Discussion
The image of the half plane \(\mathrm{Re}(z) + \mathrm{Im}(z) \gt 0\) under the map \(w = \frac{z-1}{z+i}\) is given by
Answer: D
The image is the disc \(|w| \lt 1\).
\(|w| = \frac{|z - 1|}{|z + i|}\), so \(|w| \lt 1\) exactly when \(z\) is nearer to 1 than to \(-i\). The points equidistant from 1 and \(-i\) form their perpendicular bisector:
\[ \begin{aligned} (x-1)^{2} + y^{2} &= x^{2} + (y+1)^{2} \\ -2x &= 2y \\ x + y &= 0 \end{aligned} \]
So the line \(\mathrm{Re}(z) + \mathrm{Im}(z) = 0\) maps onto the unit circle, and the half plane \(x + y \gt 0\), which contains \(z = 1\) and is on the side of 1, maps into \(|w| \lt 1\). Check: \(z = 1\) gives \(w = 0\).
\(|w| \gt 1\) is the image of the other half plane. \(\mathrm{Re}(w) \gt 0\) and \(\mathrm{Im}(w) \gt 0\) are half planes, but the boundary line \(x + y = 0\) does not pass through the pole \(z = -i\), so a Mobius map sends it to a circle, not a line, and the image must be a disc or its outside.
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