roots of complex number

**Hushai1** · November 15th 2011, 05:05 PM

In the set of complex numbers, do we have

[w=z^1/2] equivalent to [w^2=z].

If so, why (as we do not have the equivalence in the set of real numbers)

I am looking for an answer as advanced or as complete as possible!

Thank you

**TKHunny** · November 15th 2011, 06:38 PM

Some of our friends in ancient history (and surprisingly not so ancient) were perfectly happy with equations they could not solve explicitly.

**HallsofIvy** · November 16th 2011, 05:56 AM

I'm not sure I understand your question (and I certainly don't understand TKHunny's response!). Are you saying that in the real numbers $y= x^{1/2}$ and $y^2= x$ are not equivalent because $(-2)^2= 4$ is true but $-2= \sqrt{4}$ is not? It is true that, because there are so very few "single valued" functions in the complex numbers, if we tried to keep "single valued" as a condition to be a function (as it is in the real numbers) we would have very few functions to work with! In that sense $\sqrt{4}$ could be interpreted as " $\pm 2$ " and we might interpret that as saying $y= x^{1/2}$ is "equivalent" to $y^2= x$ - the set of numbers for which one equation is true is the same as the set of numbers for which the other equation is true.

**Hushai1** · November 16th 2011, 10:03 PM

Originally Posted by HallsofIvy

I'm not sure I understand your question (and I certainly don't understand TKHunny's response!). Are you saying that in the real numbers $y= x^{1/2}$ and $y^2= x$ are not equivalent because $(-2)^2= 4$ is true but $-2= \sqrt{4}$ is not? It is true that, because there are so very few "single valued" functions in the complex numbers, if we tried to keep "single valued" as a condition to be a function (as it is in the real numbers) we would have very few functions to work with! In that sense $\sqrt{4}$ could be interpreted as " $\pm 2$ " and we might interpret that as saying $y= x^{1/2}$ is "equivalent" to $y^2= x$ - the set of numbers for which one equation is true is the same as the set of numbers for which the other equation is true.

Thank you for your answer. Yes I think you understood the question well. Another way to put it would be: How is the function $\sqrt{z}$ defined in the set of complex numbers?

**Hushai1** · November 17th 2011, 07:17 PM

I would like to ask again why in the set of real number $\sqrt 4 =2$ and in the set of complex number $\sqrt4=\pm2$ . Thank you. I would really appreciate an answer as complete as possible, involving how the function $\sqrt{z}$ has been defined. Thank you.

**TKHunny** · November 17th 2011, 08:26 PM

You have it wrong.

$\sqrt{4} = 2$ , no matter where you live.

Solutions to the equation $x^{2} = 4$ , are x = 2 and x = -2. That requires only Reals.

Please rethink your question.

**Hushai1** · November 17th 2011, 09:59 PM

Originally Posted by TKHunny

You have it wrong.

$\sqrt{4} = 2$ , no matter where you live.

Solutions to the equation $x^{2} = 4$ , are x = 2 and x = -2. That requires only Reals.

Please rethink your question.

I think you're right. Sorry. Let me rephrase my question.

Typically in textbooks when they want to solve the equation
$z=x+iy=\sqrt(a+ib)$ (equation 1),
they square both sides, find two solutions, z1 and z2, and keep the two solutions as solutions to equation 1.
However equation 1 is not equivalent to $z^2=a+ib$ .
So why are they still keeping both solutions?

Keeping both solutions is like solving $x^2=4$ and saying that both $\pm2$ are solutions to $x=\sqrt{4}$ , which obviously is not true.

I also would like to know how the function $\sqrt{z}$ has been defined in complex analysis.

Thank you.

**sbhatnagar** · November 18th 2011, 02:24 AM

Let us find the square root of a complex number $z=x+iy$ .

Let $\sqrt{x+iy}=a+ib$ .

$\Rightarrow x+iy=a^2-b^2+2iab$ ..........[Square both the sides]

On equating real and imaginary parts:

$x=a^2-b^2$ ..........[1]
$y=2ab$ ..........[2]

Now, $(a^2+b^2)^2=(a^2-b^2)^2+4a^2b^2$

$\Rightarrow (a^2+b^2)^2=x^2+y^2$

$\Rightarrow a^2+b^2=\sqrt{x^2+y^2}$ ..........[3]

Add equations [1] and [3]:

$2a^2=x+\sqrt{x^2+y^2}$

$\implies a=\pm \sqrt{\frac{1}{2}(x+\sqrt{x^2+y^2})}$

Similarly solving for b:

$b=\pm \sqrt{\frac{1}{2}(-x+\sqrt{x^2+y^2})}$

Conclusion

If $z=x+iy$ then $\sqrt{z}=a+ib$ where :

$a=\pm \sqrt{\frac{1}{2}(x+\sqrt{x^2+y^2})}$

$b=\pm \sqrt{\frac{1}{2}(-x+\sqrt{x^2+y^2})}$

Let us check that in the set of complex numbers, do we have $w=z^{1/2}$ equivalent to $w^2=z$

$z=x+iy$

$\implies \sqrt{z}=\pm \sqrt{\frac{1}{2}(x+\sqrt{x^2+y^2})}\pm i\sqrt{\frac{1}{2}(-x+\sqrt{x^2+y^2})}$
__________________________________________________ _

$\sqrt{z}=\pm \sqrt{\frac{1}{2}(x+\sqrt{x^2+y^2})}\pm i\sqrt{\frac{1}{2}(-x+\sqrt{x^2+y^2})}$

Square both sides

$\implies z=\frac{x+\sqrt{x^2+y^2}}{2}-\frac{-x+\sqrt{x^2+y^2}}{2}+2i\sqrt{\frac{1}{4}(y^2)}$

$\implies z=\frac{x+\sqrt{x^2+y^2}-\sqrt{x^2+y^2}+x}{2}+2i\frac{y}{2}$

$\implies z=x+iy$

Conclusion

If w and z are complex numbers then: $w=\sqrt{z} \iff w^2=z$ .

