1 Linear Functions

A linear function has the form $f(z) = az+b$ where $a, b \in \C$ and $a \neq 0$ (if $a=0$, then $f$ is a constant function). The effect of a linear function on a set in the complex plane is to scale, rotate and translate that set. The constant term $b$ acts as a translation vector, and the modulus and argument of $a$ are responsible for producing scaling and rotation, respectively.

example 1 Find the image of the unit square under the mapping $f(z) = (1+i)z + i$.
In this case, $a = 1+i$ which has modulus $\sqrt 2$ and argument $\pi /4$, and $b = i$. The effect on the square of multiplying by $1+i$ is to scale the square by a factor of $\sqrt 2$, rotate counter-clockwise by $\pi /4$ and then translate it by $i$. Due to the order of operations, the scaling and the rotation are done first (in either order), and the translation is done last.

Here is a video solution of one part of problem 1, part iv:

2 Squaring

If $z = r \cis \theta$, then $z^2 = r^2 \cis 2\theta$. Hence, the effect of squaring is to square the modulus and double the argument of a complex number, $z$.

example 2 Find the image of the sector of an annulus given by: $$S = \left \{r\cis \theta : \frac 12 < r < 2, \;\frac {\pi }{6} < \theta < \frac {\pi }{2}\right \}$$ under the mapping $f(z) = z^2$.
The moduli of the complex numbers in the region $S$ are between $\frac 12$ and $2$. Hence, the moduli of their squares are between $\left (\tfrac 12\right )^2$ and $2^2$. Additionally, the arguments of the points in $S$ are between $\tfrac {\pi }{6}$ and $\tfrac {\pi }{2}$, and thus, the arguments of their images are between $\tfrac {\pi }{3}$ and $\pi$. Putting these facts together yields the image, $f(S)$, which is also a sector of an annulus: $$f(S) = \left \{r\cis \theta : \frac 14 < r < 4, \;\frac {\pi }{3} < \theta < \pi \right \}$$

Here is a video solution of one part of problem 2, part iv:

example 3 Find the image of vertical line $\Rep z = 1$ under the mapping $f(z) = z^2$.
Instead of a polar analysis, we will use rectangular coordinates with $x = 1$. We have $$w = u+iv = (1+iy)^2 = (1-y^2) + 2iy$$ Thus $u = 1-y^2$ and $v = 2y$, where $y \in R$. Eliminating $y$ gives $$u= 1-\frac {v^2}{4}$$ which is a parabola in the $uv$-plane which opens to the left and has vertex at $(1,0)$.

Here is a video solution of one part of problem 3, part iv:

3 Conjugates and Reciprocals

The function $f(z) = \overline {z}$ maps a set in the complex plane to its mirror image in the real axis.

example 4 Find the image of the closed disk $D(z_0, r)$ under the mapping $f(z) = 2\overline {z}$.
The image of the center of the disk is $f(z_0) = 2\overline {z_0}$ and so the image of the disk $D(z_0, r)$ is the disk of twice the radius, $D(2\overline {z_0}, 2r)$.

The function $f(z) = \frac {1}{z}$ maps the punctured complex plane, $\C \setminus \{0\}$, to itself. Moreover, it is a bijection whose inverse is itself. To understand the mapping properties of $f(z) = \frac {1}{z}$ it is useful to rewrite it in the following way: $$\frac {1}{z} = \frac {1}{z} \cdot \frac {\overline {z}}{\overline {z}}= \frac {\overline {z}}{|z|^2}$$ Thus the mapping $f(z) = \frac {1}{z}$ has both a reflecting property like the conjugate mapping and a scaling property due to the division of the positive real number $|z|^2$. As a result of the scaling, the image of a point inside the unit circle will map to a point outside the circle and vice versa. This is shown in the figure below.

