Principal square root of a complex number, why is it unique?

[mcastillo356]

TL;DR Summary

If it comes to calculate the principal square root of a complex number, it yields a unique number. Is it because it calculates two squares at the same time, that is, because includes that of a +bi and -a-bi?

This is a quote from "Calculus", by Robert A. Adams. It's a translation from spanish:
"Roots of square numbers
If $a$ is a positive real number, there exist two different real numbers whose square is $a$. They are

$\sqrt{a}\;$ (the positive square root of $a$)
$-\sqrt{a}\;$ (the negative square root of $a$)​

Every complex number $z=x+yi$ different from zero (this is, $x^2+y^2>0$) has two square roots; if $w_1=z$ is a complex number so that $w_1^2=z$ then $w_2=-w_1$ also suits $w_2^2=z$. Is interesting to point out one of this roots to call it $\sqrt{z}$.
Be $r=|z|$, so $r>0$. Be $\theta=\mbox{Arg}(z)$. Hence, $-\pi<\theta\leq{\pi}$. As

$z=r(\cos{\theta}+i\sin{\theta})$​

the complex number

$w=\left(\cos{\dfrac{\theta}{2}}+i\sin{\dfrac{\theta}{2}}\right)$​

fills clearly $w^2=z$. We will call $w$ the principal square root of $z$, and we will write it as $\sqrt{z}$. The two solutions of the equation $w^2=z$ are, thus, $w=\sqrt{z}$ and $-\sqrt{z}$.
It's remarkable that the real part of $\sqrt{z}$ is always non-negative because $\cos{(\theta/2)}\geq{0}$ in $-\pi/2<\theta\leq{\pi/2}$. At this interval $\sin{(\pi/2)}=0$ only if $\theta=0$, in which case $\sqrt{z}$ is real and positive."
The questions are: $\sqrt{z}$ is at the same time square root of $w_1$ and $w_2=-w_1$?; and why do we choose only the positive square $\sqrt{z}$?
Thanks! I'm not native, so forgive the language mistakes. Of course, if it doesn't make sense anything, ignore me. I've posted maybe too quickly.

 

[jedishrfu]

The principal square root came from the notion of taking a square root of a positive number should yield a positive result that's what would be used in practical problems where square roots are needed.

However, mathematicians realized that there exists another square root which mathematically is equally valid but in practical problems is not usually. Hence they defined the positive root of a positive number, the principal square root and then simply called it the square root with the understanding that it refers to the positive root.

The square root of complex numbers doesn't have that convention.

https://en.wikipedia.org/wiki/Square_root

 

[PeroK]

Summary:: If it comes to calculate the principal square root of a complex number, it yields a unique number. Is it because it calculates two squares at the same time, that is, because includes that of a +bi and -a-bi?

This is a quote from "Calculus", by Robert A. Adams. It's a translation from spanish:
"Roots of square numbers
If $a$ is a positive real number, there exist two different real numbers whose square is $a$. They are

$\sqrt{a}\;$ (the positive square root of $a$)
$-\sqrt{a}\;$ (the negative square root of $a$)​

Every complex number $z=x+yi$ different from zero (this is, $x^2+y^2>0$) has two square roots; if $w_1=z$ is a complex number so that $w_1^2=z$ then $w_2=-w_1$ also suits $w_2^2=z$. Is interesting to point out one of this roots to call it $\sqrt{z}$.
Be $r=|z|$, so $r>0$. Be $\theta=\mbox{Arg}(z)$. Hence, $-\pi<\theta\leq{\pi}$. As

$z=r(\cos{\theta}+i\sin{\theta})$​

the complex number

$w=\left(\cos{\dfrac{\theta}{2}}+i\sin{\dfrac{\theta}{2}}\right)$​

fills clearly $w^2=z$. We will call $w$ the principal square root of $z$, and we will write it as $\sqrt{z}$. The two solutions of the equation $w^2=z$ are, thus, $w=\sqrt{z}$ and $-\sqrt{z}$.
It's remarkable that the real part of $\sqrt{z}$ is always non-negative because $\cos{(\theta/2)}\geq{0}$ in $-\pi/2<\theta\leq{\pi/2}$. At this interval $\sin{(\pi/2)}=0$ only if $\theta=0$, in which case $\sqrt{z}$ is real and positive."
The questions are: $\sqrt{z}$ is at the same time square root of $w_1$ and $w_2=-w_1$?; and why do we choose only the positive square $\sqrt{z}$?
Thanks! I'm not native, so forgive the language mistakes. Of course, if it doesn't make sense anything, ignore me. I've posted maybe too quickly.

If you didn't have a convention for the principal square root, then the symbol $\sqrt z$ would not be a number. It would be a two-point set: $$\sqrt z = \{w, -w \}$$ which wouldn't be very useful.

