1 is equal to -i

Question

Pretty simple, but I'm sure there's some subtlety to it I'm missing.

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

Looking at an Argand diagram however, gives some reason to doubt this.

Edit: as is pointed out in the comments/answers this is all kinds of wrong:

primarily, $(-i)^2=(-1)(-1)(i)(i)=i^2=-1$.

also, $a^2=b \Rightarrow \pm a=\sqrt{b}$

In general $\sqrt{a^2}\ne a$ (in fact this is already the case for $a=-1$) — Hagen von Eitzen, Jun 13 '15 at 14:15
Ah of course, dumkopf. no mystery here after all. I think I was confused by the fact that $\sqrt{-1}=\pm i$ — Alex DB, Jun 13 '15 at 14:21
However, you could "prove" in the same way that $-i = i$. ;) The error basically is that $a^2 = b^2$ does not imply $a = b$ (unless you assume, for example, that $a$ and $b$ are positive real numbers). — fkraiem, Jun 13 '15 at 14:23

score 4 · Answer 1 · answered Jun 13 '15 at 14:22

There are two errors in this proof, which both have been mentioned in comments and answers, but it is worth putting them together, for completeness.

(1) It is not true that $(-i)^2=1$. Indeed, for any $x$, $x^2=(-x)^2$, so $(-i)^2=i^2=-1$.

(2) In generally, knowing that $x^2=y^2$ does not let you conclude that $x=y$. The most obvious example is $x=-1,y=1$.

Emilio Novati · Answer 2 · 2015-06-13T15:20:57.423

3

Hint:

$$ (-i)^2=(-i)(-i)=(-1)(i)(-1)(i)=(-1)(-1)(i)(i)=(i)(i)=i^2=-1 $$

so, you have ''missed'' the difference between $(-i)^2$ and $-(i)^2$

edited Jun 13 '15 at 15:20

answered Jun 13 '15 at 14:16

Emilio Novati

score 1 · Answer 3 · answered Jun 13 '15 at 14:20

1

The problem is that $\sqrt{x^2}=|x|,$ which is only $x$ for real values greater than or equal to zero.

In your case $|\pm i|=1,$ but NOT when you drop absolute values.

answered Jun 13 '15 at 14:20

John Molokach

2

In complex domain your first equation is not true. Otherwise we'd get $\sqrt{(1+i)^2}=\sqrt2$. – Ruslan Jun 13 '15 at 14:26
That's why I said 'real values.' The second equation is just a special case that applies to the OP's question. Plus, it does apply to strictly imaginary numbers, which is where the original question originates from. – John Molokach Jun 13 '15 at 15:06
No it doesn't: $\sqrt{i^2}=\pm i\ne1=|i|$. – Ruslan Jun 13 '15 at 18:31
True. I stand corrected. – John Molokach Jun 13 '15 at 18:57

3 Answers3