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Q: maths stuff ( Answered 5 out of 5 stars,   0 Comments )
Subject: maths stuff
Category: Miscellaneous
Asked by: mongolia-ga
List Price: $4.00
Posted: 12 Dec 2004 12:33 PST
Expires: 11 Jan 2005 12:33 PST
Question ID: 441682
what is (1+i) raised to the power of (1+i) expressed as another complex number?
Thank You

Subject: Re: maths stuff
Answered By: mathtalk-ga on 12 Dec 2004 13:43 PST
Rated:5 out of 5 stars
Hi, mongolia-ga:

The seemingly simple operation of raising a complex number to a
complex power is not well-defined in general.  This is a bit like the
fact that a nonzero complex number has two square roots, three cube
roots, etc., but in your case there turn out to be infinitely many

One can raise a positive real number to a complex power, and the
result is uniquely defined.

One can raise a nonzero complex number to a (real) integer power, and
the result is again uniquely defined.

Here we have a nonzero (complex) Gaussian integer raised to a like
power, but the result is not uniquely defined.  To see why, let's
write (1+i) as a power of e:

  1 + i = e^( ln(2)/2 + i*(pi/4) )

But for any integer k, e^(i*(2pi*k)) = 1.  In other words the
exponential function (base e) has "imaginary" period 2pi*i, and any
multiple of 2pi*i can be added to an exponent of e without changing
the result.  Therefore the general way of expressing 1 + i as a power
of e is the following:

  1 + i = e^( ln(2)/2 + i*(pi/4) + i*(2pi*k) )

for any integer k.

Therefore when one attempts to raise both sides to the power (1+i):

(1 + i)^(1 + i)

    = e^( (1+i)*(ln(2)/2 + i*(pi/4) + i*(2pi*k)) )

    = e^( (ln(2)/2 - pi/4 - 2pi*k) + i*(ln(2)/2 + pi/4 + 2pi*k) )

    = e^( (ln(2)/2 - pi/4 - 2pi*k) + i*(ln(2)/2 + pi/4) )

Even after disposing of the term i*(2pi*k) in the exponent, which
would not affect the result of exponentiation, we still have the term
-2pi*k involving an infinite number of possibilities for k.

In other words there is no unique answer, but instead a countably
infinite number of possible results.

For the particular choice k = 0, the result would be:

  e^( (ln(2)/2 - pi/4) + i*(ln(2)/2 + pi/4) )

which is approximately:

  0.27395725383... + i * 0.58370075876...

Taken together all these possibilites form a doubly-infinite geometric
progression whose common ratio is:

  e^(2pi) ~  535.49165552...

Plotted in the complex plane these points all lie on a ray extending
from the origin in the first quadrant at angle ln(2)/2 + pi/4 radians,
or 64.857204014... degrees, to the positive real axis.

regards, mathtalk-ga
mongolia-ga rated this answer:5 out of 5 stars and gave an additional tip of: $1.00

although you fully answered my question your answer has begged some
more questions. However I will submit these as new questions.
Many Thanks

There are no comments at this time.

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