## Monday, December 31, 2007

$(-z)^2 = z^2$

where z is a complex number. Hence,
$2 \log(-z) = 2 \log(z)$

So,
$\log(-z) = \log(z)$.

What is the fallacy?

Oh and congratulations to those who selected the option (b) in the previous post! :o

KTC said...

You can't just bring along your properties for real-value logarithm along to the party and use it for complex logarithm. ;-P

Beans said...

Hey KTC,

Haha, it took me a while to realise that! (Well I had spotted that paradox in the book before actually "defining exp(z)" for complex numbers.)

Although life in the reals seems more appealing than life in the complexes at the moment! (I don't care that if you can differentiate once in the complex plane, then all higher derivatives exist). I wonder though, what will I be saying after my exams?!

I think it is more correct to write that if y^2=(-x)^2, then y =|(-x)|.

Jake said...

but that isn't true.

e.g. y = -2
x = 2

then y^2 = 4 = (-x)^2

but |(-x)| = 2

Beans said...

Hi Jake,

Yes you are right--thanks for letting me know. I meant to say \sqrt{x^2}= |x| (where x is a real number), and not just x. (I think so anyway!)

So to correct what I previously wrote: If y^2=(-x)^2, then |y|= |-x|. The example I had in mind was (x-1)^2, which we then wrote as |x-1| upon "square rooting".

Jake said...

"Yes you are right--thanks for letting me know. I meant to say \sqrt{x^2}= |x| (where x is a real number), and not just x. (I think so anyway!)"

Depends on the context i.e. whether you are looking for a proper function or both roots of an equation etc. where you might say that sqrt(x^2) = x or -x

but you always have |y|=|-x|=|x|