# What is the norm of a complex number?

I'm in a number theory class and I'm trying to understand what the norm is... For some complex number $Z = a +bi$, $Z$ times the conjugate of z is equal to $(a^2)+(b^2)$.

Most of what I've read about the norm online says that the norm of $z$ would then be the square root of $(a^2)+(b^2)$. i.e. $\sqrt{(Z**z*)}$. Which I guess makes sense because that would be the absolute value/size of $z$ (which from what I understand is the purpose of finding the norm)

But I also have a number theory textbook which says that the norm of $Z$ is simply $Z**z*$ without the square root. Can anybody shed any light on this? Am I just missing something?

Thanks

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Not missing anything. Different fields use words in different, though sometimes related senses. –  André Nicolas Jan 23 '13 at 22:49
Which norm? There are different norms, so it's a bit silly to ask about the norm. –  wim Jan 24 '13 at 1:58

In number theory the word "norm" is used with a different meaning than in analysis. While in anaylsis it is important that every positive real number be equal to its own norm, for the use of norm in number theory it is much more important that (for instance) the norm of $a+bi$ be integer if $a$ and $b$ are integers. In some cases the norm may even be negative, for instance in the ring $\mathbf Z[\sqrt 3]$ one would define the norm of $a+b\sqrt 3$ to be $a^2-3b^2$, which is often negative, but it does have the property that an element is invertible if and only if its norm is so (in $\mathbf Z$, i.e., the norm is $\pm1$). In general, in a finite field extension of $K$, the norm of an element is the determinant of the $K$-linear operator defined by multiplication by that element (and the trace of that operation is called the trace of the element).

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Think of a complex number $z = (a+ib)$ as a point in the plane $(a,b)$. Think about the vector from the origin to the point $(a,b)$. The length of this vector is typically defined by $\sqrt{a^2 + b^2}$. See the Pythagorean theorem for why.

The norm of $|z|$ is just the length of this vector. When $z$ is real, the norm is the absolute value. It generalizes the notion of the absolute value being the "distance from 0" for real numbers to the distance from the origin.

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In number theory, though, the "norm" is the square of the usual normal. This is to keep the number an integer when $a,b$ are integers (and because it generalizes to number fields where the "length" interpretation doesn't make sense.) –  Thomas Andrews Jan 23 '13 at 22:40
OK thanks that makes sense –  user58437 Jan 23 '13 at 22:43

A given space or set can have a different norm defined on it for different situations. In general, a norm is mostly a function from the space in question to the non-negative reals. In your question, both of the examples you give are valid norms on $\mathbb{C}$. The one with the square root is far more common. The number theory book could have a typo or just be using a slightly different norm. http://en.wikipedia.org/wiki/Norm_(mathematics)

Thomas Andrews' comment in another answer explains why the number theory book uses a different norm. Both are valid, but unless told otherwise you should assume the one with the square root is being used.

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Except in number theory, the norm is not always non-negative. For example, in the ring $\mathbb Z[\sqrt{2}]$ the norm of $a+b\sqrt{2}$ is $a^2-2b^2$. This form of norm is less geometric than it is something algebraic. –  Thomas Andrews Jan 23 '13 at 22:42
Hmmm... clearly the word "norm" is fraught with peril. It is often presented as a more rigorous concept of magnitude and therefore non-negative, but I'm seeing some other uses of the word in mathematics that are different in many ways. I'll leave my answer as is with your comment as a caveat since I think starting with the idea of a function to the non-negative reals with three other qualities is a good transition for someone not used to more esoteric uses of the word. –  Todd Wilcox Jan 23 '13 at 22:46

Though this has already been answered, I wanted to give the "geometry" of what is going on.

If we treat multiplication by $a+bi$ as a linear transformation on the space $x+yi$, then, under the usual basis of $1,i$, multiplication by $a+bi$ is written as the matrix multiplication:

$$\left(\begin{matrix}a &-b\\b&a\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}ax-by\\ay+bx\end{matrix}\right)$$

In number theory, the "norm" is the determinant of this matrix. In that sense, unlike in analysis, the norm can be thought of as an area rather than a length, because the determinant can be interpreted as an area (or volume in higher dimensions.) However, the area/volume interpretation only gets you so far. The reality is that the determinant of a matrix is an "algebraic" quantity which has the nice property that it is independent of the basis chosen, so it is well defined without picking a basis.

In particular, if $A,B$ are two $n\times n$ matrices, then $\det AB = \det A \det B$, which means that the norm defined this way has the same property: $N(z)N(w)=N(zw)$. That nice property follows through to other cases of "number fields" and their rings of integers where the "area" interpretation is less clear-cut.

In particular, this "algebraic norm" is not measuring distance, but rather measuring something about the multiplicative behavior of $a+bi$. That it turns out to be the square of the geometric norm in this case is a deep geometric fact about the geometry of complex numbers.

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Suppose $z = x+iy$, with $x$ real part and $y$ imaginary part, using Euclidean norm

norm(z) = $\sqrt{(\overline{z}z)} = \sqrt{(x-iy)(x+iy)} = \sqrt{x^2+y^2}$

Your textbook might have misplaced the square root.

This is just a particular norm we normally use for complex number, there are infinitely other norms, but they all satisfy these axioms:

1) Positive homogeneity

2) Triangle inequality

3) zero norm iff zero vector

All norms with these axioms can be studies together, common norms are Euclidean norm (2-norm), which is what I wrote above, you sum up all the component squared and take square root of the sum. There could be 3-norm, sum up all the component cubed and take cubic root, and infinite norm, simply take the maximum component's absolute value as the norm, or 1-norm, taking the sum of absolute value of all components. All these norms fulfill the axioms and in fact are equivalent.

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