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I know that $\sqrt{1}=\sqrt{(-1)*(-1)}$ and we can't expand $\sqrt{(-1)*(-1)}$ as $\sqrt{-1}*\sqrt{-1}$ since the later part becomes $\iota$*$\iota$ which becomes -1 which is not equals the original number we are given-$1$.

But I want to know why we can't take square root of two imaginary numbers in product to be equals the product of square roots of two imaginary numbers by distributing the square root function over the two numbers under the square root function?

Gerry Myerson
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ankit
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    You just gave a counter-example though. What more are you looking for? – Hayden Sep 11 '16 at 00:12
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    In some sense we can, we just have to be very careful. Note that there are two complex numbers $z$ that fulfill $z^2=-1$, i.e. there are two values of $\sqrt{-1}$. The first $\sqrt{-1}$ in your expression becomes one of them and the last $\sqrt{-1}$ becomes the other. Then it all works out. (Don't do this at home, kids; it's not really correct.) – Arthur Sep 11 '16 at 00:13
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    These tags are strange. Why linear algebra and number theory? – Integral Sep 11 '16 at 00:15
  • @Hayden I want to know why complex numbers can't be expanded like that not through the counter-example but through the nature of imaginary numbers which makes them different from real numbers and to understand the behavior of root function. – ankit Sep 11 '16 at 00:18
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    @ankit See for example Square Root - Notes: "*Because of the discontinuous nature of the square root function in the complex plane, the following laws are not true in general...*". – dxiv Sep 11 '16 at 00:23
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    What is your definition of $\sqrt{-1}$ anyways? There are two different complex numbers whose square is $-1$... – Eric Wofsey Sep 11 '16 at 00:27
  • @dxiv and what do you mean by the discontinuous nature of the square root function in the complex plane? – ankit Sep 11 '16 at 00:31
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    If you're talking about the principal square root function, it's not continuous at any point on the negative real axis. The value of the square root "jumps" as one traverses across it. – anon Sep 11 '16 at 00:33
  • @Eric $\sqrt{-1}$ is just $\iota$ that's it we can't define it any further just as we can't define $1$. BTW what is the other complex number whose square is $-1$? – ankit Sep 11 '16 at 00:34
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    The square roots of $-1$ are $i$ and $-i$. Every nonzero complex number has two distinct square roots which are opposites of each other. Anyway, are you familiar with polar form? – anon Sep 11 '16 at 00:34
  • @arctic what is the principle square root function? I can't understand any of your sentence. what do you mean by "jumping"? Can you kindly elaborate? – ankit Sep 11 '16 at 00:36
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    Here is an example of a jump discontinuity (for a real variable function). For instance, $\sqrt{-1}$ would be $i$, but if $z$ is extremely close to $-1$ underneath the real axis then $\sqrt{z}$ is extremely close to $\color{Red}{-}i$, not $i$. If you're not familiar with principal square roots, then what do you think $\sqrt{z}$ even means when $z\in\Bbb C$ is an arbitrary complex number? What is your definition of $\sqrt{z}$? – anon Sep 11 '16 at 00:38
  • @arctic Yes, arctic I'm familiar with polar form. – ankit Sep 11 '16 at 00:41
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    @ankit The principal square root is defined on the same page that I linked before, which has further links to Branch cuts and Principal Value. Any complete answer to your question would involve those concepts. If you are not ready to read and understand them now, then until the time you do you'll have to accept the shorthand version that "it just doesn't work for arbitrary complex numbers, since it would lead to contradictions" like the one you quoted. – dxiv Sep 11 '16 at 00:44
  • @arctic to me $\sqrt{z}$ means that we are referring to a number $z_1$ which have square root times the modulus of $z$ as it's modulus but possessing half the value of it's argument. – ankit Sep 11 '16 at 00:48
  • You're wrong about the modulus being the same. And the value of its argument, are you assuming that's a value in $(-\pi,\pi]$, or $[0,2\pi)$, are any real number in $\Bbb R$? – anon Sep 11 '16 at 00:50
  • @arctic oh yes! I've corrected the value of modulus change in $\sqrt{z}$ in my preceding comment. The value of argument can be any real number, which unlike amplitude is not restricted to any domain. And yes argument is a real number since it measures the rotation. – ankit Sep 11 '16 at 00:54
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    Okay. The number $1$ can be written as either $e^{0i}$ or $e^{2\pi i}$. But $e^{0i/2}=1$ and $e^{2\pi i/2}=-1$, so in your definition does $\sqrt{1}$ refer to $1$ or $-1$? – anon Sep 11 '16 at 00:56
  • @arctic I'm confused! – ankit Sep 11 '16 at 00:59
  • @Arthur and @ arctic since $\sqrt{-1}$ has two values doesn't it means that square root is no longer a function in complex number world? – ankit Sep 11 '16 at 01:06
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    Square root is not a function in the real number world either, except that by convention we tend to take only the nonnegative square root of nonnegative reals. – Eric Wofsey Sep 11 '16 at 01:08
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    Well, every positive number has two square roots as well, but there is still a principal square root function defined on the nonnegative reals. This can be extended to a function on the whole complex plane as long as one is willing to sacrifice continuity across a branch cut. (Seriously, read the Wikipedia links.) – anon Sep 11 '16 at 01:08
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    [Naturally, "arctic tern" is familiar with "polar form"] – Gerry Myerson Sep 11 '16 at 01:14

2 Answers2

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It is a convention that when $x\geq 0,$ the notation $\sqrt x$ denotes the non-negative $y$ satisfying $y^2=x.$ The symbol $\sqrt z$ should be avoided with complex numbers as there is no convention, and so $\sqrt z$ is ambiguous. The $\sqrt x$ symbol distributes over multiplication within the non-negative reals. The reason that it doesn't do that in $\mathbb C$ is simply that it doesn't always do so, as your example shows.

DanielWainfleet
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    To be noted however that $\sqrt{z}$ is the commonly used notation for the principal value of the square root of $z \in \mathbb{C}$ (wikipedia, mathworld, Peano). – dxiv Sep 11 '16 at 01:40
  • @user254665 "The symbol $\sqrt{z}$ should be avoided with complex numbers as there is no convention, and so .....ambiguous. That is the best answer of this whole tread and I could not have put it any better! +1 – imranfat Sep 11 '16 at 03:05
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    @imranfat Along the same line, $w^z$ should not be used with complex numbers, either, though it's a common notation for $e^{z ;\text{Log}; w}$. IMHO it's all relative to the context and established conventions. – dxiv Sep 11 '16 at 03:18
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Any nonzero number $z$ has exactly two square roots, which differ only in sign. For $z$ real, we denote the positive one by $\sqrt{z}$ and the negative one by $-\sqrt{z}$. For $z$ not real, there's no canonical choice, and so the notation $z^{1/2}$ is effectively meaningless. (There are ways of resolving this issue--- branch cuts, for example, but that's not relevant here.) What we can say is that if $u^2 = a$ and $v^2 = b$, then $(uv)^2 = u^2 v^2 = ab$. There's no guarantee, though, that $uv$ will be the particular choice of the two square roots you arbitrarily have in mind for $\sqrt{ab}$.

anomaly
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