Images+and+preimages

Sketch the image of the set {1≤ |z| < 2} under the map

1) f(z)=(1+i) z + (3-2i)

The mapping does the following to any number z: it stretches it by √2, rotates the result by π/4 and shifts the result by vector (3,-2). What happens to the annulus {1≤ |z| < 2}: after stretching it becomes the annulus {√2≤ |z| < 2√2}. Rotation doesn't change the result. Shifting by (3,-2) produces the annulus {√2≤ |z-(3-2i)| < 2√2} (the way to see the minus sign algebraically: the image of the set {z| √2≤ |z| < 2√2} under w=f(z)=z+(3-2i) is {w=z+(3-2i) | √2≤ |z| < 2√2} or {w| √2≤ |w-(3-2i)| < 2√2}

A sketch:

(the center is at 3-2i, the radii are √2 and 2√2, the inside boundary is included, the outside boundary isn't)

2) f(z)=z^2+1

The image of {1≤ |z| < 2} under z->z^2 is {w=z^2|1≤ |z| < 2} or {w| 1 ≤ √|w| < 2} or {w| 1 ≤ |w| < 4}, an annulus centered at 0 with inner radius 1 and outer radius 4. Hence the image under f(z) is the annulus {w| 1 ≤ |w-1| < 4}, i.e. the same thing shifted by vector (1,0)

3) f(z)=|z|

The image is the set {|z| | 1≤ |z| < 2}, i.e. simply a segment on the real line between 1 and 2, closed at 1, open at 2

4) f(z) = Rez

The mapping is just a projection onto the x-axis, so we have to draw the "shadow" of the set onto the x-axis. This is the segment of real numbers from -2 to 2, both ends not included

5) math f(z) = \overline{z} math

The mapping is reflection in the x-axis. The image of the annulus is the same annulus, since it is symmetric about the x-axis.

6) Find the image of {z|Re z<0, Im z >0} under f(z)=z^2+1.

The easiest way is to rewrite the set as {z| pi/4z^2 is {w=z^2| pi/4z^2 "opens" the wedge a0} under f(z)=z^3.

Similar. The answer: all the plane except {Re w ≤ 0, Im w ≤ 0} 8) Find the image of {z| z ≠1} under f(z)=z^2. The image is {w=z^2|z≠1}, i.e. all complex numbers which can be represented as a square of some number not equal to 1. This includes all numbers, even w=1: indeed, 1=(-1)^2 and -1 ≠1. Hence the image is all the complex plane.