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
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 $\displaystyle y= x^{1/2}$ and $\displaystyle y^2= x$ are not equivalent because $\displaystyle (-2)^2= 4$ is true but $\displaystyle -2= \sqrt{4}$ 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 $\displaystyle \sqrt{4}$ could be interpreted as "$\displaystyle \pm 2$" and we might interpret that as saying $\displaystyle y= x^{1/2}$ is "equivalent" to $\displaystyle y^2= x$- 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.
I would like to ask again why in the set of real number $\displaystyle \sqrt 4 =2$ and in the set of complex number $\displaystyle \sqrt4=\pm2$. Thank you. I would really appreciate an answer as complete as possible, involving how the function $\displaystyle \sqrt{z}$ has been defined. Thank you.
I think you're right. Sorry. Let me rephrase my question.
Typically in textbooks when they want to solve the equation
$\displaystyle z=x+iy=\sqrt(a+ib)$ (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 $\displaystyle z^2=a+ib$.
So why are they still keeping both solutions?
Keeping both solutions is like solving $\displaystyle x^2=4$ and saying that both $\displaystyle \pm2$ are solutions to $\displaystyle x=\sqrt{4}$, which obviously is not true.
I also would like to know how the function $\displaystyle \sqrt{z}$ has been defined in complex analysis.
Thank you.
Let $\displaystyle \sqrt{x+iy}=a+ib$.Let us find the square root of a complex number $\displaystyle z=x+iy$.
$\displaystyle \Rightarrow x+iy=a^2-b^2+2iab$..........[Square both the sides]
$\displaystyle x=a^2-b^2$..........[1]On equating real and imaginary parts:
$\displaystyle y=2ab$..........[2]
Now, $\displaystyle (a^2+b^2)^2=(a^2-b^2)^2+4a^2b^2$
$\displaystyle \Rightarrow (a^2+b^2)^2=x^2+y^2$
$\displaystyle \Rightarrow a^2+b^2=\sqrt{x^2+y^2}$..........[3]
$\displaystyle 2a^2=x+\sqrt{x^2+y^2}$Add equations [1] and [3]:
$\displaystyle \implies a=\pm \sqrt{\frac{1}{2}(x+\sqrt{x^2+y^2})}$
$\displaystyle b=\pm \sqrt{\frac{1}{2}(-x+\sqrt{x^2+y^2})}$Similarly solving for b:
Conclusion
If $\displaystyle z=x+iy$ then $\displaystyle \sqrt{z}=a+ib$ where :
$\displaystyle a=\pm \sqrt{\frac{1}{2}(x+\sqrt{x^2+y^2})}$
$\displaystyle b=\pm \sqrt{\frac{1}{2}(-x+\sqrt{x^2+y^2})}$
$\displaystyle z=x+iy$Let us check that in the set of complex numbers, do we have $\displaystyle w=z^{1/2}$ equivalent to $\displaystyle w^2=z$
$\displaystyle \implies \sqrt{z}=\pm \sqrt{\frac{1}{2}(x+\sqrt{x^2+y^2})}\pm i\sqrt{\frac{1}{2}(-x+\sqrt{x^2+y^2})}$
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$\displaystyle \sqrt{z}=\pm \sqrt{\frac{1}{2}(x+\sqrt{x^2+y^2})}\pm i\sqrt{\frac{1}{2}(-x+\sqrt{x^2+y^2})}$
Square both sides
$\displaystyle \implies z=\frac{x+\sqrt{x^2+y^2}}{2}-\frac{-x+\sqrt{x^2+y^2}}{2}+2i\sqrt{\frac{1}{4}(y^2)}$
$\displaystyle \implies z=\frac{x+\sqrt{x^2+y^2}-\sqrt{x^2+y^2}+x}{2}+2i\frac{y}{2}$
$\displaystyle \implies z=x+iy$
Conclusion
If w and z are complex numbers then:$\displaystyle w=\sqrt{z} \iff w^2=z$.
So where do I get it wrong?
($\displaystyle \sqrt{z}=-2$) $\displaystyle \Longrightarrow (z=4) $$\displaystyle \Longrightarrow (\sqrt{z}=\pm2) $
So it really looks as if $\displaystyle (\sqrt{z}=-2)$ IS NOT EQUIVALENT TO $\displaystyle \(z=4) $
the problem is that:
$\displaystyle \sqrt{z}, z \in \mathbb{C}$ and $\displaystyle \sqrt{x}, x \in \mathbb{R}$ 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" $\displaystyle \sqrt{-1}$, 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.
that's because there isn't "one" square root function. it helps to write z in polar form:
$\displaystyle z = re^{i\theta} = r(\cos\theta + i\sin\theta)$
then $\displaystyle \sqrt{z} = z^{1/2} = \sqrt{r}(\cos \frac{\theta}{2} + i\sin \frac{\theta}{2})$
why is this not a single-valued function? because
$\displaystyle z = re^{i(\theta + 2\pi)} = r(\cos(\theta + 2\pi) + i \sin(\theta + 2\pi))$,
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 = $\displaystyle \cos(\pi) + i \sin(\pi)$ is:
$\displaystyle \cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2}) = 0 + i = i$.
but note that -1 is also: $\displaystyle \cos(3\pi) + i\sin(3\pi)$, which gives the square root:
$\displaystyle \cos(\frac{3\pi}{2}) + i\sin(\frac{3\pi}{2}) = 0 + i(-1) = -i$.
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 $\displaystyle \theta = 0 = 2\pi$).
if we let $\displaystyle \psi = \frac{\theta}{2}$ and apply DeMoivre's theorem, we obtain:
$\displaystyle (\sqrt{r}(\cos\psi + i\sin\psi))^2 = r(\cos(2\psi) + i\sin(2\psi)) = r(\cos\theta + i\sin\theta) = z$
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 $\displaystyle \sqrt{z} = e^{log(z)/2}$ doesn't improve the situation any, as log is multi-valued on $\displaystyle \mathbb{C}$ as well.
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 $\displaystyle \sqrt{z}$, or, more generally, how $\displaystyle z^{1/q}$, q being an integer, is defined. I have looked at a few complex analysis textbooks and they "seem" to miss that part!