# What is the mathematical nature of $i$? [duplicate]

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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.

## marked as duplicate by John B, 3SAT, Kamil Jarosz, Community♦Feb 12 '16 at 16:45

• $i$ 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
• 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 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: en.wikipedia.org/wiki/Gaussian_integer – Gyro Gearloose Feb 12 '16 at 16:43

$i$ is an abbreviated form for $(0,1)$.
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 $(a,b)\dot (c,d) = (ac-bd, ad+bc)$
• 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 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