1 + y)) + x^2 = -(1 - y^2) + x^2 = x^2 + y^2 - 1$. Thus, $\operatorname{Re} \left( \frac{iz + 1}{iz - 1} \right) = \frac{x^2 + y^2 - 1}{x^2 + (y + 1)^2} = 2$. This simplifies to $x^2 + y^2 - 1 = 2(x^2 + y^2 + 2y + 1)$, which is $x^2 + y^2 - 1 = 2x^2 + 2y^2 + 4y + 2$. Rearranging gives $x^2 + y^2 + 4y + 3 = 0$. This is the equation of a circle $x^2 + (y + 2)^2 = 1$.

The point $z$ in the Argand plane moves such that $\operatorname{Re} \left( \frac{iz + 1}{iz - 1} \right) = 2$. Then the locus of $z$ is:

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