Wrong implication in Complex number

Question

Actually, I am facing a problem with a wrong implication/fallacy in complex number, which is

$\frac{1}{i}=\frac{1\times i}{i\times i}=\frac{i}{i^2}=\frac{i}{-1}=-i\implies i=-i$

I know I should not ask this kind of question here, but I want to learn the reason behind this type of unwanted fallacies.
Can anybody help me? Thanks for your assistance in advance.

Answer 1

Your argument is not a fallacy: it just proves that $i^{-1}=-i$, which is also clear from the fact that $$ (-i)i=-i^2=-(-1)=1 $$

However it's easy to make a true fallacy out of it: $$ \sqrt{-1}=\sqrt{\frac{1}{-1}}\color{red}{=} \frac{\sqrt{1}}{\sqrt{-1}}=\frac{1}{\sqrt{-1}}\tag{⚡} $$ Together with your argument this seems to prove that $i=-i$.

However, the step marked in red is fallacious. And it's very bad practice using the ambiguous symbols $\sqrt{-1}$: in the complex numbers there is $i$, such that $i^2=-1$, but writing it the other way may lead to false arguments. The problem is that it is not possible to define a function $z\mapsto\sqrt{z}$ such that, for all complex numbers $z_1$ and $z_2$, $$ \sqrt{z_1z_2}=\sqrt{z_1}\sqrt{z_2} $$ and the fallacious ⚡ uses such a (non existent) function.

Answer 2

What makes you think it's incorrect? :)

It happens to be true that $$\frac1i = -i$$

All your first computations are justified and correct. However, we have that $1/i\neq i$. I think you're confusing the two. Remember that in general we don't have $1/x = x$, either.

More generally for a complex number $z = a+bi$, we have that $$\frac{1}{z} = \frac{1}{a+bi} = \frac{a-ib}{(a+ib)(a-ib)} = \frac{a-ib}{a^2+b^2}. $$ Here the denominator is real, and you can see when $a = 0$ and $b =1$ (which is your example):

$$\frac1{a+ib} = \frac{1}{0+i\cdot 1} = \frac{1}{i} = \frac{0-i\cdot 1}{0^2+1^2} = -i. $$

Answer 3

From $\frac{1}{i}=-i$ we obtain $1=i\cdot -i$, and in no way that $i=-i$. So your last step is incorrect, and you also give no computation to show it.

For any complex number with $z=-z$ we obtain $2z=0$, and since $2\neq 0$ this implies $z=0$. Again, $i\neq 0$.