Why Does the Modulus Only Take Positive Values?

[flyingpig]

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?

 

[summerwind]

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).

 

[SammyS]

...

$$\sqrt{25}$$ = 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.

 

[flyingpig]

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

Wow what?

 

[statdad]

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. $$-\sqrt{25} = -5$$.

 

[SammyS]

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: $$x=\pm5\,.$$​