If I'm asked whether a polynomial has any imaginary root, what does it mean?

Does it mean whether the polynomial has any root which is purely imaginary?


Does it mean whether the polynomial has any complex root, where real part is not zero?

  • $\begingroup$ It means a purely imaginary root (i.e. a number $x = i \alpha$ such that $\alpha \in \mathbb{R}$) $\endgroup$
    – Thomas
    Dec 30, 2017 at 16:28
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    $\begingroup$ I would take it to mean the first, but you really need to ask whoever is asking you the question what they mean. $\endgroup$ Dec 30, 2017 at 16:28
  • $\begingroup$ It depends on the question. For $x^2 = -1$ the roots are purely imaginary. For $x^2 + x + 1 = 0$ the roots are complex. $\endgroup$ Dec 30, 2017 at 16:32
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    $\begingroup$ By imaginary most people mean complex, because if they said complex then that would also include real and that would still be confusing. $\endgroup$
    – user541686
    Dec 31, 2017 at 7:47

2 Answers 2


Both meanings are used (for instance, both are mentioned in the introduction of the Wikipedia page https://en.wikipedia.org/wiki/Imaginary_number). So lacking any additional context, there is no way to know (though I would consider your first interpretation more likely). You will have to figure out what makes sense from context or else ask whoever posed the question.

  • $\begingroup$ +1. The other answer stating that imaginary only means purely imaginary does not accord with any of my experience, which is that imaginary can also be synonymous to complex unless expressly stated otherwise. $\endgroup$
    – Nij
    Dec 30, 2017 at 22:49
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    $\begingroup$ +1. It is most unfortunate, but even mathematical giants occasionally use "imaginary" when they really mean complex. $\endgroup$ Dec 31, 2017 at 0:16
  • $\begingroup$ @PaulSinclair Correct terminology strictly though is what should be discussed. One may say "potato" and mean a "tomato" but what's the correlation ? The set of imaginary numbers $\mathbb I$ is called set of imaginary numbers for a reason. It also means the numbers it consists are termed as imaginary which have the known property. What is used to be said with what should be said or meant is not the same thing. $\endgroup$
    – Rebellos
    Dec 31, 2017 at 1:19
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    $\begingroup$ @Rebellos: It's not a matter of imprecision; it's a matter of differing conventions. Some people define "imaginary number" to mean "non-real complex number" and use it that way consistently, not just as a sloppy abuse of terminology. Personally, I simply don't use the term "imaginary number" at all (I would only ever say "purely imaginary number" for real multiples of $i$) $\endgroup$ Dec 31, 2017 at 4:57
  • $\begingroup$ @Rebellos - the OP is asking about the meaning of a phrase asked of him by someone else. "Correct terminology" is NOT what is important in understanding that phrase. What IS important is what the person saying the phrase means. The OP needs to know that they cannot simply assume that "imaginary" means multiples of $i$. Unlike Eric Wolsey apparently, I have never seen anyone officially define "imaginary" to include all non-real complex numbers, but I have heard several respected mathematicians use it in that way. $\endgroup$ Dec 31, 2017 at 22:54

As complex number, we consider the number $z$, such that :

$$z= x + yi$$

where $x,y \in \mathbb R$.

An imaginary number, is a number of the form above, but with $x=0$. More specifically, the number $z$ is considered imaginary, when :


where $y\in \mathbb R$. You may also come across that called as a purely imaginary number.

The imaginary numbers lie on the vertical coordinate axis on the complex plane and thus the definition of a complex number stated above is derived, as real numbers lie on the horizontal axis.

The set of imaginary numbers is obviously a subset of the complex numbers : $\mathbb I \subset \mathbb C.$ The set of the real numbers is a subset of the complex numbers : $\mathbb R \subset \mathbb C$ since all the real numbers are derived if $y=0$.

Thus, if you're asked to determine whether a polynomial (or generally an equation) has imaginary roots, you should consider it as checking if it has purely imaginary roots, else you would be asked about complex roots.

There is an important differentiation between purely imaginary and complex on many fields of mathematics and one example is the type of a stationary point while discussing dynamical systems : If the eigenvalues of the matrix of the system/linearised-system are complex then the stationary point is a focus (with some properties regarding the complex number...) but when the eigenvalues are purely imaginary (or imaginary simply as one may write) then there's a different case, as the stationary point is then considered a center.

Exercise examples :

The roots of the equation $x^2 = -4$ are the purely imaginary numbers $x = \pm 2i$.

The roots of the equation $x^2+x+1=0$ are the complex numbers $x= -\frac{1}{2} \pm \frac{\sqrt{3}}{2}i$.

  • $\begingroup$ The set of imaginary numbers is commonly denoted $\mathrm i\mathbb R$. $\endgroup$ Dec 31, 2017 at 8:01
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    $\begingroup$ Sometimes I've heard "imaginary" used to refer to purely imaginary numbers, as you interpret it in your answer, but other times I've heard "imaginary" used to mean simply "non-real", i.e. $x+iy$ with $y\ne0$. I don't think it's clear from the question which interpretation is intended. $\endgroup$
    – Carmeister
    Dec 31, 2017 at 10:49
  • $\begingroup$ In my opinion language in mathematics has strict definitions. Where I study at, these two terminologies are differentiated from "common" terms. $\endgroup$
    – Rebellos
    Dec 31, 2017 at 10:52
  • $\begingroup$ In any given work, in well-written mathematics, the language has strict definitions. But the language used in one context is not always the same in one work as in another. Your answer is no doubt correct where you study, but might be incorrect at some other institution. $\endgroup$
    – David K
    May 7, 2018 at 13:36

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