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Name: Kevin
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Question:
I was messing around with my calculator one day and I have stumbled onto this little problem (i is the imaginary number, square root of negative 1) i^i^i^i entered like that roughly equals 4.810477381. How does the calculator figure this out? Oh and this also happens when you do i^i^i^i^i^i^i^i^i= 4.810477381 and when you have any number of i's that are divisible by 4 (8,12,32,40, etc.)



Replies:
Maybe the easiest way to see this is to write i as follows:
i = cos(pi/2) + i * sin(pi/2)
or, equivalently (though not obviously)
i = e^(i*pi/2)

Now if we raise both sides to the power i,
i^i = (e^(i*pi/2))^i = e^(i*i*pi/2) = e^(-pi/2)
continuing...
i^i^i = (e^(-pi/2))^i = e^(-i*pi/2)
and so
i^i^i^i = (e^(-i*pi/2))^i = e^(-i*i*pi/2) = e^(-pi/2) = 4.810477...

Tim Mooney


So called "complex numbers" or "imaginary numbers" really have a bad reputation that is the result of history, since they are neither particularly "complex" nor "imaginary". They just obey different rules -- in particular the rule for multiplication -- which of course is closely related to multiplication. The multiplication of two complex numbers (a+ib)*(c+id) = (ac-bd) + i*(bc + ad). You can verify this by multiplying the numbers like they are binomials and using the definition that: i^2 = -1. The web site:

http://home.att.net/~srschmitt/script_complex_power.html

is very useful in explaining what happens when the exponent is a complex number itself. It makes use of the identity: e^ix = cos(x) + i*sin(x). This particular web site uses the letter "j" instead of the letter "i" which is commonly used by electrical engineers because they use the letter "i" to mean the electric current, so that letter is "reserved". So i^i (or j^j) = e(-pi/2) = 0.20788... The web site gives a "calculator" and also provides a clear derivation of complex numbers raised to complex powers, which often turn out to be "real" numbers.

After calculus, probably the most important and useful branch of mathematics is complex variables. Any student considering a career in science or engineering should take a course in complex variables ASAP, because without it you won't be able to go very far in science or engineering without it.

Vince Calder


Kevin,

This is based on two principles. The first is presented in algebra but often forgotten: (a^b)^c=a^(bc). This yields, calculated from the inside out:
(((i^i)^i)^i)=(((i^(i^2))^i)=((i^-1)^i)=(i^-i).

To continue requires one more principle arrived at through calculus and properties of complex numbers: e^(ix)=cos(x)+i*sin(x), where x is in radians. When x=/2, e^(i/2)=i. This can then lead to: i^-i=(e^(i/2)^-i=e^-(i^2*/2)=e^(/2) If you calculate this quantity, you will find it to equal 4.81. You can use the procedure from the first paragraph for any number of exponents. You will find that having any multiple of four "i"'s will result in this answer.

Dr. Ken Mellendorf
Physics Instructor
Illinois Central College


Kevin,

Facts: (i) = (pi/2)i and Ln(a^b)=e^(b*Ln(a)).
So i^i = e^(i*Ln(i))=e^(i*(pi/2)i)=e^(-pi/2).
So (i^i)^i=[e^(-pi/2)]^i=e^(i*Ln(e^(-pi/2)))=e((-pi/2)i) = -i.
Hence x = ((i^i)^i)^i = e^(pi/2).
Since, x^i = e^(i*Ln(x))=i, and the pattern is cyclic.

Paul Beem



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