Doubt about modulus in complex numbers

[dunkelheit]

Lets suppose we need to find the roots of [MATH]z^2+2z+2[/MATH], ignoring the fact that we can use the formula for quadratic equations we get that

[MATH]z^2+2z+2=0\Rightarrow (z+1)^2+1=0 \Rightarrow (z+1)^2=-1 \Rightarrow |z+1|=i \Rightarrow z+1=\pm i \Rightarrow z=-1 \pm i[/MATH]​

My doubt is the following: why we take the modulus of [MATH]z+1[/MATH] like we were in [MATH]\mathbb{R}[/MATH]? It seems strange to me because in [MATH]\mathbb{R}[/MATH] we use the modulus because of the sign of the expression in the root, but we know that [MATH]\mathbb{C}[/MATH] is not an ordered field and so it is impossible to know if [MATH]z+1[/MATH] is positive or negative; so it seems like the modulus in [MATH]\mathbb{C}[/MATH] is a different thing respect to the modulus in [MATH]\mathbb{R}[/MATH], but we use it the same if we have to extract a root with an even index. Can someone explain me this thing? Thanks.

 

[pka]

Lets suppose we need to find the roots of [MATH]z^2+2z+2[/MATH], ignoring the fact that we can use the formula for quadratic equations we get that
[MATH]z^2+2z+2=0\Rightarrow (z+1)^2+1=0 \Rightarrow (z+1)^2=-1 \Rightarrow |z+1|=i \Rightarrow z+1=\pm i \Rightarrow z=-1 \pm i[/MATH]My doubt is the following: why we take the modulus of [MATH]z+1[/MATH] like we were in [MATH]\mathbb{R}[/MATH]? It seems strange to me because in [MATH]\mathbb{R}[/MATH] we use the modulus because of the sign of the expression in the root, but we know that [MATH]\mathbb{C}[/MATH] is not an ordered field and so it is impossible to know if [MATH]z+1[/MATH] is positive or negative; so it seems like the modulus in [MATH]\mathbb{C}[/MATH] is a different thing respect to the modulus in [MATH]\mathbb{R}[/MATH], but we use it the same if we have to extract a root with an even index.

Why not just use the basic quadratic equation? Look at this.
There is no mystery there. Is there?

 

[dunkelheit]

Yes, it is clear by the quadratic equations since it has [MATH]\pm[/MATH] sign that leads to the two solutions, but I would actually know why it works the same by considering the modulus with the reasoning and the doubt I've said in my first post. Thanks!

 

[HallsofIvy]

Lets suppose we need to find the roots of [MATH]z^2+2z+2[/MATH], ignoring the fact that we can use the formula for quadratic equations we get that

[MATH]z^2+2z+2=0\Rightarrow (z+1)^2+1=0 \Rightarrow (z+1)^2=-1 \Rightarrow |z+1|=i \Rightarrow z+1=\pm i \Rightarrow z=-1 \pm i[/MATH]​

My doubt is the following: why we take the modulus of [MATH]z+1[/MATH] like we were in [MATH]\mathbb{R}[/MATH]? It seems strange to me because in [MATH]\mathbb{R}[/MATH] we use the modulus because of the sign of the expression in the root, but we know that [MATH]\mathbb{C}[/MATH] is not an ordered field and so it is impossible to know if [MATH]z+1[/MATH] is positive or negative; so it seems like the modulus in [MATH]\mathbb{C}[/MATH] is a different thing respect to the modulus in [MATH]\mathbb{R}[/MATH], but we use it the same if we have to extract a root with an even index. Can someone explain me this thing? Thanks.

You should doubt! Going from [math](z+ 1)^2= -1[/math] to [math]|z+ 1|= i[/math] is wrong! The absolute value, of any complex number is a non-negative real number. What you should have is [math]z+1= \pm i[/math]. That is NOT the same as saying |z+1|= I. For a real number, x, |x|= a means either x= a or x= -a. For a complex number, x |x|= a (and a must be a non-negative real number) means that x could be any point on the circle in the complex plane with center at (0, 0) and radius a.

