Proof that 1 = 5?

Question

Let i = sqrt(-1) (that is, let i = (-1)^(1/2)
i = i^5
log (i) = log (i^5)
log (i) = 5*log (i)
1 = 5

Q.E.D

What's wrong with this?

Answer 1

The problem with that proof is that the logarithm of a complex number has infinitely many values.

i =
... = e^(-9i*pi/2) = e^(-5i*pi/2) = e^(i*pi/2) = e^(5i*pi/2) = e^(9i*pi/2) = ...

To say, "log i = 5 log i," isn't quite true. It's better to say something like, "There exist complex numbers z and w such that e^z = e^w = i and z = 5w." Then it is clear that you're talking about two different values of log i.

Answer 2

Now i^4 = 1 so log(i^4) = log(1) = 0

Write i^5 = i*i^4
So i = i^5 = i*i^4
log(i) = log(i) + log(i^4) = log(i) + 0 = log(i)

This looks good but it also has a flaw. The imaginary part of the log is being ignored just as in yours as is shown below.

Actually what you said will be true for any power 1+4k where k=0,1,2,3,...

But as someone else said there are infinite solutions to log(i) and they are:
log(i) = i(pi/2 + 2kpi)
Also:
log(i^5) = 5 [i(pi/2 + 2mpi) ]
Set these equal and get:
1/2 + 2k = 5/2 + 10m
2k = 2 + 10m
k = 10m + 1 so m=0, k=1 or m=1, k=11 and so on

Answer 3

What is wrong with this is this:

log(i) = log [sqrt(-1)] = 1/2 * log (-1)

but logarithms are not defined for negative numbers.

edit:
I like Brent's solution, too. I didn't try to convert to polar; "infinitely many solutions" is just as bad as "undefined".

Answer 4

1 is 1 and 5 is 5. It's impossible to equate one with the other as 1 is one thing and 5 is another. You need to add some more expansion or parameter to your problem equation.

Answer 5

It doesn't make any sense.