We know that in real numbers $\sqrt{x^2}=|x|$. But in complex numbers my query is we can have two square roots. For example in my book the question is to find the value of

$$z=\sqrt{i}-\sqrt{-i}$$ I did squaring on both sides and got

$$z^2=(\sqrt{i})^2+(\sqrt{-i})^2-2\sqrt{i \times -i}=-2\sqrt{1}=-2$$

since $z^2=-2$ we have $z=\pm i\sqrt{2}$

But my book answer is only $z=i\sqrt{2}$.

I am confused why the other value is not taken

  • 3
    $\begingroup$ does substituing the other value into the equation work? $\endgroup$ – Arjang Apr 5 '16 at 13:35
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    $\begingroup$ Your book is quite strange. Firstly it says that square root is not well defined, but then it gives a special value to something undefined. If I read this book, I would be confused like you. $\endgroup$ – Crostul Apr 5 '16 at 13:39
  • $\begingroup$ math.stackexchange.com/search?q=[complex-numbers]+square+root $\endgroup$ – Hans Lundmark Apr 5 '16 at 13:58

It is true that the equation $z^2 = -2$ has both $z = i\sqrt2$ and $z = -i\sqrt2$ as solutions. But the original problem was to find the value of (i.e., simplify) $z = \sqrt i - \sqrt{-i}$. You did this by squaring both sides. This is fine as long as you remember that squaring both sides of an equation could potentially introduce extraneous solutions.

You may recall from precalculus that this can happen with equations where we only care about real solutions. For example, $\sqrt{x+2} = x$ has only $x = 2$ as its solution, but if we solve by squaring both sides first then we get $x+2 = x^2$, which has both $x=-1$ and $x=2$ as solutions. $x=-1$ is extraneous because it does not satisfy the original equation $\sqrt{x+2} = x$, because $\sqrt{-1+2} = 1 \ne -1$. This is the same thing that happened in your case, just with imaginary numbers instead of real numbers.

I don't know if you were allowed to use the exponential form of complex numbers but that's how I would've done it:

\begin{align*} \sqrt i - \sqrt{-i} &= \sqrt{e^{i\pi/2}} - \sqrt{e^{-i\pi/2}}\\[0.3cm] &= e^{i\pi/4} - e^{-i\pi/4}\\[0.3cm] &= \left(\cos\frac\pi4 + i\sin\frac\pi4\right) - \left[\cos\left(-\frac\pi4\right) + i\sin\left(-\frac\pi4\right)\right]\\[0.3cm] &= \cos\frac\pi4 + i\sin\frac\pi4 - \underbrace{\left(\cos\frac\pi4 - i\sin\frac\pi4\right)}_{\text{$\cos$ is even, $\sin$ is odd}}\\[0.3cm] &= 2i\sin\frac\pi4\\[0.3cm] &= i\sqrt2 \end{align*}

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    $\begingroup$ But I don't see how this answers the OP's concern that $\sqrt{i}$ (for instance) has two values: $\text{cis } \pi/4$ and $\text{cis } 5\pi/4$ (where $\text{cis } \theta = \cos \theta + i \sin \theta$). A priori there's no reason to prefer one over the other, and yet one branch can yield one value, and the other branch can yield another value. $\endgroup$ – Brian Tung Sep 9 '16 at 16:29
  • $\begingroup$ @BrianTung, perhaps we disagree on what OP's concern was. OP explicitly said, "I am confused why the other value is not taken." I interpreted this as OP wondering why $z = -i\sqrt{2}$ is not also a valid solution to the problem, and I believe I sufficiently explained why. Also I disagree that $\sqrt{i}$ has two values, much like how, e.g., $\sqrt 4$ has only one value. If that convention doesn't apply to imaginary numbers then it's news to me. $\endgroup$ – tilper Sep 9 '16 at 16:59
  • $\begingroup$ The difference is not one of what the OP's concern is; I think we agree on that. I would adhere to the convention described in user 254665's answer, that the square root has one well-defined value (the non-negative number whose square is the argument) when the argument is a non-negative real. Otherwise, it may take on multiple values. I understand that you may disagree with it, but apparently the OP has been taught that it can take on multiple values, and yet the book gives an answer that can only be justified if it has just one. OP already understands where that one comes from. $\endgroup$ – Brian Tung Sep 9 '16 at 17:14
  • $\begingroup$ @BrianTung, I'm not convinced OP has been taught it can take multiple values. I interpreted "my query is we can have two square roots" as "my query is can we have two square roots?" because otherwise the word "query" doesn't make sense there. I see how that can be taken in the other direction, i.e., "query" is out of place and the OP did learn that imaginary numbers can have two square roots. But that interpretation already had an answer. $\endgroup$ – tilper Sep 9 '16 at 17:38
  • $\begingroup$ I think "query" indicates OP might not be a native English speaker; I don't think it quite makes sense either way. I'm not faulting any of your mathematics, mind you. Now that I know what question you're answering, it makes perfect sense. $\endgroup$ – Brian Tung Sep 9 '16 at 17:50

Here's another way of looking at the problem. What do we mean by $\sqrt{i}$? Well, there are two numbers $z$ for which $z^2 = i$; it's not too hard to show that they are $z = \pm (1 + i)/\sqrt{2}.$ Similarly, there are two numbers for which $z^2 = -i$; they are $z = \pm (1 - i)/\sqrt{2}$. Thus, we have $$ \sqrt{i} - \sqrt{ -i} = \pm \frac{1 + i}{\sqrt{2}} - \left( \pm \frac{1 - i}{\sqrt{2}} \right) $$

The book's answer of $i \sqrt{2}$ corresponds to picking the positive sign in both cases. Your other answer of $-i \sqrt{2}$ corresponds to picking the minus sign in both cases. Both are valid answers, depending on what you mean by "the" square root of $i$. You could define "the" square root to be the one with the positive real part, for example; this would be most closely analogous to what we do on the real number line. But this doesn't make the other answer wrong, per se; it just means that we've made a choice to focus on one particular choice of the square root.


The arithmetic convention is that $\sqrt A$ is the non-negative solution $x$ of $x^2=A$ only when $A$ is a non-negative real number, and it is also a convention that $\sqrt B$ is meaningless when $B$ is not a non-negative real number.

This is because there is no linear ordering on the complex numbers that has the relationships to complex arithmetic that the order on the reals has (That is, $a<b\to a+c<b+c$ and $[a<b\land 0<c]\to a c<b c.$) So there is no reason to choose one of the solutions of $x^2=B$ to the other, when $\neg (B\geq 0).$

We speak of a square root of $i$, not the square root of $i$. Of course there are $2$ solutions to $x^2=i$, which are $\pm(1+ i)/\sqrt 2.$

I would say that a book written after about 1750 or so that uses $\sqrt i$ needs a re-write.

  • $\begingroup$ Nitpick: the solutions to $x^2 = i$ are $\pm(1 + i)/\sqrt{2}$, not $(1 \pm i)/\sqrt{2}$. $\endgroup$ – Michael Seifert Sep 9 '16 at 14:17
  • $\begingroup$ @MichaelSeifert. Typo. Thank you. $\endgroup$ – DanielWainfleet Sep 9 '16 at 16:26

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