The color coding in the figure is intended to convey reflection in the real axis. A blue point in the upper half plane inside the disk maps to a blue point in the lower half plane outside the disk. Similarly, a red point in the lower half plane inside the disk maps to a red point outside the disk.

example 5 Find the image of the vertical ray $\Rep z = \frac 12, \Imp z > 0$ under the mapping $f(z) = \frac {1}{z}$.
The equation $w = 1/z$ is equivalent to the equation $z = 1/w$ where $w$ and $z$ are both non-zero. We write $1/w$ in rectangular form: $$z = \frac {1}{w} = \frac {\overline {w}}{|w|^2} = \frac {u-iv}{u^2 + v^2}$$ Thus, $$x = \frac {u}{u^2+v^2} \quad \text {and} \quad y = -\frac {v}{u^2+v^2}$$ Later, we will use the fact that $v$ is negative since $y$ is positive. First, we substitute $x = 1/2$ into the first equation, yielding: $$\frac 12 = \frac {u}{u^2+v^2}$$ Rearranging gives: $$u^2 - 2u + v^2 = 0$$ Finally, we complete the square by adding $1$ to both sides to obtain: $$(u-1)^2 + v^2 = 1$$ This is the equation of a circle in the $w$-plane with center at $(1,0)$ and radius $1$, i.e., $C(1,1)$.

Recalling that $v <0$, we conclude that the image of the ray $x = 1/2, y >0$ is a semicircle in the lower half plane with center at $1$ and radius $1$. See the figure below.

Here is a video solution of problem 5, part iii:

3.1 Stereographic Projection

The mapping $f(z) = \frac {1}{z}$ in some sense inverts the unit disk, with the singularity at the origin corresponding to a ring of points far from 0. In the complex plane this entire ring is considered to be “$\infty$”. Stereographic projection enables us to identify the complex plane with a punctured sphere $S\setminus N$ where $N$ is the north pole of the sphere (sorry Santa) and further, to identify “$\infty$” with the north pole, $N$.

At the heart of stereographic projection lies a one-to-one, onto map $\varphi : \C \to S\setminus N$ where $S$ is the sphere with center at $\left (0,0,\tfrac 12\right )$ and radius $\tfrac 12$, so that $N = (0,0,1)$.

To find the precise formula for the mapping $\varphi$, recall that a line segment in three dimensions can be expressed in vector form as $$\gamma (t) = {\bf u} + t{\bf v}, \; 0 \leq t \leq 1$$

This segment goes from $\bf u$ to ${\bf u} + {\bf v}$ as $t$ goes from $0$ to $1$:

In our case, the vector $\bf u$ is $(x,y,0)$ and the vector $\bf v$ goes from $(x, y, 0)$ to $(0, 0, 1)$, so ${\bf v} = (-x, -y, 1)$. Now, we wish to find $t$ so that the point $${\bf u}+ t{\bf v} = (x- tx,y- ty, t) = (x(1-t), y(1-t), t)$$ lies on the sphere, $S$, given by

$$x^2 + y^2 + \left (z-\frac 12 \right )^2 = \left (\frac 12\right )^2$$ Substituting gives $$x^2(1-t)^2 + y^2(1-t)^2 + \left (t-\frac 12 \right )^2 = \left (\frac 12\right )^2$$ which is equivalent to $$x^2(1-t)^2 + y^2(1-t)^2 + t(t-1) = 0$$ Dividing by $t-1$ gives $$x^2(t-1) + y^2(t-1) + t = 0$$ Solving this linear equation for $t$ we have $$t= \frac {x^2 + y^2}{1+x^2+y^2}$$ Noting that $\displaystyle 1-t = \frac {1}{1+x^2 +y^2}$, we can define $\varphi$: $$\varphi (x+iy) = \left ( \frac {x}{1+x^2 +y^2}, \frac {y}{1+x^2 +y^2}, \frac {x^2 + y^2}{1+x^2+y^2} \right )$$

The bijection $\varphi$ can be extended naturally to include $\infty$ in its domain: $\varphi (\infty ) = N$.

