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# imaginary unit

The *imaginary unit* $i:=\sqrt{-1}$. Any imaginary number $m$ may be written as $m=bi$, $b\in\mathbb{R}$. Any complex number $c\in\mathbb{C}$ may be written as $c=a+bi$, $a,b\in\mathbb{R}$.

Note that there are two complex square roots of $-1$ (i.e. the two solutions to the equation $x^{2}+1=0$ in $\mathbb{C}$), so there is always some ambiguity in which of these we choose to call “$i$” and which we call “$-i$”, though this has little bearing on any applications of complex numbers.

Related:

Imaginary,Complex

Synonym:

i

Type of Math Object:

Definition

Major Section:

Reference

## Mathematics Subject Classification

12D99*no label found*

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## Recent Activity

Jul 5

new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

Jun 23

new question: A trascendental number. by Ron Castillo

Jun 19

new question: Banach lattice valued Bochner integrals by math ias

new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

Jun 23

new question: A trascendental number. by Ron Castillo

Jun 19

new question: Banach lattice valued Bochner integrals by math ias

## Comments

## imaginary unit

I would rather say something like:

We define "i" to be a root of f(X)=X^2-1=0. With this choice, the polynomial f has two roots, namely i and -i.

But I guess the entry is fine as it is.

Alvaro

## Re: imaginary unit

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

sqrt(-1)sqrt(-1)=i.i=-1

What do you think?

## Re: imaginary unit

Well, sqrt(ab)= sqrt(a)sqrt(b) is only valid for a,b>0?

## Re: imaginary unit

Of course not.