Discussion
The straight lines \(L_{1} : x = 0,\ L_{2} : y = 0\) and \(L_{3} : x + y = 1\) are mapped by the transformation \(w = z^{2}\) into the curves \(C_{1}, C_{2}\) and \(C_{3}\) respectively. The angle of intersection between the curves at \(w = 0\) is
Answer: D
The curves meet at \(w = 0\) at an angle of \(\pi\).
Only \(L_{1}\) and \(L_{2}\) pass through \(z = 0\), the point that maps to \(w = 0\); the line \(L_{3}\), \(x + y = 1\), misses the origin, so \(C_{3}\) does not pass through \(w = 0\). On \(L_{1}\), \(z = iy\) gives \(w = -y^{2}\), the negative real axis. On \(L_{2}\), \(z = x\) gives \(w = x^{2}\), the positive real axis. The two image curves leave \(w = 0\) in opposite directions, so the angle between them is \(\pi\).
The map \(w = z^{2}\) is not conformal at \(z = 0\), where \(dw/dz = 2z = 0\): there it doubles angles, which is why the right angle between the axes becomes a straight angle. Answering \(\pi/2\) assumes conformality at a critical point; 0 and \(\pi/4\) do not match either image.
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