Many students question the importance of complex numbers in real life. We can find many important applications of imaginary numbers in Engineering field and physics. This question is not related to that kind of general issue. Here what I want to know is how we can emphasize the importance of imaginary numbers when we are dealing with real numbers by using examples where we need knowledge of complex numbers to work with real numbers.
Here you have one example, there is a formula derived in finding roots of depressed cubic equation $x^3 - 3px - 2q = 0$ and it can be stated as ,
$x = \root3\of{q+ \sqrt{q^2 - p^3}}+ \root3\of{q-\sqrt{q^2 - p^3}}$
Now if you use the formula to obtain roots of $x^3 - 15x - 4 = 0$ you have to substitute $p = 5$ and $q = 2$ , which gives roots in complex form such as ,
$x = \root3\of{2+ 11\sqrt{-1}}+ \root3\of{2-11\sqrt{-1}}$ which can be simplified using the knowledge of complex numbers to obtain the real root 4 of the above mentioned cubic equation. Here we need to know $(2+i)^3 = 2 + 11i$ and $(2-i)^3=2-11i$
Actually this real root 4 can be directly found by applying factor theorem and without having idea of complex numbers we cannot verify that formula. I think this kind of awareness may be helpful for students to understand the value of studying complex numbers. What would you suggest in parallel to this?

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    $\begingroup$ In your example, stronger students may be interested to know that this more of a rule than an exception: whenever $x^3-3px-2q$ has three distinct real roots $x_1,x_2,x_3$, then by Viete's formulas the discriminant $((x_1-x_2)(x_2-x_3)(x_3-x_1))^2$ (which is positive) equals $108 \cdot (p^3-q^2)$. This means that our formula involves imaginary numbers. $\endgroup$ Commented Nov 8, 2023 at 9:13
  • $\begingroup$ @MichałMiśkiewicz thanks for the important information, when equation has imaginary roots, it's something usual to have knowledge of complex numbers to simplify but as you see in this suggested example you need knowledge of complex numbers to obtain a real root. Sometimes this can be unusual happening for many who question the importance of knowing how to work with imaginary numbers . $\endgroup$ Commented Nov 8, 2023 at 10:35
  • $\begingroup$ Concerning the comment by @Michał Miśkiewicz, there are real number solutions to integer-coefficient cubic equations (i.e. algebraic numbers with algebraic degree (over the rationals) of degree $3)$ that cannot be expressed in finite form, beginning with the integers (or equivalently, beginning with the rationals) using the 4 basic arithmetic operations and/or positive integer root extractions (cube roots, 4th roots, 5th roots, etc.). So in the sense described, these are real numbers that CANNOT be expressed without the use of non-real complex numbers. (continued) $\endgroup$ Commented Nov 8, 2023 at 20:31
  • $\begingroup$ See Algebraic numbers expressible in terms of real-valued radicals and this 26 September 2005 sci.math post. Note that the constructible real numbers are a proper subset of the real radical numbers (e.g. $\sqrt [3] 2$ is real radical but not constructible), (continued) $\endgroup$ Commented Nov 8, 2023 at 20:31
  • $\begingroup$ and the real radical numbers are a proper subset of the real numbers that can be expressed in terms of radicals (e.g. each of the $3$ real roots of $2x^3 - 9x^2 - 6x + 3 = 0$ is an example), and the explicit algebraic real numbers are a proper subset of the algebraic real numbers (e.g. the real root of $x^5 - x - 1 = 0$ is an example). $\endgroup$ Commented Nov 8, 2023 at 20:31

12 Answers 12


Another topic that uses complex numbers in service of real numbers, and is immediately accessible to a high schooler who has learned or is learning calculus, is integration using Euler's formula.

Here's a decent example from that article. (There are also some simpler examples and some more advanced examples in that article.)

integration using Euler's formula

Some extra references on this topic, courtesy of Dave L Renfro:

Paul Glaister, Complex integrals, International Journal of Mathematical Education in Science and Technology 22 #3 (1991), pp. 476−477

Charles Otto Gunther (1879−1958), Integration by Trigonometric and Imaginary Substitution, D. Van Nostrand Company, 1907 (2nd edition -- only a corrected reprint -- is 1920), vi + 79 pages (internet archive copy and HathiTrust copy and Google-Books copy).

