Strange Imaginary Numbers Make 1=-1

Question

Let $i=\sqrt{-1}$

$$\frac{1}{\sqrt{-i}}=\frac{1}{\sqrt{(-1)i}}=\frac{1}{i\sqrt{i}}=\frac{-i}{\sqrt{i}}$$

And

$$\frac{1}{\sqrt{-i}}=\sqrt{\frac{1}{-i}}=\sqrt{\frac{-1}{i}}=\frac{i}{\sqrt{i}}$$

However,

$$\frac{-i}{\sqrt{i}}\ne\frac{i}{\sqrt{i}}$$

Where did I go wrong?

The problem is with "$\sqrt{ab}=\sqrt{a}\sqrt{b}$" that you use in the second equation. — quid, Jul 27 '17 at 13:10
Yes, This is a duplicate of that question. This Answer was perfect. — Astor Florida, Jul 27 '17 at 13:18

score 3 · Accepted Answer · answered Jul 27 '17 at 13:12

You are applying some well-known properties of the square root: $$\sqrt{ab}=\sqrt{a}\sqrt{b} \quad \mbox{and} \quad \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$$ but these do not hold in general. You probably saw and maybe even proved these for positive real numbers $a$ and $b$, but now you're applying them in a more general context - probably without asking yourself it these properties are still valid.

See also: Square roots of negative numbers.

score 1 · Answer 2 · answered Jul 27 '17 at 13:35

To complement the other answer,

$\sqrt z$ is a multivalued function. Once you choose a branch of $\log$, say the principal branch $\log z= \ln |z| + i \arg z$, $\arg$ is the principal argument, taking values in $(-\pi,\pi]$, you define:

$$\sqrt z = e^{\frac12 \log z}$$

Then $\sqrt 1 = 1$ but $\sqrt{-1} \sqrt{-1} = i \times i = -1 \neq \sqrt{-1 \times -1}$, so this function (square root) is not multiplicative, i.e. in general we don't have $\sqrt{ab} = \sqrt a \sqrt b$.

If you take any other branch you can similarly show that the resulting function is not multiplicative.

Strange Imaginary Numbers Make 1=-1

2 Answers2