If z= x+ iy then |z| is defined as \(\displaystyle \sqrt{zz*}= \sqrt{x^2+ y^2}\) which is, again, a non-negative real number. Geometrically |z| can be defined as the distance of z from the origin in the complex plane. And "distance" is always a non-negative real number.

 

[dunkelheit]

@Hallsoflvy Thanks for the clarification! I Now another question is raising inside my head: so if it is different from real numbers, why then from [MATH](z+1)^2=-1[/MATH] we deduce that [MATH]z+1= \pm i[/MATH]? In real numbers it follows, as you've said, from the modulus, but in complex numbers from what follows? It is clear to me that we square both sides and we get [MATH]i[/MATH] at the right hand side, but the rest of the process from which we get [MATH]z+1[/MATH] from [MATH]\sqrt{(z+1)^2}[/MATH] without caring, differently from the real case, of the fact that we are extracting an even index root of an even power and the [MATH]\pm[/MATH] is not that clear.

 

[pka]

@Hallsoflvy Thanks for the clarification! I Now another question is raising inside my head: so if it is different from real numbers, why then from [MATH](z+1)^2=-1[/MATH] we deduce that [MATH]z+1= \pm i[/MATH]? In real numbers it follows, as you've said, from the modulus, but in complex numbers from what follows? It is clear to me that we square both sides and we get [MATH]i[/MATH] at the right hand side, but the rest of the process from which we get [MATH]z+1[/MATH] from [MATH]\sqrt{(z+1)^2}[/MATH] without caring, differently from the real case, of the fact that we are extracting an even index root of an even power and the [MATH]\pm[/MATH] is not that clear.

Lets go back to the ninth/tenth grade. Could you find the two roots of \(z^2+2z+2=0\)? If yes then what is your problem?

 

[dunkelheit]

Yes I can, but to prove that formula (if I don't recall wrong) we use the absolute value when we extract the square root of the term [MATH]\left(x+\frac{b}{2a}\right)[/MATH]; here it isn't like this, according to what Hallsoflvy (if I've not interpreted wrong what he said).
I think it's nice to understand better things by trying other ways I don't know perfectly (as it is perfectly clear that if I have a doubt I would use the way I already know it is correct, but I'm here for learning now!).

 

[Dr.Peterson]

I Now another question is raising inside my head: so if it is different from real numbers, why then from [MATH](z+1)^2=-1[/MATH] we deduce that [MATH]z+1= \pm i[/MATH]? In real numbers it follows, as you've said, from the modulus, but in complex numbers from what follows? It is clear to me that we square both sides and we get [MATH]i[/MATH] at the right hand side, but the rest of the process from which we get [MATH]z+1[/MATH] from [MATH]\sqrt{(z+1)^2}[/MATH] without caring, differently from the real case, of the fact that we are extracting an even index root of an even power and the [MATH]\pm[/MATH] is not that clear.

There is no modulus/absolute value involved there, and I don't think HallsofIvy said there is.

Yes, you can express it that way (that is, [MATH]\sqrt{a^2} = |a|[/MATH]) if all the numbers are real, but you are really saying simply that if [MATH]w^2 = a[/MATH], then there are two possible values of [MATH]w[/MATH], which are "opposites", called [MATH]\pm\sqrt{a}[/MATH], because [MATH](\pm\sqrt{a})^2 = (\pm1\sqrt{a})^2 = (\pm1)^2(\sqrt{a})^2 = 1a = a[/MATH].

In your example, we know that [MATH]i^2 = -1[/MATH], and therefore [MATH](\pm i)^2 = -1[/MATH]; and there is no other number whose square is -1. So if [MATH](z+1)^2 = -1[/MATH], we know that [MATH]z+1 = \pm i[/MATH].