I am not sure about this:
$$\frac{1}{\sqrt{-1}}=\frac1i=-i $$
or
$$\frac{1}{\sqrt{-1}}=\sqrt{-1}=i $$
?
I would say the second but Wolfram Mathematica says
$$\frac{1}{\sqrt{-1}}=\frac1i=-i $$
What is the convention?
What is the convention: $\frac{1}{\sqrt{-1}}=-i $ or $i$?
-1
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complex-numbers
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0Strictly speaking, $\sqrt{i}$ defines globally the pair of solutions to equation $z^2+1=0$. Computer Algebra Systems make a choice, but this choice is not engraved in marble stone. – 2017-01-14
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1@JeanMarie you mean Sqrt[-1] ? – 2017-01-14
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1If you say $\dfrac{1}{\sqrt{-1}}=\sqrt{-1}$ so $1=\sqrt{-1}\sqrt{-1}=ii=i^2=-1$ – 2017-01-14
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0@Mencia Yes my bad Sqrt[-1] – 2017-01-14
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2Could you explain why you would choose the second? – 2017-01-14
7 Answers
2
Let $i=\sqrt {-1}$.
From $i\cdot i=-1$ divide both sides by $i$ to obtain $i=-\frac 1i$
You need to make a consistent choice of sign, because there are two possible square roots $i$ and $-i$. To avoid errors, use a notation which enforces this.
2
This question strikingly shows what calamity is the notation $\sqrt{-1}$. There is no such mathematical object, and the notation $\sqrt{\phantom{a}}$ should be used exclusively for its original aim: denote the positive square root of a positive real number. In all other contexts, we cannot distinguish between the two square roots of a complex number
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0According to you, what should Wolfram Mathematica do when there is something like $1/ \sqrt{-f^2}$ for real $f$ ? – 2017-01-14
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0I really don't know, as this depends very much on the context: in Number theory, one often meets things like $\mathbf{Q}(\sqrt{-3})$, but there is no ambiguity,as it denotes any element satisfying he equation $x^2+3=0$, and it is more of a symbolic, not to say metaphoric, notation. I suppose Mathematica (which I don't use) knows this usage). – 2017-01-14
1
Note that the powers of $i$ are cyclic.
$$\color{red}{i^0=1}$$ $$\color{blue}{i^1=i}$$ $$\color{green}{i^2=-1}$$ $$\color{orange}{i^3=-i}$$ $$\color{red}{i^4=1}$$ $$\color{blue}{i^5=i}$$ $$...$$
This pattern also applies for negative powers.
$$\color{red}{i^{-4}=1}$$ $$\color{blue}{i^{-3}=i}$$ $$\color{green}{i^{-2}=-1}$$ $$\color{orange}{i^{-1}=-i}$$
Therefore, it is easy to deduce that the result of $i^{2017}=i$ is for example, as well as $\frac{1}{i}=-i$ just by dividing the power by $4$ and looking at the remainder obtained.
1
The typographical picture $\sqrt{-1}$ does not define a mathematical entity, but is the formulation of a problem. Therefore it makes no sense to write things like ${1\over\sqrt{-1}}$ or even $e^{\sqrt{-1}\,\phi}$.
When we write "Let $i=\sqrt{-1}$" this is a colloquial expression of the agreement that henceforth $i$ is a constant satisfying $i^2=-1$.
In any case one has ${1\over i}=-i$. This is not a convention but a stringent consequence from $i^2=-1$ and the field axioms valid in the "complex environment".
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0Thank you for your answer. When I asked about the convention, I was not referring to 1/i=-i. I was referring to 1/Sqrt[-1]=i vs 1/Sqrt[-1]=-i. – 2017-01-14
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0According to you, what should Wolfram Mathematica do when there is something like $1/ \sqrt{- f^2}$ for real $f$ ? – 2017-01-14
1
Whenever we have some negative term in square root. Its not good practice to solve it like you are solving in method 2.
You have to replace -1 into $i^2$.
Edit -
Product Rule (extended) -
$\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$
where a ≥ 0, b ≥ 0 Or a ≥ 0, b < 0
But NOT a < 0, b < 0
So $\sqrt{-1} \cdot \sqrt{-1} \ne 1$
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0Thank you for your answer. Apparently that is what Wolfram Mathematica does. Is that something well established among mathematicians? – 2017-01-14
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0See my updated answer. Hope now its clears to you. – 2017-01-14
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Addition to what other said, we can say for $i=0+i=(0,1)\in\mathbb{C}$: $$\dfrac{1}{\sqrt{-1}}=\frac{1}{0+i}=\frac{1}{0+i}\times\frac{0-i}{0-i}=\frac{-i}{(-1,0)}{\color{red}\equiv}\frac{-i}{-1}=i$$
0
$\mathbb{C}$ is a field, so exist $a , b \in \mathbb{R}$ such that $\frac{1}{i} = a + i b$. We will show that $a = 0$ and $b = - 1$. We have that $$ \frac{1}{i} = a + i b \Longleftrightarrow 1 = i (a + i b) = i a + i^2 b = - b + i a\mbox{,} $$ so, taking real and imaginary parts, $$ Re (a i - b) = Re 1 \Longleftrightarrow - b = 1 \mbox{ and } Im (a i - b) = Im 1 \Longleftrightarrow a = 0\mbox{,} $$ because of this $a = 0$ and $b = - 1$.