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Complex Square Root

February 14, 2008 – 1:45 am

Proof that 1 = -1:

$\frac{1}{\sqrt{i}} = \sqrt{\frac{1}{i}} = \sqrt{-i} = \sqrt{-1} \sqrt{i} = i \sqrt{i}$
$\Rightarrow 1 = i \sqrt{i} \sqrt{i} = i^2 = -1$

This function should be taken outside and shot.

2 Responses to “Complex Square Root”
uh-oh. I think it might be spreading to the reals! Proof that 1 = -1:

sqrt(1) = 1
sqrt(1) = -1

1 = sqrt(1)/sqrt(1) = 1/-1 = -1

By jascha on Mar 24, 2009
Yes of course. Perhaps the post was not clear. My only point is this: With the reals, the mistake is easy to spot. More importantly, one can work with a definition (sqrt = positive root), using which algebra behaves as you expect.

With the complex numbers the mistake is more subtle, and there is no such convenient definition for sqrt. Seemingly innocent manipulations often conceal mistakes, as one must always be careful when crossing the branch cut.

By Guy Gur Ari on Mar 24, 2009