4 Exponential and Logs

Next, we turn to the exponential function, $f(z) = e^z$ and its mapping properties. Recall that $e^z$ is $2\pi i$ periodic. As a result, the image of the band of complex numbers satisfying $-\pi < \Imp z \leq \pi$ is its full image, $\C \setminus \{0\}$.

example 6 Find the image of the vertical segment $x = 2, -\pi < y \leq \pi$ under the mapping $f(z) = e^z$.
For $z$ on this segment, $$e^z = e^2 \cis y$$ and since $y$ takes on all values in an interval of length $2\pi$, $\cis y$ will take on all values on the unit circle. Hence, the image of the segment is the circle centered at the origin with radius $e^2$.

An alternative method to arrive at the conclusion of the last example is to write $$u = e^2\cos y \quad \mbox {and} \quad v = e^2 \sin y$$ and use the Pythagorean trigonometric identity $\cos ^2 \theta + \sin ^2 \theta = 1$ to obtain: $$u^2 + v^2 = (e^2)^2 = e^4$$ To see that the image is this entire circle, note that since $-\pi < y \leq \pi$ we have $-e^2 \leq u \leq e^2$ and similarly for $v$.

Here is a video solution of one part of problem 6, part ii:

example 7 Find the image of the horizontal line $y = \frac {\pi }{4}$ under the mapping $f(z) = e^z$.
For $z$ on this horizontal line, $$e^z = e^x \cis \left (\frac {\pi }{4}\right )$$ where $-\infty < x < \infty$. Such a point lies on the polar ray $\theta = \pi /4$ at a distance $e^x$ units from the origin. Since $e^x$ goes from $0$ to $\infty$ as $x$ goes from $-\infty$ to $\infty$, the image of the horizontal line is the polar ray, $\theta = \pi /4$ with $r>0$.

An alternative method to arrive at the conclusion of the last example is to write $$u = e^x\cos \left (\frac {\pi }{4}\right ) = \frac {e^x}{\sqrt 2} = e^x\sin \left (\frac {\pi }{4}\right ) = v$$ where $u$ and $v$ are both strictly positive. This gives the portion of the line $u = v$ in the first quadrant.

The next example involves the principal logarithm which we recall is defined by: $$\Log z = \ln |z| + i\Arg z$$

example 8 Find the image of the semicircle $z = 3\cis \theta , \; \pi /2 < \theta \leq 3\pi /2$ under the mapping $f(z) = \Log z$.
On the given semicircle, $|z| = 3$, so $$w = \Log z = \ln 3 + i\Arg \theta$$ Thus $u = \ln 3$ and so the image lies on a vertical line. By definition of $\Log z$, the values of $v$ must lie in the interval $\left (-\pi , \pi \right ]$. For $\pi /2 <\theta \leq \pi$ we have $\pi /2 < \Arg z \leq \pi$ and for $\pi < \theta \leq 3\pi /2$ we have $-\pi /2 \geq \Arg z > -\pi$. Hence, the range of values for $v$ is $(\pi /2, \pi ] \cup [-\pi /2, -\pi )$

Here is a video solution of one part of problem 8, part ii:

Below is a Geogebra applet that shows the image of a line under $1/z, e^z$ and $z^2$. Move the red line and observe the effects! (The Geogebra activity can be accessed by going to geogebra.org and typing Ptnx5Z3m into the search bar.)

Below is a Geogebra applet that shows the image of a circle under $1/z, e^z$ and $z^2$. Move the red circle and observe the effects! (The Geogebra activity can be accessed by going to geogebra.org and typing WuXRYxUm into the search bar.)

5 Sine and Cosine

example 9 Find the image of the horizontal line $\Imp z =1$ under the mapping $f(z) = \cos z$.
Recall $$\cos z = \cos x \cosh y - i \sin x \sinh y$$ Writing $w = u+iv = \cos z$ and substituting $y = 1$ gives $$u = \cos x \cosh 1 \quad \mbox {and} \quad v = -\sin x \sinh 1$$ We square $u$ and $v$ to take advantage of the Pythagorean Identity, $$\frac {u^2}{\cosh ^2 1} + \frac {v^2}{\sinh ^2 1} = \cos ^2 x + \sin ^2 x = 1$$ This is the equation of an ellipse with $u$-intercepts $u = \pm \cosh 1$ and $v$-intercepts $v = \pm \sinh 1$.

Here is a video solution of one part of problem 9, part iii: