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It is well known that $i$ is unit imaginary part of any complex number, but many uses of $i$ show that has others mathematical properties, for example in integration area, if I want to compute integral of $ix$ I will get $i \frac{x²}{2} $ then here $i$ is considered a constant. Also if I want to check divisors of $i$ I got only $1$ then here $i$ has divisors however it is not integer.

Really I would like to know more about the nature of "unit imaginary part" $i$, or what the Mathematical nature of $i$ is?

Thank you for any help.

Bobson Dugnutt
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zeraoulia rafik
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    $i$ is just a number and should be treated as any other numbers. – N. S. Feb 12 '16 at 16:36
  • if it is a number , is it integer ? – zeraoulia rafik Feb 12 '16 at 16:38
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    No, it is not an integer, nor a real number. It is a "complex number." It is something called an "algebraic integer," but that is not the same as being an integer. $\sqrt{2}$ is an algebraic integer, for example, but it is not an integer. Similarly, there is a class of numbers called "Gaussian integers," and $i$ is a Gaussian integer, but again, not an integer. – Thomas Andrews Feb 12 '16 at 16:39
  • Look at https://en.wikipedia.org/wiki/Field_extension for better understanding. $\mathbb C$ is isomorphic to $\mathbb R[X]/(X^2+1)$. (And be warned, this is a big thing, don't be frustrated if you don't getunderstanding in 10 minutes). – Gyro Gearloose Feb 12 '16 at 16:40
  • Integers are defined to be real, but there's an extension: https://en.wikipedia.org/wiki/Gaussian_integer – Gyro Gearloose Feb 12 '16 at 16:43
  • @user51189 no, $i$ is not an integer. It is a complex number which is neither real, nor integer nor rational. – N. S. Feb 12 '16 at 16:44
  • @ThomasAndrews sounds like your comment should be the answer. – lisyarus Feb 12 '16 at 16:46

1 Answers1

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$i$ is an abbreviated form for $(0,1)$.

Its introduction is particularly useful because it lets you remember what the rules of complex product are.

E.g. $i^2=-1$ is more simple to remember than $(0,1) \dot (0,1) = (-1, 0)$ Also $(a+ib)(c+id)=ac-bd+i(ad+bc)$ is more simple to remember than $(a,b)\dot (c,d) = (ac-bd, ad+bc)$

trying
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    That's certainly one way to look at the complex numbers, but it is far from the only way to look at them. – Thomas Andrews Feb 12 '16 at 16:42
  • It is worth mentioning that this representation of complex numbers might be circular or even misleading. Is there an intuitive way , a natural way to introduce this kind of operation between vectors in $\mathbb R ^2$? – Ranc Feb 12 '16 at 16:43
  • @Ranc My favourite way to introduce the complex numbers is a the set of matrices of the form $$\begin{bmatrix} a & b \ -b & a \end{bmatrix}$$ This set with standard matrix addition and multiplication is a field which is isomorphic to $\mathbb C$. Moreover, conjugation is just transpose of matrices, and $|z|^2$ is just the determinant. You can then link the inverse formula for complex numbers with the standard formula of calculating the inverse of a $2 \times 2$ matrix. – N. S. Feb 12 '16 at 16:47
  • @Thomas Andrews I completely agree with you, but the question, I think, is on the nature of $i$ not on the nature of complex numbers. – trying Feb 12 '16 at 16:48
  • @Ranc And there is a natural way to discover this set of matrices: For a complex number $w \in \mathbb C$ define $T_w: \mathbb C \to \mathbb C$ as $T_w(z)=wz$. Now threat $\mathbb C$ as a vector space over $\mathbb R$ and write the corresponding matrix with respect to the basis ${1,i }$. You'll get exactly these matrices. The fact that this behaves perfectly wrt addition and multiplication is just the observation$$T_{w_1+w_2}=T_{w_1}+T_{w_2} \T_{w_1 w_2}=T_{w_1} \circ T_{w_2}$$ – N. S. Feb 12 '16 at 16:51
  • @N.S. ok now this isomorphism seems a whole lot more natural to me. This form indeed encompasses alot of intuition regarding complex numbers and even the Cauchy-Riemann Eqs. can be derived from this (as the differential of a complex function must be a conformal map).

    I was trying to say that the above way - showing how to multiply vectors in $\mathbb R ^2$ introduces a non intuitive operation. Where as matrix multiplication and summation is more intuitive and meaningful (after a short class in linear algebra).

     – Ranc Feb 12 '16 at 16:55
  • @trying That's not my point. My point is that the representation of $i$ as $(0,1)$ is only one way to look at $i$, based on only one way to look at the complex numbers. There are many other ways to look at $i$, and I thought your answer was too definitive: "is an abbreviated form of..." rather than "can be seen as an abbreviated form of...." – Thomas Andrews Feb 12 '16 at 17:00
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    @Ranc I agree, it looks artificial too me too, unless the students took a class in Ring theory. Then, you can explain the $\mathbb R^2$ construction easily. Basically $\mathbb C$ is just $\mathbb R[X]/ <X^2+1>$. Now, every element in here can be identified with a vector in $\mathbb R^2$ in the standard way: $1,X$ is a basis and you can replace each polynomial by its coordinates with respect to this basis. The complex multiplication is just multiplication in the factor ring $\mathbb R[X]/ <X^2+1>$ translated to coordinates. So it is very natural, if students know about factor rings ;) – N. S. Feb 12 '16 at 17:00