 

[mcastillo356]

Ok. Now I must think. It's 20.08 PM round here. I must walk the dog...We will meet tomorrow. What about the translation? Comprehensive? Please tell me the mistakes. Answer as soon as I can

 

[Stephen Tashi]

What about the translation?

"Roots of square numbers

Square roots of numbers

If $a$ is a positive real number, there exist two different real numbers whose square is $a$. They are

$\sqrt{a}\;$ (the positive square root of $a$)
$-\sqrt{a}\;$ (the negative square root of $a$)​

Every complex number $z=x+yi$ different from zero (this is, $x^2+y^2>0$) has two square roots;

(The author is using the term "square root"in the sense of "n-th root" in complex analysis, so he is permitted to say that a number may have two square roots. This is confusing to students who have only studied algebra on the real numbers and have been drilled in the fact that "A number has only one square root". )

if $w_1=z$ is a complex number so that $w_1^2=z$ then $w_2=-w_1$ also suits $w_2^2=z$.

also satisifies $w_2^2 = z$.

Is interesting to point out one of this roots to call it $\sqrt{z}$.

It is useful to select one of these roots and denote it $\sqrt{z}$

Be $r=|z|$, so $r>0$. Be $\theta=\mbox{Arg}(z)$.

Let $r = |z|$, so $r > 0$. Let $\theta = \mbox{Arg}(z)$.

Hence, $-\pi<\theta\leq{\pi}$. As

$z=r(\cos{\theta}+i\sin{\theta})$​

the complex number

$w=\left(\cos{\dfrac{\theta}{2}}+i\sin{\dfrac{\theta}{2}}\right)$​

fills clearly $w^2=z$.

clearly satisifies $w^2 = z$.

We will call $w$ the principal square root of $z$, and we will write it as $\sqrt{z}$. The two solutions of the equation $w^2=z$ are, thus, $w=\sqrt{z}$ and $-\sqrt{z}$.

Instead of the syntax "are, thus", it would be more usual to say:
"Thus the two solutions of the equation $w^2 = z$ are $w = \sqrt{z}$ and $w = -\sqrt{z}$."

( It would be clearer to emphasize that "the two solutions" is a set and write "The solution set of the equation $w^2= z$ is $\{\sqrt{z}, -\sqrt{z}\}$.")

It's remarkable that the real part of $\sqrt{z}$ is always non-negative because $\cos{(\theta/2)}\geq{0}$ in $-\pi/2<\theta\leq{\pi/2}$.

? "It is notable that " or "It is clear that" or "It is obvious that".
To say "It is remarkable that" suggests "It is surprising that".

At this interval $\sin{(\pi/2)}=0$ only if $\theta=0$, in which case $\sqrt{z}$ is real and positive.

In this interval $\sin{(\pi/2)}= 0$ ...

 

[Stephen Tashi]

We should keep in mind that the definition of $\sqrt{z}$ as a unique number does not guarantee that familiar properties of $\sqrt{}$ for non-negative real numbers will apply.

For example: $\sqrt{ z w} = \sqrt{z} \sqrt{w}$?

$1 = \sqrt{1} = \sqrt{ (-1)(-1))}\ne \sqrt{-1} \sqrt{-1} = (i)(i) = i^2 = -1$

( Of course we must verify that the definition of "$\sqrt{}$" makes $\sqrt{-1} = i$ instead of $\sqrt{-1} = -i$. )

 

[DaveE]

Please tell me the mistakes.

Well... There is a typo: $w=\sqrt{r}\left(\cos{\dfrac{\theta}{2}}+i\sin{\dfrac{\theta}{2}}\right)$

This all seems kind of trivial to me. If you want to define a principle value for $w=\sqrt{z}$ that is single valued, then you just have to pick one.

 

[jedishrfu]

Well... There is a typo: $w=\sqrt{r}\left(\cos{\dfrac{\theta}{2}}+i\sin{\dfrac{\theta}{2}}\right)$

This all seems kind of trivial to me. If you want to define a principle value for $w=\sqrt{z}$ that is single valued, then you just have to pick one.

but the question then becomes: which one?

 

[DaveE]

but the question then becomes: which one?

I think the real issue is understanding that there are two roots (on the defined branch) and communicating clearly which you are talking about.

But, for $z=re^\left(i\theta\right)$, then it seems like $\sqrt{z}=\sqrt{r}e^\left(i\frac{\theta}{2}\right)$ is a good choice.

I seems simpler (for this branch cut) than
$\sqrt{z}= \left\{ \begin{array}{cc} \sqrt{r}e^\left(i\frac{\theta}{2}-\pi\right), &\mbox{ if } \theta>0 \\ \sqrt{r}e^\left(i\frac{\theta}{2}+\pi\right), &\mbox{ if } \theta\leq0 \end{array} \right.$

Of course, anyone in their right mind would just say $\sqrt{z}=-\sqrt{r}e^\left(i\frac{\theta}{2}\right)$ instead, and let someone else figure out the branch issues.