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    $\begingroup$ Thanks for sharing your knowledge, unfortunately I can't explain your example to my students because they don't have polar form of the complex number in their syllabus. But we had it before 2000 . $\endgroup$ Commented Nov 8, 2023 at 12:39
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    $\begingroup$ Thanks for the extra references @DaveLRenfro; I've added them to the bottom of the answer. $\endgroup$ Commented Nov 9, 2023 at 16:04

we need knowledge of complex numbers to work with real numbers

We need to be honest that at the level of your students, there are no examples of this: in all sufficiently elementary uses of complex numbers, one can get by without them.

Consider your example: if $p,q$ are rational and $q^2<p^3$, then either your polynomial has a rational root (in which case it can be found directly using Eisenstein rule, for example), or evaluating the third root of a complex number will proceed via calculating its argument, and dividing by 3. Hence, your final answer will involve $\arccos$ or some other similar function. It is a theorem that it does not simplify further, e.g., it's cannot be expressed in real radicals. But, you can also reduce your equation to the form $$ 4\cos^3t-3\cos t =q/p^\frac{3}{2}, $$ by substituting $x=2 p^\frac12 \cos t$. Since the left-hand side is now $\cos 3t$ and the right-hand side is $\leq 1$ in absolute value, we find three distinct solutions, corresponding to $$t_1=\frac13 \arccos(q/p^\frac{3}{2}), \quad t_{2,3}=\frac13 \arccos(q/p^\frac{3}{2})\pm\frac{2\pi}{3}.$$ This is of course the same solution in disguise, and the same $\arccos$ that you would have calculated in the approach with the complex numbers. So, in this case, not only aren't complex numbers needed, they actually only complicate matters.

That said, there are many elementary examples where complex numbers are sufficiently useful as a notational device to motivate their introduction:

  • Euler's formulae for $\cos t$ and $\sin t$ in terms of $e^{it}$ reduce trigonometric identities to algebraic ones. One example is given in Justin's answer; another one: calculate $$ \sin \frac{\pi}{100}+\sin\frac{2\pi}{100}+\dots+\sin\frac{50\pi}{100}. $$ Of course, you can do it by any number of tricks, but if you use Euler's formula, no tricks are needed: this is just geometric series. The entire scope of "application" of complex numbers in electrical engineering, signal processing, etc., boils down to this. Another example: find $\int\cos x e^{x}$. You can do it by integrating by parts two times, but Euler's formula immediately gives $\Re(\frac{1}{1+i}e^{(1+i)x}).$
  • Polynomial factorization, in particular, e.g., as applied to partial fraction decomposition. Suppose you want to integrate $$\int \frac{2x^2+x+1}{x^3+x}.\,dx$$ The usual way to do it is to decompose this into simple fractions. Over the reals, the decomposition reads $\frac{1}{x}+\frac{x}{x^2+1}+\frac{1}{x^2+1},$ but if you don't know that, you will need to solve a system of equiations. Then, you will have to remember how to integrate each of the terms. Over the complex numbers, on the other hand, we know a priori that the decomposition will be $\frac{a}{x}+\frac{b}{x-i}+\frac{c}{x+i}$, with $a$ real and $b,c$ complex conjugate. You can now observe that, e.g., $b$ is the limit of the above expression, multiplied by $(x-i)$, as $x\to i$. So, it can be found simply by removing $(x-i)$ from the denominator of the original fraction and plugging in $x=i$: $$b=\frac{3i^2+i+1}{i(i+1)}.$$ No solving linear system is needed. To calculate integrals, you only need to remember that $\int \frac{dx}{x-a}=\log (x-a)+C$ for any complex $a$; of course if you want an answer not involving complex logarithms, you will need to work a bit in the end.

If series and radius of convergence are within the reach of your syllabus, then a standard example of "needing to go through the complex numbers" is the radius of convergence of the series expansion of $$ f(x)=\frac{1}{x^2+1} $$ around 0. Said radius is 1, even though $x^2+1$ has no (real!) roots.

