Complex Functions and the Complex Plane as a Domain
Treat a function of a complex variable as a map of the plane to itself, and set up limits and continuity in that setting.
What a learner can do afterwards
- Describe the image of a line or circle under a simple complex map
- Define a limit of a complex function over its domain, explain why the epsilon-delta condition controls every sufficiently near input, and state what continuity at a point demands
- Split a complex function into its real and imaginary parts as two real functions of two variables
1 · Read
A complex function f takes a domain D, a set of points in the plane, and sends each z in D to one output f(z). Write z = x + iy and f(z) = u(x,y) + iv(x,y). The real part is u and the imaginary part is v, not iv. Both u and v are real functions of the two inputs x and y.
Split w = z² by expansion. With z = x + iy, z² = (x + iy)² = x² + 2ixy + (iy)² = x² - y² + 2ixy, using i² = -1. So u(x,y) = x² - y² and v(x,y) = 2xy. Check: at (1,2), u = 1 - 4 = -3 and v = 2·1·2 = 4.
For example, under w = z + i the real axis z = t maps to w = t + i, and every point with imaginary part 1 is attained as t runs over the reals.
To map a whole set, do not sample a few points. Parametrize the input with one real parameter t, compute the image, then show every point of the proposed image is attained. Under w = 2z + i, the circle |z| = 1 maps to |w - i| = 2, and z = (w - i)/2 shows full attainment. Under w = z², the line z = 1 + it gives w = 1 - t² + 2it, so u = 1 - t² and v = 2t, giving the parabola v² = 4(1 - u). Every point is attained since t = v/2 recovers the parameter.
An accumulation point of D has points of D other than itself arbitrarily nearby. A complex limit controls every near input. The limit of f(z) as z tends to z0 is L when for every epsilon > 0 there is delta > 0 such that 0 < |z - z0| < delta implies |f(z) - L| < epsilon, for z in D with z0 an accumulation point of D.
Polynomials are continuous: the limit of z^2 + 1 at any point equals its value there. For a punctured quotient, z^2 - a^2 = (z - a)(z + a), so (z^2 - a^2)/(z - a) = z + a for z ≠ a, and the limit as z tends to a is 2a.
Two conflicting paths kill a limit. For f(z) = xy/(x² + y²) with z = x + iy, along the real axis (y = 0) the value is 0, but along the diagonal y = x the value is x²/(2x²) = 1/2. The two paths disagree, so the limit at 0 does not exist. A few agreeing paths cannot prove a limit. For an all-path bound, |z|² tends to 0 because |z|² - 0 = |z|² < epsilon whenever |z| < sqrt(epsilon), so the limit is 0 from every direction at once. Continuity at z0 needs f(z0) defined and equal to the limit.
Split maps into u and v, prove whole images with parameters or moduli, and demand one limit that survives every nearby input.
All ten extra tasks (x1-x10) and three readings are original authorship created for this topic. No source prose, textbook examples, or pictures were copied or relabelled. The previous enrichment carried unsupported CC0 credits on custom summaries that overstated source scope; those credits are replaced with original null-license entries. The ten PG records and three textbook records were read as design studies only: their shapes (complex multiplication, powers of i, real continuity conditions) informed task design but no text was imported. The two-path counterexample xy/(x^2+y^2) and the all-path bound |z|^2 are standard mathematical examples, not source text.
2 · Watch
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Where it sits
18 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.