**Hushai1** · November 18th 2011, 04:13 AM

So where do I get it wrong?

( $\sqrt{z}=-2$ ) $\Longrightarrow (z=4)$ $\Longrightarrow (\sqrt{z}=\pm2)$

So it really looks as if $(\sqrt{z}=-2)$ IS NOT EQUIVALENT TO $\(z=4)$

**Deveno** · November 18th 2011, 05:52 AM

the problem is that:

$\sqrt{z}, z \in \mathbb{C}$ and $\sqrt{x}, x \in \mathbb{R}$ mean two different things, even though they use the same symbol.

with real numbers, we have a "preferred" square root, the positive (actually "non-negative") one.

in the complex numbers, there is no consistent way to define "positive", so we take both. from an algebraic standpoint,

there is no way to decide if i or -i is the "true" $\sqrt{-1}$ , so we choose one arbitrarily

(there's no real way to tell if we have the complex numbers "upside down", a+ib and a-ib behave the same way,

it's purely a matter of orientation, and orientations are arbitrary).

so it's more proper to say i is "a" square root of -1, rather than THE square root of -1. understand?

or to use your earlier example: the complex square roots of 4 are 2 and -2. the real square root of 4 is simply 2.

**Hushai1** · November 18th 2011, 05:58 PM

I would like to thank the people who provided answers. The answers were good, however, I still don't understand the exact definition of $\sqrt{z}$ on $\mathbb{C}$ .

**Deveno** · November 18th 2011, 06:22 PM

that's because there isn't "one" square root function. it helps to write z in polar form:

$z = re^{i\theta} = r(\cos\theta + i\sin\theta)$

then $\sqrt{z} = z^{1/2} = \sqrt{r}(\cos \frac{\theta}{2} + i\sin \frac{\theta}{2})$

why is this not a single-valued function? because

$z = re^{i(\theta + 2\pi)} = r(\cos(\theta + 2\pi) + i \sin(\theta + 2\pi))$ ,

as well, and applying the above formula to this form of z gives a 2nd square root of z.

for example, one square root of -1 = $\cos(\pi) + i \sin(\pi)$ is:

$\cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2}) = 0 + i = i$ .

but note that -1 is also: $\cos(3\pi) + i\sin(3\pi)$ , which gives the square root:

$\cos(\frac{3\pi}{2}) + i\sin(\frac{3\pi}{2}) = 0 + i(-1) = -i$ .

as you might have guessed, this difficulty is because sine and cosine are periodic, so to find all square roots on the unit circle, we have to "go around the circle twice" (because our half-angle may be almost 2pi, so our original angle might be almost 4pi:

the real case comes from where $\theta = 0 = 2\pi$ ).

**Hushai1** · November 18th 2011, 06:37 PM

Originally Posted by Deveno

that's because there isn't "one" square root function. it helps to write z in polar form:

$z = re^{i\theta} = r(\cos\theta + i\sin\theta)$

then $\sqrt{z} = z^{1/2} = \sqrt{r}(\cos \frac{\theta}{2} + i\sin \frac{\theta}{2})$

why is this not a single-valued function? because

$z = re^{i(\theta + 2\pi)} = r(\cos(\theta + 2\pi) + i \sin(\theta + 2\pi))$ ,

as well, and applying the above formula to this form of z gives a 2nd square root of z.

for example, one square root of -1 = $\cos(\pi) + i \sin(\pi)$ is:

$\cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2}) = 0 + i = i$ .

but note that -1 is also: $\cos(3\pi) + i\sin(3\pi)$ , which gives the square root:

$\cos(\frac{3\pi}{2}) + i\sin(\frac{3\pi}{2}) = 0 + i(-1) = -i$ .

as you might have guessed, this difficulty is because sine and cosine are periodic, so to find all square roots on the unit circle, we have to "go around the circle twice" (because our half-angle may be almost 2pi, so our original angle might be almost 4pi:

the real case comes from where $\theta = 0 = 2\pi$ ).

In line 3 of your post you use Demoivre theorem that applies only with interger powers. Shouldn't we use instead as definition of $\sqrt{z}=e^(1/2*log{z})$ ?

Thank you

**Deveno** · November 18th 2011, 08:02 PM

Originally Posted by Hushai1

In line 3 of your post you use Demoivre theorem that applies only with interger powers. Shouldn't we use instead as definition of $\sqrt{z}=e^(1/2*log{z})$ ?

Thank you

if we let $\psi = \frac{\theta}{2}$ and apply DeMoivre's theorem, we obtain:

$(\sqrt{r}(\cos\psi + i\sin\psi))^2 = r(\cos(2\psi) + i\sin(2\psi)) = r(\cos\theta + i\sin\theta) = z$

justifying our original formula. it is not true that DeMoivre's theorem only holds for integers, in fact it holds for any rational number.

using $\sqrt{z} = e^{log(z)/2}$ doesn't improve the situation any, as log is multi-valued on $\mathbb{C}$ as well.

**Hushai1** · November 18th 2011, 08:09 PM

Deveno, thank you very much for your answer. Actually I know that Demoivre's theorem holds for rational numbers and that log is a multi valued function on C. I was just looking for an "official" definition of $\sqrt{z}$ , or, more generally, how $z^{1/q}$ , q being an integer, is defined. I have looked at a few complex analysis textbooks and they "seem" to miss that part!

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