The z² Mapping

Complex Analysis · Axiom Academy

Track a horizontal line through the squaring map and watch it bend into a parabola. Find the image of the horizontal line y = 1 under the squaring map w = z^2 . In other words, as z runs along every point with imaginary part 1 , what curve does w trace out in the w -plane? The straight line on the left maps to the rightward-opening parabola on the right. We will derive its equation below. Nice work. You pushed a whole line through w = z^2 by parametrizing it, applying the map, and eliminating the parameter — the standard recipe for finding the image of any curve. The recipe: parametrize the source curve, substitute into w = z^2 , separate and , then eliminate the parameter to get a relation between u and v . This result: the line y = 1 maps to the parabola v^2 = 4(u + 1) , opening right with vertex (-1, 0) . Lines become parabolas: more generally a horizontal line y = c (with ) maps to v^2 = 4c^2(u + c^2) , and a vertical line x = c maps to a parabola opening the other way with vertex (c^2, 0) . Polar form is cleaner: writing gives , so the modulus is squared and the angle is doubled. A circle |z| = r becomes the circle |w| = r^2 , and a ray becomes the ray . Doubling angles folds regions: because angles double, the first quadrant ( ) opens out to the entire upper half-plane, and the unit circle wraps around itself twice.

This is the written version of the interactive lesson above. See the full Complex Analysis course.