English in math, complex numbers

Il Pitagorico
Hello,
I have never written in this forum section, this is the first time that I try talk about math in English. I have some problems in English and I can't understand very well when someone speaks to me in English (and I do so many mistakes when I write, Correct me please!). What about talk complex numbers? That is a very interesting matematical proof for me, but I don't know nothing about complex numbers. Can you talk about the basis of complex numbers?
Thank you very much.
The Pitagorico

Risposte
Zero87
"WalterLewin90":
"Seguent"? I'm pretty sure that the world "seguent " doesn't exist. Did you mean "following"?

I remember a "sketch" of Aldo, Giovanni & Giacomo, in which Giacomo traslated all in english and he used often the word "seguent". :lol:

WalterLewin90
[quote=Zero87]Hello world, my english is awful, but I try to explaining something useful for you. ;-)

Complex numbers may be defined in several ways, but I think the simplest definition is the seguent:
/quote]

"Seguent"? I'm pretty sure that the world "seguent " doesn't exist. Did you mean "following"?

Caenorhabditis
"Zero87":

It is a Theroem which has a nice and simple proof for people who have knowledge of complex analysis... :roll:

Courant and Robbins (What is mathematics?) gave a very simple proof of this, based on the fixed-point theorem.

Zero87
"Caenorhabditis":
A very nice fact about complex numbers is that every equation of $n$-th degree in the complex field has got exactly $n$ solutions. So, the equation
$z^456- \frac{24}{23}z^118+\sqrt{39}z^66-12.5=0$
has got precisely 456 solutions!

"Delirium":
...counted with multiplicity.

It is a Theroem which has a nice and simple proof for people who have knowledge of complex analysis... :roll:

Then, I want to underline this previous passage of Caenorhabditis (hi! :D :smt039 )
"Caenorhabditis":
So, the equation
$z^456- \frac{24}{23}z^118+\sqrt{39}z^66-12.5=0$
has got precisely 456 solutions!

I remember a nice result which says that if we have $m\ge n+1$ zeroes of a polynomial equation of degree $n$, the polynomial must be equal to zero ($P(x)\equiv 0$).

Caenorhabditis
"Delirium":
...counted with multiplicity.

Yes, of course.

Sk_Anonymous
...counted with multiplicity.

Caenorhabditis
A very nice fact about complex numbers is that every equation of $n$-th degree in the complex field has got exactly $n$ solutions. So, the equation
$z^456- \frac{24}{23}z^118+\sqrt{39}z^66-12.5=0$
has got precisely 456 solutions!

Stellinelm
"Zero87":

Have a nice weekend forumists. :smt039

Thanks . Catch you soon :-D
007 , ladies & gentlemen , i wish you a good weekend :wink:

Zero87
"Pianoth":
[quote="Zero87"]$ (1) $ There is another way to indicate $ Re(z) $ and $ Im(z) $ but I don't know LaTeX code for them...
Do you mean like \(\mathfrak{R}(z)\)? Anyway you can also avoid the italic $Re$: \(\operatorname{Re}(z)\)[/quote]
:smt041
\(\mathfrak{R}(z)\)

:smt039

Pianoth
"Zero87":
$ (1) $ There is another way to indicate $ Re(z) $ and $ Im(z) $ but I don't know LaTeX code for them...
Do you mean like \(\mathfrak{R}(z)\)? Anyway you can also avoid the italic $Re$: \(\operatorname{Re}(z)\)

Pianoth
Let me first give you some grammatical advices:
"Il Pitagorico":
Hello,
I have never written in this forum section, this is the first time that I try talk$text()^1$ about math in English. I have some problems in English and I can't understand very well when someone speaks to me in English (and I do so many mistakes when I write, Correct me please!). What about talk complex numbers?$text()^2$ That is a very interesting matematical proof$text()^3$ for me, but I don't know nothing$text()^4$ about complex numbers. Can you talk about the basis of complex numbers?
Thank you very much.
The Pitagorico

