The following visualization example in Fig7.2 shows powers for arbitrary positive or negative, integer or fractional exponents: $u={z}^{n}$. Thus we have for example $u={z}^{2};\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}u={z}^{-3}=\frac{1}{{z}^{3}};\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}u={z}^{1,5}={z}^{\frac{3}{2}}=\sqrt[3]{{z}^{2}}$.

The controls elements for the different simulations of conformal mappings are mostly identical. We describe them for this first example in detail and refer only to differences later. Extensive details are given in the respective description pages of the simulation.

A quadratic point-grid with preset side length is located in the $z$-plane. One of the corners is marked in red and connected to the origin by a vector. With the mouse the square can be moved in $z$-plane while maintaining its orientation by grabbing it on the red corner. While the other coordinate stays exactly the same you can change one of the coordinates with the sliders $x,y$. Very accurate values can also be defined in the number fields $x,y$. One can also enter values, that go beyond the range of the sliders. Points with different imaginary parts are differentiated by colour in order to be able to follow the mapping point by point. The colour coding becomes most clear, if you pull the window to full screen size. With a slider side length the square can be grown or shrunk to a point.

In addition a circular colour coded point grid is located around the origin with a radius that depends on the function. The center of the circle is marked in blue; it can be moved with the mouse. The points of the circle, that are initially on the real axis and mirror images of each other, are highlighted. The right point, that is marked by a red disk is connected to the origin by a vector, which can be pulled with the mouse. With a second slider radius the circle can be grown or pulled together to a point.

Via collapsing the square or circle to a point you can plot the other function more clearly and study the mapping of a single point.

The scale can be adjusted separately in both windows in the number fields scale${}_{z}$ and scale${}_{u}$.

In the $u$-plane you see the mapping of the individual points of the square or circle via the chosen function. Accurate coordinates are shown, if you click on the points. With the Play button an animation is started, that moves the corner point of the square arrays step by

Even during the animation that coordinates of the corner points can be changed with the mouse, the sliders or by entering numbers, such that the whole plane can be scanned in strips.

With Pause/Play the animation is stopped. With the initialization button you can reset the grid, the circle and scale to its original state.

For the power function of Fig7.2 you may enter an arbitrary positive or negative power $n$, also rational numbers (now comma, but point must be used!). The changes become effective on pressing the Enter button. We have

$$u={z}^{n}={\left(r{e}^{i\phi}\right)}^{n}={r}^{n}{e}^{in\phi}={r}^{n}\left(cosn\phi +isinn\phi \right)$$

The mapping ${z}^{n}$ rotates a point $z=r{e}^{i\phi}$ from preimage plane to the image plane by $n-1$ times its angle. Its absolute value increases to ${r}^{n}$ for $n>1$ and decreases for $n<1$. The unit circle is mapped to itself under rotation.

Due to the angular rotation it follows, that for $n>1$ the simple $u$-plane is not sufficient to accommodate the mapping of all $z$-values. For $n=2$ or $n=3$ the mapping provides for a two-or threefold coverage. In the theory of complex functions one refers to $n$ Riemannian sheets of the $u$-plane. On these sheets not only the Riemann

In the simulation picture the Riemannian sheets are superimposed, as one can easily see from the mapping of the circle: the two loops belong to different sheets.

For fractional exponents $n$ and negative real values the mapping splits the point grid in two parts, which is initially surprising. On one part lie the transformed points from the positive imaginary half plane, on the second part the points of the negative imaginary one. Whether the $u$-plane is covered only partially or many times depends on whether $n$ is larger or smaller than 1. Fig.7.3 shows the picture for $n=0.5$ ($u={z}^{0.5}=\sqrt[2]{z}$).

With a bit of calculation it is easy to see that the splitting has to be as observed for $n=0.5$, i.e. for $u={z}^{\frac{1}{2}}=\sqrt[2]{z}$. The point $z=i$ (angle of 90${}^{\circ}$) is mapped to the point $u=\sqrt{i}=\frac{1+i}{\sqrt{2}}$ with the angle of 45${}^{\circ}$ as we show via inverting the function.

$${\left[\frac{1}{\sqrt{2}}\left(1+i\right)\right]}^{2}=\frac{1}{2}\left(1+2i+{i}^{2}\right)=\frac{1}{2}\left(1+2i-1\right)=i\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}q.e.d$$

How is the point $-i$ transformed. We assume, that it is mapped to point that is complimentary to the one obtained above, (same real part, opposite sign of the imaginary part) and again prove this via the inverse function:

$${\left[\frac{1}{\sqrt{2}}\left(1-i\right)\right]}^{2}=\frac{1}{2}\left(1-2i+{i}^{2}\right)=\frac{1}{2}\left(1-2i-1\right)=-i\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}\phantom{\rule{0em}{0ex}}q.e.d$$

Thus we indeed have the following situation: the point $-i$ and all other points with negative imaginary part are mapped to the part of the plane where the imaginary part is negative and all point with positive imaginary part are mapped to a part with positive imaginary component which is its mirror image.

For the circle around the origin the situation can be most easily understood. For $n=2$ the circle is mapped to two segments, as soon as individual points have a negative imaginary part. The two partial curves lie on different Riemannian sheets. Via counting you may convince yourself, that there are equally many points on the two partial curves if the setup is symmetric to the origin.

For suitable parameters the conformal mapping yields very interesting symmetries. Fig7.4 shows on the left hand side for the 17th and on the right hand side for the 60th power the superimposed mapping of 100 points of a circular array or radius 1 on as many Riemannian sheets. The array is slightly shifted from the origin. Remember, that the unshifted unit circle is mapped to itself, and thus the shifted one will be mapped to its immediate vicinity.

The simulation gives you many opportunities for experiments, that you access with the interactive simulation in Fig7.2. The description pages contain further details and suggestions for experiments.