High School Understanding Absolute Value and the Symbol ||: What Does it Mean?

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The discussion focuses on understanding the absolute value and modulus represented by the symbol ||, particularly in relation to the complex number i. It clarifies that |i| equals 1, derived from the definition of absolute value in the complex plane, where |z| is calculated as the distance from the origin using the formula |z| = √(Re(z)² + Im(z)²). Participants emphasize that the modulus can also be expressed as |z| = √(zz*), confirming that |i| = √(i(-i)) = √1 = 1. The conversation highlights the importance of proper definitions and understanding of complex numbers to avoid confusion. Overall, the thread provides insights into the mathematical principles behind absolute value and modulus in complex analysis.
mcastillo356
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TL;DR
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!
 
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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##.
 
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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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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##.
 
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Thank you. I've got it!
 
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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##
 
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  • #12
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!
 
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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.
 

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