$text()^1$ Fortunately that is correct, but pay attention, because that is an infinitive, namely a verb form that is used as a noun, so it's "I try talking/to talk/talk" (the last form is used the least actually, but it's still correct)
$text()^2$ Same thing here, talk is an infinitive so it is correct, but the sentence in this case is not: What about talk about complex numbers? This should be the correct form (otherwise it means "Che dite di parlare i numeri complessi" and that doesn't mean anything). As you can see it sounds bad because you need to repeat two times the word "about"... Just say Why don't we talk about complex numbers?
$text()^3$ I think you mean topic/subject/argument, since you haven't proved anything :lol:
$text()^4$ The double negative is almost never used in English, and in this case it's even wrong. Correct is I don't know anything or I know nothing.
___________________________________________________

Zero87 has already explained the basis, so I'll just add some details:
We started from natural numbers, like $1, 2, 3, 4, \ldots$. If we want the natural number system $NN$ to be extended to the integer number system $ZZ$, we need to create negative numbers. If we want to extend $ZZ$ we need to create rational numbers (like $3/5$) so we can create the set $QQ$, and to extend even more we must create irrational numbers, such as $sqrt(2)$, $log_2 12$,$\phi$, ecc. (algebraic, this set is actually indicated with symbol $\bar(QQ)$, if I am not wrong) and $pi$, $e$, ecc. (trascendental). This way we can represent any quantity on a continuous line, and we name the set that includes every quantity $RR$ (the set of the real numbers).
Now, if we want to expand this set we need to create imaginary numbers, we define them as numbers that can be written as real numbers multiplied by the imaginary unit $i$, which is defined as $i^2=-1$. This can create some problems, so we must also create the principal square root of a complex number. I won't go into details anyway, because this could be very confusing. However, now we can extend the set of real numbers $RR$: we define a complex number as the sum of a real number and an imaginary number ($a + bi$ where $a$ and $b$ are real numbers and $i$ is the imaginary unit).
If we usually represent a general natural number with the letter $n$, a real number with the letter $x$, or $a$, well, we usually represent a complex number with letter $z$.
Since we can already represent all real numbers on a continuous line, to represent a complex number we need two dimensions. So, we can say that a complex number can be viewed as a point or a position vector in a two-dimensional Cartesian coordinate system called complex plane, where horizontal axis is the real axis (indicated usually by "Re" or $RR$) and the vertical axis is the imaginary axis (indicated usually by "Im").
Hope that is clear enough.

Meringolo1
"Zero87":

It doesn't have real solutions (in high school we call him "falso quadrato" ;-) ),



Well,I think it's more accurate to say "...we call it "falso quadrato"", cause we talking about an object. :wink:
But this is just find "the hair in the egg" :D

Zero87
Hello world, my english is awful, but I try to explaining something useful for you. ;-)

Complex numbers may be defined in several ways, but I think the simplest definition is the seguent:

A complex number $z$ is a number of the form $z=x+iy$ in which
- $x,y\in \RR$ are the real and complex part of $z$
- $i$ is the imaginary unit, $i=\sqrt(-1)$.

So, there are many ways to explain why $i$ is $\sqrt(-1)$, but I think is better go on. ;-)
But we have, for example, $\sqrt(-25)= \sqrt(25\cdot (-1))= \sqrt(25)\cdot \sqrt(-1)= \pm 5i$.

A little notation hint:
- $x$ is the real part of $z$ and it can be indicated by the symbol $Re(z)$ (or $Rez$, but I don't like this form)$(1)$
- $y$ is the imaginary part of $z$ and it can be written as $Im(z)$.

We can note that complex numbers are an extension of the real ones, in fact a real number is a complex number without imaginary part (or with null imaginary part):
$25= 25+0i$
$\sqrt(2)= \sqrt(2)+0i$
and so on...

Now we can use complex number to give a sense to the roots of polynomials.

Example 1.
$x^2+1=0$
It doesn't have real solutions (or real roots).
$x^2=-1$ implies $x=\pm \sqrt(-1)$
and so
$x=\pm i$
This equation has two complex roots.

Example 2.
$x^2 + 2x+4=0$
It doesn't have real solutions (in high school we call it "falso quadrato" ;-) ), but
$x_(1,2)= -1 \pm \sqrt{1-4}= -1 pm \sqrt(-3)=-1 \pm i \sqrt(3)$.

___
$(1)$ There is another way to indicate $Re(z)$ and $Im(z)$ but I don't know LaTeX code for them... :roll:

EDIT
Thanks to Meringolo for correction. :)
Have a nice weekend forumists. :smt039

Stellinelm
well done!

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