Absolute value of complex numbers $|a+bi|$

Question

I didn't understand why the absolute value of $(a+bi)$ is equal to $\sqrt{a^2+b^2}$ but not $\sqrt{(a+bi)^2}$ like $|x|=\sqrt{x^2}$

if $|x|=\sqrt{x^2}$ is right and if we give $x = a+bi$ it should be $\sqrt{(a+bi)^2}$

and why this proof is wrong:

$z=a+bi$
$|z|^2=z^2$
$|z|^2=(a+bi)^2$
$|z|=\sqrt{(a+bi)^2}$

Note: I don't want "Cartesian representation of a complex number" I don't see it as proof.

So I wrongly generalized the equation of the absolute value of a real number:

my generalization:
$|x|=\sqrt{x^2}$ (only right for real numbers)
right definition: (by my understanding so far)
$|a_1x_1+a_2x_2...a_nx_n| = \sqrt{\Sigma_n{(a_kx_k)^2}}$

Answer 1

For a complex number $z$, we don't speak about the "absolute value of $z$" but we speak about the modulus of $z$ denoted by $|z|$. If the algebraic form of $z$ is $a + ib$, then by definition you set $|z| = \sqrt{a^2 + b^2}$. Note that this definition extends the absolute value on $\mathbb R$.

Also, if $z = i$, how do you define $\sqrt{i^2}$ ? More generally, how do you define the square root of a complexe number ? How can you distinguish the two roots ?

Answer 2

Starting from the definition of the absolute square $|z|^{2}$ of a complex number $z = a + i\,b$ while keeping the definition of the complex conjugate in mind, i.e., $z^{\ast} = a - i\,b$, such that overall,

\begin{equation} |z|^{2} = z z^{\ast} = \left(a + i\,b\right)\left(a + i\,b\right)^{\ast} = \left(a + i\,b\right)\left(a - i\,b\right) = a^{2} + b^{2}\,. \end{equation}

Obviously, the absolute value is given by

\begin{equation} |z| = \sqrt{a^{2} + b^{2}} \end{equation}

Answer 3

Absolute value gives you distance of a real number from the origin regardless of sign. The concept of a modulus generalizes this to complex numbers. In the complex plane, numbers may not just be on the real line, but anywhere in the plane. The concept of distance from the origin is still maintained and is given by the Pythagorean Theorem, $\sqrt{a^2+b^2}$. This reduces to the absolute value for complex numbers on the real line.

While $z=a+ib$ is the Cartesian representation of a complex number, it can also be represented as $z=re^{i\theta}$ where $r$ is the modulus and $\theta$ is the argument, $\theta = \arctan (b/a)$ and in general $e^{i\theta}=\cos \theta + i \sin \theta$.

Essentially, the definition of the modulus is $|z|=\sqrt{z \bar z}=\sqrt{(a+ib)(a-ib)}=\sqrt{a^2+b^2}$.

$a+ib$ is not a distance, but may have multiple distances associated with it. One might have reason only to care about the real part, or just the imaginary part. Each of these gives you some concept of distance to associate with the complex number.

Answer 4

Why should $|x|=\sqrt{x^2}$? That's a consequence of the definition. Not the definition itself. If $x\in \mathbb R$ then $x = x + 0i$ and $Re(x) = x$ and $Im(0) = 0$. And $|x| = \sqrt{Re(x)^2 + Im(x)^2} = \sqrt{x^2 + 0^2} = \sqrt{x^2}$.

The idea is that $|z|$ is a norm which can be thought of as "the distance" from the origin". It can have a more formal and precise definition but that's the "organic" definition. But some of the fundamental aspects of a norm/distance for $0$ are: its a real, non-negative value and that only the origin has a zero norm.

For real numbers, a positive number is its own value distance from $0$. That is $7$ is $7$ units from $0$ and $\pi$ is $\pi$ units from $0$. So if $x > 0$ then $|x| = 0$. And negative values or their values but in the opposite direction. $-19$ is $19$ units away from $0$ but in the negative direction. And $0$ is $0$ distance from $0$.

So we can describe the $|x|; x\in \mathbb R$ as $|x| =\begin{cases}|x|=x&if x>0\\ |x|=-x&if x< 0\\|x|=0&if x=0\end{cases}$. But that's only a description and it only works if $x\in \mathbb R$.

And some people don't like step-wise definitions. They might note that if $x \in R$ then $x^2 \ge 0$. and for every $y>0$ there are two real numbers, one positive, $x_1$, and one negative, $x_2$, so that $x_1^2 =x_2^2 = y$ and they notice $x_2 = -x_1$. They call the positive one $\sqrt y$.

With that we can note. If $x > 0$ than $\sqrt{x^2} = x$. If $x=0$ then $\sqrt{0^2} = 0$. and if $x < 0$ then $-x > 0$ and $x^2 = (-x)^2$ and $\sqrt{x^2} = -x$. In these three cases we always have $\sqrt{x^2} = |x|$ (as described above) and so they can use $|x| = \sqrt{x^2}$ as description for absolute value.

BUT THIS ONLY WORKS IF $x$ IS A REAL NUMBER!!!!!

Complex numbers we need a concept distance from origin. $1 = 1 + 0i$ is clearly one unit away, as is $i = 0 + 1i; -1=-1 +0i;$ and $-i=0 +(-1)i$. But what of $3 + 4i$? Well.... if we think of this geometrically (I'm kind of kicking myself and and debating whether I should bring this up) then $3+4i$ is a $3$ units in the positive real direction and $4$ units in the up-ward $i$ direction and so a sensible norm would be to use the pythogorean theorem. The square of the distance of the $3$ units plus the square of the distance of the $4$ units, is equal to the square of the hypotenuse distance from $3+4i$ to $0=0+0i$. So $|3+4i|^2 = 3^2 + 4^2$ and $|3+4i| =\sqrt{3^2 + 4^2}=\sqrt{25} =5$.

In general: $|z| = \sqrt{Re(z)^2 + Im(z)^2}$.

And what's nice about it is it works for reals as we already defined it. If $z \in \mathbb R$ then $Re(z) = z$ and $Im(z) = 0$ and $|z| =\sqrt{Re^2(z) + Im^2(z)} =\sqrt{z^2 + 0^2} = \sqrt{z^2}$. But only if $z$ is real.

.....

All this brings up the issue when you are expanding from one set of numbers to another, from whole numbers to include the negative integers; from the integers to include the rationals; from the rationals to include the irrationals; from the reals to include the imaginary and than to include all the complex-- we will always have to expand or definitions and concepts. The important thing is to recognize and keep and modify what is basic and necessary from that which is coincidental and incidental. That $\sqrt{x^2} = |x|$ is incidental as real numbers are always positive, negative, and $0$. But that $|x|$ is a normed non-negative real distance is fundamental. As we move from reals to complex where numbers are not always positive, negative, and $0$ having $\sqrt{z^2}=|z|$ will not required or possible but having $|z|$ be a normed non-negative real distance will.