  • $\begingroup$ A better example would be $\frac{1}{e^x+1}$ or $\frac{x}{e^x-1}$, so that you actually do need (or at least, benefit from) complex numbers. $\endgroup$
    – Kostya_I
    Commented Nov 8, 2023 at 21:11
  • $\begingroup$ I'm missing the point here - if we're interested in finding the radius of convergence, it's just $1$, no complex numbers involved. (I know that for complex power series there's always something fishy happening on the boundary of the ball of convergence, but that's another story) @Kostya_I, I understand you have a similar issue? What are your examples showing exactly? $\endgroup$ Commented Nov 9, 2023 at 10:09
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    $\begingroup$ @MichałMiśkiewicz, If you happen to be interested in the radius of convergence of the Taylor series of $\frac{1}{e^x+1}$ at the origin, then complex analysis tells you it is $\pi$, because of singularities at $x=\pm \pi i$. I don't immediately see how to prove it by purely real means. $\endgroup$
    – Kostya_I
    Commented Nov 9, 2023 at 11:21
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    $\begingroup$ @MichałMiśkiewicz Perhaps complex numbers are not necessary to find and/or prove RoC. But without them RoC on reals only seems arbitrary. With complex numbers you can (begin to) understand why. See link above, also answer by KCd, and proof (sketch) of analyticity of holomorphic functions. $\endgroup$
    – Pablo H
    Commented Nov 9, 2023 at 17:14

Consider the real initial value problem

$$ \begin{cases} y'' + 2y' + 10y = 0\\ y(0) = 0\\ y'(0) = 1 \end{cases} $$

One nice way to do it is to first finding generic solutions with real inputs but complex outputs! Guessing that we may have some solutions of the form $y = e^{\alpha t}$ for $\alpha$ complex, we obtain

$$ \begin{align*} &y'' + 2y' + 10y = 0\\ &(\alpha^2 + 2\alpha + 10) e^{\alpha t} = 0\\ &\alpha^2 + 2\alpha + 10 = 0\\ &\alpha^2 + 2\alpha + 1 + 9 = 0\\ &(\alpha + 1)^2 = -9\\ &\alpha + 1 = \pm -3i\\ &\alpha = -1 \pm -3i \end{align*} $$

So we have solutions

$$e^{(-1 - 3i)t} = e^{-t}(\cos(3t) - i\sin(3t))$$


$$e^{(-1 + 3i)t} = e^{-t}(\cos(3t) + i\sin(3t))$$

These are complex valued solutions, but we can see that adding/subtracting and rescaling them leads to two real valued solutions:

$$ \begin{align*} &e^{-t}\cos(3t)\\ &e^{-t}\sin(3t) \end{align*} $$

So our general solution is

$$ ae^{-t}\cos(3t) + be^{-t}\sin(3t) $$

Using our initial values we find that the solution to our IVT is

$$ y = \frac{1}{3}e^{-t}\sin(3t) $$

This is an important example of the solution to a problem with real numbers being more easily solved when we expand our notion of solution to permit complex values.

  • $\begingroup$ Thanks for sharing another issue relevant to the post , doesn't y = {e^α}^t give 1 when t = 0 ? , it's doubtful for me. $\endgroup$ Commented Nov 10, 2023 at 14:16
  • $\begingroup$ Have I made a mistake there when typing y = e to the power αt , is it y = e^{αt} ? $\endgroup$ Commented Nov 10, 2023 at 14:35
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    $\begingroup$ @JanakaRodrigo I am finding a basis for the space of solutions of the problem $y'' +2y' +10y = 0$ without imposing initial value constraints initially. Then the hope is to find a solution to the IVT which is a linear combination of those particular solutions. $\endgroup$ Commented Nov 10, 2023 at 14:58
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    $\begingroup$ This is a standard method for solving constant coefficient linear ODEs that one finds in most every textbook. A similar application that is not so standard is the following from pp. 50-51 of Mechanics by Keith R. Symon (1971 3rd edition; also with slightly different font on p. 51 of the 1960 2nd edition) when solving $m\frac{d^2x}{dt^2}+b\frac{dx}{dt}+kx=F_0\cos(\omega t + \theta_0)$: (continued) $\endgroup$ Commented Nov 11, 2023 at 15:27
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    $\begingroup$ "The algebra is simpler, however, if we write the force as the real part of a complex function: $F(t)=\text{Re}(\mathbb F_0e^{i\omega t}),$ [where] $\mathbb F_0=F_0e^{i\theta_0}.$ Thus if we find a solution $[\ldots]$ then, by splitting the equation into real and imaginary parts $[\ldots]$" (Here I've used $\mathbb F$ for the bold-face font F in the text.) Note that the non-physical notion of a complex-valued force is introduced as a mathematical artifice to solve this forced harmonic oscillator problem. $\endgroup$ Commented Nov 11, 2023 at 15:27

In the setting of single-variable calculus, here is one of the most striking uses of complex numbers: the radius of convergence of the power series at $x = 0$ of a rational function $p(x)/q(x)$, written in reduced form, is the distance from $0$ to the nearest root of $q(x)$ in $\mathbf C$.

