# Complex number to a complex power may be real

Not only that, but if we look at the quintessential complex number, the imaginary i, then you can verify that i^{ i} is real.

To prove this, recollect Euler's formula:

e^{it} = cos(t) + i·sin(t)

where e is, of course, the base of natural logarithms. Substitute

cos(p/2) = 0 and sin(p/2) = 1,

we get

i = e^{ip/2}

From here, i^{ i} = e^{i·i·p/2} = e^{-p/2}, which is not large but still very real.

(For a more serious discussion of the properties of complex numbers, see the collection of pages starting with Algebraic Structures of Complex Numbers.)

As can be seen from the correspondence below, the assertion that ^{ i} = e^{i·i·p/2} = e^{-p/2}*The Advanced Geometry of Plane Curves and Their Applications*, we read (he uses j instead of i, as is customary in engineering science)

Raise both sides of the equation ^{j} = exp(-p/2)^{j} is real. A Frenchman once called this formula "la plus belle formule de la mathématique".

To my chagrin, I could not identify this person of taste.

In a more recent, but no less distinguished *Impossible?* by J. Havil, we find a quote from the famous Augustus De Morgan:

Imagine a person with a gift of ridicule. [He might say] First that a negative quantity has no logarithm; secondly that a negative quantity has no square root; thirdly that the first nonexistent is to the second as the circumference of a circle to the diameter.

The quote serves to introduce *The Power of Complex Numbers* chapter, where the author also refers to a 1921 article from *American Mathematical Monthly* (28 (3), pp. 114-116) by H. S. Uhler who gave an approximate value for the constant

i^{ i} ≈ 0.207 879 576 350 761 908 546 955 ...

Subject: | Some ponderings |
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Date: | Thu, 17 Jun 2000 01:32:31 +0200 |

From: | Tobias Brandt |

Hi,

congratulations on an excellent web-site. I just have some thoughts to add to the contents
of the page entitled "Complex number to a complex power may be real". I enjoyed reading that
article and was quite fascinated by the fact that i^{ i} = e^{- p/2}. I mean there is nothing magical in the derivation but these curious results of complex numbers always seem awe inspiring. After some thinking I realized that this is actually just another formulation in disguise of an expression which has dazzled me for years, namely

e^{i·p} + 1 = 0;

since this implies that

e^{i·p} = -1.

Taking square roots on both sides

e^{i·p/2} = (-1)^{(1/2)}

in other words

e^{i·p/2} = i

Taking the i-th power on both sides yields the sought result

i^{ i} = [ e^{i·p/2} ]^{i} = e^{-p/2}

Cheers

Tobias Brandt

P.S. Looking over this again I realized that this is essentially the same as you had done. I'm sorry to waste your time but to me the explicit statement of ^{i·p} + 1 = 0

Subject: | Infinite number of values |
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Date: | Mon, 01 Jan 2001 01:57:22 -0800 |

From: | Robert D. Rasmussen |

Alexander,

I've been enjoying your web site. Well done!

Your page titled "Complex number to a complex power may be real" caught my attention, because the example you use, while simple, is still not well known. What is even less well known is that i^{ i} actually has an infinite number of real values. This follows by choosing the more general value of

t = p·(2·n + 1/2)

in your derivation, where n may be any integer. The result is

i = e^{ i·p·(2·n + 1/2)} ,

from which it follows that

i^{ i} = e^{ -p·(2·n + 1/2)} .

Each value of n produces a different real value.

So which value for n is right? They all are. Complex valued expressions often have the property of being multiple valued. We typically adopt a so-called "principle value" for practical purposes (the value for n = 0 in this case). People familiar with functions like arcsine will recognize this.

It's important to note, however, that even simple real valued expressions can have multiple values. When we write

x^{1/2} = ±sqrt(x)

(sqrt() always takes the principle, positive value),

we mean that x^{1/2} has two values. The expression i^{ i} just happens to have infinitely many values.

These two examples of powers, and the number of values that result, are actually very closely related. In fact, they both arise from the same source. To see this, consider the expression x^{y} in general, where x and y are any two complex numbers. How many values does it have?

When x = 0, the result is trivial. There is only one value, since ^{y} = 0

For non-zero x, a method similar to that used for i^{ i} can be used to find out how many values x^{y} has.

I'll use the formula

ln x = ln |x| + i·(2·n·p + arg(x) )

where ln |x| is the real-valued natural log of the real number |x|, the magnitude of x. arg(x) is the angle of the ray from the origin to x in the complex plane relative to the positive real axis.

This new formula is just an application of Euler's formula in reverse, as seen here:

e^{ ln x} | = e^{ ln |x| + i·(2·n·p + arg(x) )} |

= |x| e^{ i·2·n·p }e^{i arg(x)} | |

= |x| e^{ i·arg(x)} | |

= |x| cos(arg(x)) + i·|x| sin(arg(x)) | |

= Re(x) + i·Im(x) | |

= x |

Now back to x^{y}. We have

x^{y} | = e^{ y·ln x} |

= e^{ y·(ln |x| + i·(2·n·p + arg(x) ) )} | |

= e^{ y·(ln |x| + i·arg(x) )} e^{ i·y·(2·n·p)} |

There are two terms in this result. The first exponential gives a single complex number. It can be evaluated further using Euler's formula. The second term, in general, has infinitely many values, as before. Therefore, the product of these two terms has infinitely many values.

But notice what happens in the special case where y is real and rational. In this case, the sequence of values with respect to n eventually repeats after a finite number of steps (predicting how many leads to other interesting questions). There are only a finite number of values for x^{y} in this case.

One example of the special case occurs with y = 1/2 (the square root, again). The second term above alternates between +1 and -1, yielding the two values of opposite sign (infinitely repeated) that we all know, of course. Isn't that interesting?

Bob Rasmussen

Subject: | Erratum in "Complex number to complex power ..." |
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Date: | Sat, 13 Apr 2002 22:35:02 EDT |

From: | David Cantrell |

Hello,

Congratulations on your excellent site!

While browsing, I found a mistake which I thought you might wish to correct. In the article "Complex number to a complex power may be real", Bob Rasmussen said "To see this, consider the expression x^{y} in general, where x and y are any two complex numbers. How many values does it have?

When x = 0, the result is trivial. There is only one value, since 0^{y} = 0 for any y (but 0)."

Although it may be trivial, I'm certain that we do not want to claim that 0^{y} = 0 if, say, y = -1. However, I believe that a reasonable argument can be given for taking 0^{y} = 0 when the real part of y is positive.

Regards,

David Cantrell

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