In remedial algebra, we learn that the graph of $y=(\sqrt x)^2$ is only in the first quadrant. We know this is the correct graph for the equation. This is because we know $y=x$ and $x \ge 0$.

However, a student in a higher level algebra class got the full line where $y=x$, $x$ is all real numbers. Her logic was that if you take $x=-4$, you'd take the square root and get $2i$, and then you'd square that to get $-4$ because $i^2 = -1$ and $2^2 = 4$. This would give you a line that goes into the negatives, which is different from the correct answer we are told, which is a line only in the first quadrant.

This student did not thrive in class, however she made a very interesting argument. She did not see the point of what she came up with and just wanted to pass the class. How would I, as a teacher, try to celebrate this students solution? Although we don’t know if it’s really right or wrong, how could a teacher try and justify her answer and what could a teacher say to encourage her thinking and show a connection to how complex numbers were made?

  • $\begingroup$ Welcome to ME.SE. I added some Latex formatting, paragraph breaks and the complex numbers tag. Please check I did not change the intended meaning of the question. $\endgroup$
    – Tommi
    Commented Apr 17, 2019 at 7:42
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    $\begingroup$ The problem I see is having a clear understanding of what is intended when functions-defined-by-formulas are not provided with explicit domains. One of the things needed when defining a function is to define its domain. Since what you're talking about was not given an explicit domain, it comes down to a notation convention, which isn't really a true mathematical issue. For example, $(\sqrt x)^2 = \sqrt x \sqrt x$ --- what if we use $2i$ for the first $\sqrt x$ and $-2i$ for the second $\sqrt x,$ and if we're not allowed to do this, why? $\endgroup$ Commented Apr 17, 2019 at 8:12
  • $\begingroup$ Yeah, it turns out that you can use $\sqrt{x}$ as a symbol for both, as a certain (branch of a) function, but trying to do both at the same time yields big problems (encountered since at least the 16th century, implicitly). Still, a good question. $\endgroup$
    – kcrisman
    Commented Apr 17, 2019 at 13:03
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    $\begingroup$ @DaveLRenfro But $2i\times -2i$ wouldn't be the same as passing $2i$ to the squaring function, which is surely what one does in $(\sqrt{x})^2$. $\endgroup$
    – J.G.
    Commented Apr 19, 2019 at 19:40
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    $\begingroup$ Side point: I find it extremely common in these classes to encounter a student on the first day of class who seems to ask "interesting" questions, and then very much "not thrive in class" (i.e., not be able to focus or learn any new definitions or standard procedures). I have to practice tempering my expectations and not cycle through over-excitement/time-expense/disappointment in these cases. You might also find this to be fairly common in the future. $\endgroup$ Commented May 17, 2019 at 13:51

3 Answers 3


I don't see any good way to talk about this without explicitly addressing the issue of domains and the fact that the same square root symbol is used for two different functions depending on context.

In the first context, the square root symbol refers to the single-valued, nonnegative real square root. In the second context, the square root symbol refers to the multi-valued, complex square root. These are different functions.

So, I think the answer is to clearly distinguish between the two functions. You can do so by specifying which one you're using, and giving a precise definition of square root in both contexts:

  • Given a real number $x \geq 0$, the nonnegative real square root of $x$ is the unique real number $\sqrt{x} \geq 0$ such that $(\sqrt{x})^2 = x$.
  • Given a complex number $z$, the complex square root of $z$ is the set of complex numbers $\{\sqrt{z}, -\sqrt{z}\}$, where $\sqrt{z}$ is any complex number such that $(\sqrt{z})^2 = z$.

Once you've provided these definitions, you can give a simple answer to a student who gives an answer for the complex square root: That's the correct answer for the complex square root, but not for the nonnegative real square root (because the domain is different).


After squaring she just added some extra roots to the equation $y=-4$ : $\sqrt{y}=2i=2\textrm{e}^{in\pi/2}$

Squaring again we get $y=4\mathrm{e}^{in\pi}$ in which $-4$ is one of the solution.

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    $\begingroup$ Hi and welcome to the site. Be careful when you post an answer, that you are answering the actual question that was posted. The question asked here is "how could a teacher try and justify her answer and what could a teacher say to encourage her thinking?" You haven't answered that question. You can click "edit" below your answer to improve it. $\endgroup$ Commented May 11, 2019 at 14:19

Suppose $z \neq 0$. Define $z^{1/2} = \{ \sqrt{z}, - \sqrt{z}\}$ where $\sqrt{z}$ is defined by $\sqrt{z} = \sqrt{|z|}e^{i\text{Arg}(z)/2}$ and $\text{Arg(z)}$ is the standard angle of $z$ in $(-\pi, \pi]$. We can study $w \in (z^{1/2})^2$ in the sense that $w$ is in the set formed by squaring the the set $z^{1/2}$ element-wise. In particular, $$ (z^{1/2})^2 = \{ (\sqrt{z})^2, (-\sqrt{z})^2 \} = \{ (\sqrt{z})^2 \}$$ but, the square of $\sqrt{z}$ can be calculated directly: $$ (\sqrt{z})^2 = \left(\sqrt{|z|}e^{i\text{Arg}(z)/2} \right)^2 = (\sqrt{|z|})^2 \left(e^{i\text{Arg}(z)/2} \right)^2 = |z|e^{i\text{Arg}(z)} = z.$$ So, if $w \in (z^{1/2})^2$ then $w \in \{ z \}$ so we might as well say $w=z$. I'm sure someone else can find a better way to communicate this to the level of student you intend.


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