Inversion of Complex Functions

When manipulating functions of complex variables, one must always keep in mind that complex structure can lead to results that differ surprisingly from the same operations on functions of real variables. This can be exhibited neatly in evaluating inverses of complex functions. For example, for real x it is straightforward that

$ln e^{x} = x$

which can be confirmed visually by plotting the left-hand side as a function of x:

Now consider including an explicit imaginary unit. While it is true on paper that

$- i ln e^{i x} = x$

the direct numerical evaluation appears quite different. The imaginary part of the left-hand side of this equation is identically zero as expected, but the real part is not continuous:

This occurs because $e^{2 π i} = 1$ , so that the numerical evaluation is modulus $2 π i$ . The complex function can be made numerically continuous by explicitly including this phase shift between branches of the complex function,

$- i ln e^{i x + 2 π i k} = - i ln e^{i x} + 2 π k$

where k is the index for the branch. The resulting plot then matches the paper result:

This difference is even more dramatic in three dimensions. A direct plot of the function

$ln e^{x} - i ln e^{i y} = x - i ln e^{i y}$

has a similar discontinuous appearance,

but if the phase shift between branches is included, one obtains the expected planar behavior:

The general takeaway is: watch out for branches of complex functions!

A somewhat more complicated example occurs in taking powers of complex numbers and inverting them with roots. Since all complex numbers can be written in the form

$z = ρ e^{i φ} = e^{ln ρ + i φ}$

use the exponential function as a proxy for the operation under consideration. For real x it is straightforward that

$(e^{n x})^{1 / n} = e^{x}$

which can again be confirmed visually by plotting the left-hand side as a function of x,

where changing value of n has no effect on the plot as expected. Including an explicit imaginary unit, while it is true on paper that

$(e^{i n x})^{1 / n} = e^{i x}$

the direct numerical evaluation again appears quite different. This time there are both real and imaginary parts of the left-hand side, which on paper are the functions cosine and sine. Numerically it looks like this,

with the real part in blue and imaginary in red. When $n = 1$ the trigonometric functions appear, but as this value increases it is clear that only parts of these functions remain. The period of these remaining parts is also decreased relative to that of the trigonometric functions, patently to the value

$T (n) = \frac{2 π}{n}$

How does one make sense of the situation? The numerical evaluation here is again modulus $2 π i$ , and this applies not to a particular value of x but its product with n. The period is thus determined by the value of the modulus alone, which after the root acquires the indicated denominator.

To find the missing pieces of each function, again include an index for branches of the complex function:

$(e^{i n x + 2 π i k})^{1 / n} = e^{2 π i k / n} (e^{i n x})^{1 / n}$

The logarithm function above had a simple linear dependence on the branch index. The dependence here is more complicated, meaning that the regions over which the function is evaluated need to be determined separately from the index. Here is a visualization in purple of the real part of each branch compared to the principal branch,

followed by the imaginary part in magenta, again compared to the principal branch:

Et voilà! The missing parts of each function reappear as the branch index is altered.

A branch recapitulates the principal branch $k = 0$ whenever $\frac{k}{n}$ is an integer. For a power n with one decimal place this generally requires ten times this number of branches, unless there is a simplification in this ratio. For a power with two decimal places this generally requires one hundred times this number of branches, and so on for more decimal places.

Knowing that all the pieces of the functions are there, how does one put them back together into a single continuous function? This is simpler than expected: use another index m to indicate on which part of the principal branch to begin, and the regions of evaluation then follow consecutive branch indices.

In applying this procedure to the central three pieces of the principal branch, a small gap is deliberately included to indicate locations of change of branch. For the real part of the function one has

and for the imaginary part

Piece of pie! Easy as cake! Get the reference?

And again things are much more dramatic in three dimensions. Consider a direct plot of the function

$(e^{n (x + i y)})^{1 / n} = e^{x} (e^{i n y})^{1 / n}$

for a restricted range of x and colored by complex argument. First the real part,

then the imaginary part:

Now repeat these two graphics with the inclusion of the branch index k to show the pieces of the complex function, again in comparison with the principal branch. First the real part,

followed by the imaginary part:

Coloring by complex argument helps one to see how the various branches connect to one another.

Lastly, one can reconstruct continuous surfaces in exactly the same way as above for one-dimensional lines. A small gap is again deliberately included to indicate locations of change of branch. For the real part of the function one has

and for the imaginary part

Easy as cake! Piece of pie!

The methods used above can naturally be extended to additional functions. Since the examples given illustrate the procedure sufficiently, this will be left as a future possible addition to this presentation.