BTW, this definition also works well for the [0, 2π) branch.

 

[mcastillo356]

Hi, PF

If you didn't have a convention for the principal square root, then the symbol $\sqrt z$ would not be a number. It would be a two-point set: $$\sqrt z = \{w, -w \}$$ which wouldn't be very useful.

The principal square root came from the notion of taking a square root of a positive number should yield a positive result that's what would be used in practical problems where square roots are needed.

However, mathematicians realized that there exists another square root which mathematically is equally valid but in practical problems is not usually. Hence they defined the positive root of a positive number, the principal square root and then simply called it the square root with the understanding that it refers to the positive root.

The square root of complex numbers doesn't have that convention.

https://en.wikipedia.org/wiki/Square_root

I'm a little bit confused. I've been working hard, but can't get my head around it: are you talking about the same word when you talk about convention, Jedishrfu, PeroK?

I think the real issue is understanding that there are two roots (on the defined branch) and communicating clearly which you are talking about.

But, for $z=re^\left(i\theta\right)$, then it seems like $\sqrt{z}=\sqrt{r}e^\left(i\frac{\theta}{2}\right)$ is a good choice.

I seems simpler (for this branch cut) than
$\sqrt{z}= \left\{ \begin{array}{cc} \sqrt{r}e^\left(i\frac{\theta}{2}-\pi\right), &\mbox{ if } \theta>0 \\ \sqrt{r}e^\left(i\frac{\theta}{2}+\pi\right), &\mbox{ if } \theta\leq0 \end{array} \right.$

Of course, anyone in their right mind would just say $\sqrt{z}=-\sqrt{r}e^\left(i\frac{\theta}{2}\right)$ instead, and let someone else figure out the branch issues.

BTW, this definition also works well for the [0, 2π) branch.

For $z=re^\left(i\theta\right)$, then it seems like $\sqrt{z}=\sqrt{r}e^\left(i\frac{\theta}{2}\right)$ is a good choice. I've been working at it, but the expression "branch cut" is extremely hard at Wikipedia. Is there any way to understand it easily? do you mean just interval, or is it as difficult as I suspect?
Greetings. If I'm being boring, I beg you a pardon.

 

[DaveE]

I've been working at it, but the expression "branch cut" is extremely hard at Wikipedia. Is there any way to understand it easily? do you mean just interval, or is it as difficult as I suspect?

Yes, it can get difficult, and yes, I should have said domain (or interval) really, I was being sloppy.

The domain and branches are similar because you can have different polar representations for the same complex number (like $e^\left(-i\frac{\pi}{2}\right)$ or $e^\left(i\frac{3\pi}{2}\right)$) so you need to define a domain if you want a single representation of each number. Branches are a similar concept used for complex operations that may return multiple values; if you want to treat those as functions, you must define a way to make them single valued. I think this video is a good explanation:

edit: In it's simplest form it's the same problem as the inverse trig functions that could return multiple values. How do we deal with the fact that sin(π/2) = sin(π/2 + 2π)? We define a principal branch that excludes all of the values of sin^-1(1) except π/2.

 

[mcastillo356]

I'm now studying a quick tutorial of this issue. Your post has raised the flag ¡Lots of things to learn! But I will be able to understand all the concepts included in this thread, included Euler's identity (Hope so), and Euler's formula. I'll be back this weekend.

 

[mcastillo356]

Hi, PeroK, DaveE, jedishrfu
Well, here is my personal opinion: $$\sqrt[n]{z}=\sqrt[n]{|z|}\;e^{i\left(\dfrac{\theta}{n}+\dfrac{k2\pi}{n}\right)}, k=0,1,2...(n-1)$$
If the book says $w=\sqrt{r}\left(\cos{\dfrac{\theta}{2}}+i\sin{\dfrac{\theta}{2}}\right)$ is the principal square root, it's ok, but trivial, having little importance. It is not a convention, it is just a choice. They just pick one, I don't know why.

 

[DaveE]

Hi, PeroK, DaveE, jedishrfu
Well, here is my personal opinion: $$\sqrt[n]{z}=\sqrt[n]{|z|}\;e^{i\left(\dfrac{\theta}{n}+\dfrac{k2\pi}{n}\right)}, k=0,1,2...(n-1)$$
If the book says $w=\sqrt{r}\left(\cos{\dfrac{\theta}{2}}+i\sin{\dfrac{\theta}{2}}\right)$ is the principal square root, it's ok, but trivial, having little importance. It is not a convention, it is just a choice. They just pick one, I don't know why.

It's the same as what you said when n=2, k=0. They just pick one because that's the point of having a principal root otherwise you would have roots. Sometimes you want to define a single valued function, other times you just want to have jargon to aid communication.