# A basic question about ##i##

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• mcastillo356
In summary, the conversation discusses the meaning and application of the symbol || in mathematics, particularly in relation to complex numbers. It is used to indicate absolute value and modulus, which can be calculated using various formulas such as ||z|| = √(zz*) or ||z|| = √(Re(z)^2 + Im(z)^2). The conversation also touches on the geometric interpretation of || as the distance from the origin on a complex plane. The conversation ends with one participant expressing their intent to review the relevant material and study further.

#### mcastillo356

Gold Member
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!

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,?

What else could it be?

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.

mcastillo356 said:
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##.

mcastillo356
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##.

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mcastillo356
Besides, the norm doesn't involve "i".

haushofer said:
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.

mcastillo356 said:
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!

mcastillo356 said:
##|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
mfb said:
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!

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.