Functions that remap the whole plane

Mathematics · Complex Analysis · ages 19-20

Name ______________________ Date ____________

Under w = 3iz, what is the modulus of the image of a point with |z| = 4?
Answer: ______________
For w = iz + 2 with z = x + iy, what are u(x,y) and v(x,y)?
- u = 2 + y, v = x
- u = 2 - y, v = x
- u = 2 - x, v = y
For w = z² with z = x + iy, what is the imaginary part v(x,y)?
- x² - y²
- x² + y²
- 2xy
Under w = z + i, the real axis maps to the line Im(w) = 1. True or false?
Circle one: True False
For f(z) = z² + 1, what is the real part of the limit as z tends to 1 + i?
Answer: ______________
Which statement gives the full defining requirement for a finite complex limit over the punctured domain?
- For every epsilon > 0, all such z close enough to z0 give |f(z) - L| < epsilon for one finite L
- For some epsilon > 0, all such z close enough to z0 give |f(z) - L| < epsilon for one finite L
- For every epsilon > 0, all such z close enough to z0 give |f(z) - f(z0)| < epsilon
Under w = 3z + 2, the circle |z| = 1 maps to which circle?
- Centre 2, radius 1
- Centre 2, radius 3
- Centre 3, radius 2
If f(z) tends to the same value along the real and imaginary axes, the complex limit exists. True or false?
Circle one: True False
Under w = z², the line z = 2 + it maps to which curve?
- v² = 16(4 - u)
- v² = 4(4 - u)
- v² = 16(u - 4)
Under w = (t + i)² with t real, what is the full image?
- Parabola v² = 4(u + 1)
- Circle |w| = 1
- Line Im(w) = 2

Answer key

For grown-ups. Fold this page away before handing over the rest.

12 · |w| = |3iz| = 3|i||z| = 3 × 4 = 12.
u = 2 - y, v = x · iz = i(x + iy) = -y + ix, so w = (2 - y) + ix, giving u = 2 - y and v = x.
2xy · Expanding (x + iy)² gives real part x² - y² and imaginary part 2xy.
True · Adding i lifts every real point one unit up to imaginary part 1.
1 · Polynomials are continuous, so the limit is f(1 + i) = (1 + i)² + 1 = 2i + 1, whose real part is 1.
For every epsilon > 0, all such z close enough to z0 give |f(z) - L| < epsilon for one finite L · The definition requires one finite L that works for every positive epsilon over all sufficiently close inputs in the punctured domain, regardless of f(z0).
Centre 2, radius 3 · |w - 2| = |3z| = 3|z| = 3, so the image is the circle with centre 2 and radius 3.
False · The diagonal path can disagree even when both axes agree, as xy/(x²+y²) shows.
v² = 16(4 - u) · Expanding (2 + it)² gives u = 4 - t² and v = 4t, so v² = 16(4 - u).
Parabola v² = 4(u + 1) · Expanding gives u = t² - 1 and v = 2t, so v² = 4(u + 1), a parabola opening right.