A basic question about ##i##

[mcastillo356]

TL;DR Summary

I don't understand a basic statement of maths: $|i|=1$

Hi, a question...Well, two (stupid, I guess):
1. Why $|i|=1$
2. The symbol $||$, what does it mean? Absolute value, modulus,...?
Greetings!

 

[Office_Shredder]

Both absolute value and modulus are used to describe ||.

What mathematical definition of || are you using? Or are asking for us to tell you what it is,?

 

[PeroK]

What else could it be?

 

[mcastillo356]

I must try to give a solution;
$|i|=\sqrt{i^2+i^2}=\sqrt{-2}$; so this is not; moreover, it's absurd: $i$ is a number, not a vector.
If $|i|$ refers to the measure in the complex axis, how can I deduce it's 1?.
The symbol $||$ is also the distance between two numbers, so $|i|=|i-0|$, but this leads me nowhere.

 

[PeroK]

I must try to give a solution;
$|i|=\sqrt{i^2+i^2}=\sqrt{-2}$; so this is not; moreover, it's absurd: $i$ is a number, not a vector.
If $|i|$ refers to the measure in the complex axis, how can I deduce it's 1?.
The symbol $||$ is also the distance between two numbers, so $|i|=|i-0|$, but this leads me nowhere.

Hmm!

If all we know about $i$ is that $i^2 = -1$, then $|i^2| = |i|^2 = |-1| = 1$.

And so: however we define the modulus of the complex numbers, we should expect $|i| = 1$.

 

[Office_Shredder]

From a geometric perspective, if you draw the complex plane, the point $i=0 + 1i$ corresponds to the point (0,1) on the plane. The length of that vector is $\sqrt{0^2+1^2} = 1$.

 

[haushofer]

Besides, the norm doesn't involve "i".

 

[mcastillo356]

Besides, the norm doesn't involve "i".

The norm is the modulus, isn't it?. What does $||$ apply for in this case?; the absolute value?; the distance?.
Excuse my poor english.

 

[PeroK]

What does $||$ apply for in this case?; the absolute value?; the distance?.

Both. It's the absolute value, which for a complex number $z$ is given by $|z| = \sqrt{zz^*}$.

And, if you view complex numbers as points in the complex plane it's the distance from the origin: $|z| = \sqrt{x^2 + y^2}$.

Where $z = x + iy$, and $z^* = x - iy$.

 

[mcastillo356]

Thank you. I've got it!

 

[mfb]

$|i|=\sqrt{i^2+i^2}$

I don't know where you get this from but it's not correct.

Using $|z| = \sqrt{zz^*}$ we get $|i|=\sqrt{i (-i)} = \sqrt 1 = 1$
Using the equivalent $|z|=\sqrt{Re(z)^2+Im(z)^2}$ we get $|i| = \sqrt{0^2+1^2} = \sqrt 1 = 1$

 

[mcastillo356]

I don't know where you get this from but it's not correct.

Using $|z| = \sqrt{zz^*}$ we get $|i|=\sqrt{i (-i)} = \sqrt 1 = 1$
Using the equivalent $|z|=\sqrt{Re(z)^2+Im(z)^2}$ we get $|i| = \sqrt{0^2+1^2} = \sqrt 1 = 1$

Thank you! I must read again the book where I started. The formula you refer to is "inspired" by the Theorem of Pytagoras, but insanely applied, in a desperate try to show my efforts to face the question. I say I must read from the start, just to check if the formulas you mention are mentioned.
Greetings!

 

[mcastillo356]

Yes, they are, mbf.
Here in Bilbao it's 1.00 AM. No responsibilities until 6.00 AM. I'm going to try to study just a little bit, and then sleep for a while.