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Imaginary Numbers to Imaginary Powers
Name: Kevin
Status: student
Age: N/A
Location: N/A
Country: N/A
Date: N/A
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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Update: June 2012
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