Do modulus and absolute value operations use the same sign?

Question

Do modulus and absolute value operations use the same sign? If so, do we always assume that a modulus is intended when the number is complex? If an expression says $|a+bi|$, this means I should interpret it as $\sqrt{a^2+b^2}$ and not $\sqrt{a^2+2abi-b^2}$.

Answer 1

Do modulus and absolute value operations use the same sign?

Yes, they do.

And this is fine- when $x \in \mathbb{R}$, the modulus of $x$ is the same as the absolute value of $x$.

Why? I'll prove it.

Let $|x|$ be the absolute value of $x$ and let $\textrm{mod(x)}$ be the modulus of $x$.

Claim: $|x|=\rm{mod(x)}$ for $x \in \mathbb{R}$.

Proof:

$\textrm{mod(x)}=\rm{mod(x+0i)}:=\sqrt{x^2+0^2}=\sqrt{x^2}=: |x|.$

$\square$

If so, do we always assume that a modulus is intended when the number is complex?

The modulus function is a generalisation of the absolute function, so, if in doubt, use the modulus!

The absolute value makes sense only for real numbers, whereas, the modulus is defined for all (complex) numbers, and all real numbers are complex!

If an expression says $|a+bi|$, this means I should interpret it as $\sqrt{a^2+b^2}$ and not $\sqrt{a^2+2abi-b^2}$ [?]

"By definition" would be my cop-out answer, but, geometrically, we can think of the modulus or absolute value of a complex number as its distance from the origin.

For real numbers, the modulus is the same as the absolute value, as it's just the length of the number without taking into account its sign.

For a strictly-complex number (i.e. one of the form $x+yi$ where $x,y \in \mathbb{R} \setminus \{0\}$), the absolute value function has no meaning, as a strictly-complex number like $2+i$ is not negative or positive.

If we continue our idea of the absolute value being the distance from a number to the origin, we can extend the absolute value function to complex numbers, and now call this the modulus.

As an intuitive answer, more than anything, consider plotting a general point in the complex plane.

Now, you should see from the above picture that $|z|=\sqrt{a^2+b^2}$ follows immediately from Pythagoras' Theorem, since we have a right-angled triangle.

It wouldn't make sense to have $\sqrt{a^2+2abi-b^2}$, since distances are real and this is not real ( for all $a,b \in \mathbb{R}$).

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