Why Does the Modulus Only Take Positive Values?

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SUMMARY

The modulus of a complex number z, defined as |z|, is calculated using the formula |z| = √(z z'), where z' is the conjugate of z. For the complex number z = 4 + 3i, the modulus is |z| = √(25) = 5, as only the positive square root is considered by definition. This definition aligns with the geometric interpretation of modulus as the length of the vector represented by the complex number in the Cartesian plane. The principal square root, denoted by the radical symbol, always yields a non-negative result, reinforcing that the modulus cannot take negative values.

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Homework Statement



Find the modulus of |z| for z = 4 + 3i






The Attempt at a Solution



z' = 4 - 3i

z'z = 25

\sqrt{25} = plus or minus 5

My book and Mathematic only take positive roots, why?
 
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The modulus of a complex number z is defined to be the positive square root of z z'. That's it - it's just a definition.

The motivation for this definition is that the modulus is supposed to be the length of z if you think of z as a vector in the x-y plane. In the case of z = 4 + 3i, the vector points from (0,0) to (4,3) and so its length is 5 (certainly not -5).
 
flyingpig said:
...

\sqrt{25} = plus or minus 5

My book and Mathematic only take positive roots, why?

\sqrt{25} = +5. That radical symbol denotes the principal square root, which is positive.
 
SammyS said:
\sqrt{25} = +5. That radical symbol denotes the principal square root, which is positive.

Wow what?
 
flyingpig said:
Wow what?

That simply means that when you write \sqrt{25}, or \sqrt{16}, or
\sqrt{2729275.5839}, by definition the positive square root is the one
that is intended. If you want the negative value you need to indicate it, viz. -\sqrt{25} = -5.
 
flyingpig said:
Wow what?
That wasn't very nice of me, was it?

Now, if you're solving x2=25, for example, then there are a couple of ways to show that the solution is: x = ±5. (I could have said: x = ±√(25) just as correctly.)

Method 1:
If x2 = 25, then x2 - 25 =0 . Factoring the LHS gives: (x+5)(x-5)=0

The zero product property of real numbers gives the solutions: x = 5 or x= -5.​

Method 2:
If x2 = 25, then taking the (principal) square root of both sides gives:

\sqrt{x^2}=\sqrt{25}

\sqrt{25}\text{ is }5\,, \text{ (That's positive 5 .) and }\sqrt{x^2}\text{ is }|x|\,.

So we have: \text{ }|x|=5 \ .

Therefore: x=\pm5\,.
 

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