Wild About Math! Making Math fun and accessible

9May/091

MMM #31: Winner!

This was a very popular contest. 57 of you sent in your solutions and almost every answer was correct.

Buddha Buck was selected by random.org as the winner of this contest. Here is Buddha's solution:

Problem: Simplify (1/i)^(1/i) - i^i (where i is sqrt(-1))

Solution:

i*i = -1, -i*i = 1, so 1/i = -i, therefore

(1/i)^(1/i) = (1/i)^(-i)

a^(-b) = (1/a)^b, for all a, b, therefore

(1/i)^(-i) = i^i

So, putting this all together...

(1/i)^(1/i) - i^i
= (1/i)^(-i) - i^i
= i^i - i^i
= 0

There were various ways to approach the problem. Aside from Buddha's, many of you used Euler's formula.
Her's one such example from Vladimir Filimonov:

The answer for the contest is 0

Using Euler formula one can write: i=exp(i*pi/2). The following steps become obvious:
i^i=exp(i*pi/2*i)=exp(-pi/2)
(1/i)^(1/i)=1/exp(i*pi/2/*(1/i))=1/exp(pi/2)=exp(-pi/2)
and finally:
(1/i)^(1/i)-i^i=exp(-pi/2)-exp(-pi/2)=0

Alec Cooper provided more detail on how to use Euler's formula:

The definition of a complex number to a complex exponent (note that this is the principal branch of the complex logarithm):
[1] a^b = e^(b * log(a))

[2] (1/i)^(1/i) - i^i = e^(1/i * log(1/i)) - e^(i * log (i)) from [1]

Representing a complex number a in polar form (note that theta is on the interval (-pi, pi] ):
a = r * e^(i * theta)
Then taking the (principal branch of the) logarithm of both sides
log(a) = log(r) + i * theta

Expanding 1/i as a complex number:
1/i = a + b*i
==> 1 = a*i + b * i^2
==> 1 + 0*i = -b + a*i
==> 1 = -b, a = 0
[3] ==> 1/i = -i

-i = 0 + i*(-1)
= cos(-pi/2) + i*sin(-pi/2)
= e^(i * -pi/2) from Euler's formula
[4] log(-i) = -i * pi/2

i = 0 + i*(1)
= cos(pi/2) + i*sin(pi/2)
= e^(i * pi/2) from Euler's formula
[5] log(i) = i * pi/2

e^(1/i * log(1/i)) - e^(i * log (i)) = e^(-i * (-i * pi/2)) - e^(i * (i * pi/2)) from [3], [4], and [5]
= e^(-pi/2) - e^(-pi/2)
= 0

So we are done, with (1/i)^(1/i) - i^i = 0

Søren Furbo (and one other person, I believe) pointed out that Google's calculator can be used to verify the answer.

http://www.google.com/search?hl=en&client=opera&rls=en&hs=cAU&q=%281%2Fi%29%5E%281%2Fi%29-i%5Ei%3D&btnG=Search

People have asked me from time to time to acknowledge all of the correct solutions. I'd love to do that and the question is how to do it without it taking lots of time. One simple way to do this would be to ask people to post their solutions as comments on a blog article which I would moderate so that they don't get posted until the contest is over. Unfortunately, this doesn't allow for the many PDF attachments I receive. Another possibility is to write some web-based code that allows you guys to submit solutions, with optional attachments, and then I would post all the correct submissions at the end of the contest. I welcome your ideas on how to largely automate this acknowledgment process.

Stay tuned for the next MMM at Blinkdagger on Monday.

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  1. There is a problem with this: i^i does not have a unique real valued solution. Whilst i = e^[i*pi/2], it also equals e^[i*(pi/2 + 2*pi*k)] for every integer k (just because pi/2 is the principal argument doesn’t mean it is the only value that works), meaning i^i equals any of the values e^[-pi/2 + 2pi*k].

    So, whilst 0 is an answer to (1/i)^(1/i) – i^i, so is any number of the form
    e^[3*pi/2] * (e^[2*pi*k] – e^[2*pi*j]),
    where k and j are integers.


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