Example. Consider $1/(x^2+x+2)$. The denominator has no real roots, but it has two complex roots, both with absolute value $\sqrt{2}$, so the power series at $x = 0$ of $1/(x^2+x+2)$ has radius of convergence is $\sqrt{2}$. Good luck trying to explain that in a simple way using only real numbers.

Complex numbers also reveal the tight connection between trigonometric functions and exponential functions. Through the power series of $e^x$, $\cos x$, and $\sin x$, we are led to define $e^z$, $\cos z$, and $\sin z$ at complex $z$ and then are led to Euler's remarkable formula $e^{i\theta} = \cos\theta + i\sin\theta$, which reveals the imaginary periodicity of the exponential function. This is totally opaque when you only work with real numbers.


AThe book 'Visual Complex Analysis' by Tristan Needham highlights the significance of complex numbers, emphasizing their historical importance in mathematics. It details how the real solutions of cubic equations were originally discovered using complex numbers."

book excerpt: enter image description here

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    $\begingroup$ Thanks for joining with the post to share much knowledge relevant to the issue. $\endgroup$ Commented Nov 9, 2023 at 1:38

The complex numbers are really just the scale-rotations in 2D geometry. So you can get lots of examples from geometry.

We start with two particular scale-rotations that form a basis for building up the rest. These are:

$1=\pmatrix{1 & 0 \\ 0 & 1}$, $i=\pmatrix{0 & 1 \\ -1 & 0}$

Then a general scale-rotation can be written:

$a+bi=\pmatrix{a & b \\ -b & a}$

Scale-rotations obey all the usual rules of algebra. We can add them. We can multiply them. Multiplication distributes over addition. There are additive inverses for all of them, and multiplicative inverses for all but zero. And they're better than matrices, because multiplication is commutative!

Rotation matrices are obviously of this form.

$\cos \theta + i \sin \theta =\pmatrix{\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta}$

Take a simple problem from geometry, add up the three angles shown here:

Add angles of lines from origin to three points (1,1),(2,1),(3,1)

We can transform the unit line segment shown in purple to any of the three other lines with a scale-rotation, as shown. (We can thus label every point of the plane with the particular scale-rotation that maps a selected unit line segment to it.) Each scale-rotation rotates its input through one of the three angles illustrated. So if we do the three scale-rotations in succession, the result will be a scale-rotation that rotates through the sum of the three angles.

$\pmatrix{1 & 1 \\ -1 & 1} \pmatrix{2 & 1 \\ -1 & 2} \pmatrix{3 & 1 \\ -1 & 3}=\pmatrix{0 & 10 \\ -10 & 0}$

We can write this more briefly and clearly as:

$(1+i)(2+i)(3+i)=10 i$

This implies that the sum of the three angles is $90^\circ$.

We have turned geometry into algebra! We can use all the normal algebraic rules we just spent years learning, and adapt them to simplifying geometry problems. Geometry only involves real numbers - real number lengths, real number angles, real number areas, and so on. Scale-rotations are easy enough to grasp geometrically, but they have this unfamiliar property that $i^2=-1$, which we have previously always said is impossible. We now discover that it is impossible for the scaling transformations, but not impossible for the scale-rotations. You can build a $180^\circ$ rotation from two identical $90^\circ$ rotations, but you can't build it from two identical scalings (i.e. by squaring real numbers).

Historically, the mathematicians found the algebra first, and only spotted the geometrical connection to scale-rotations much later. This misled everyone into thinking the rotations part of the algebra were made up of nonsensical, or 'imaginary' numbers. They worked, but people didn't know why.

Mathematics has lots of situations like this: where two completely separate areas of study turn out to be about exactly the same thing, in disguise. Then all the tools we developed for solving problems in one subject can suddenly be used to solve problems in the other subject, and vice versa. This often gives us new avenues for solving problems that have previously resisted all efforts.


This might be the oldest question in mathematics education! But you are probably not going to like Euclid’s answer.

A youth who had begun to read geometry with Euclid, when he had learnt the first proposition, inquired, "What do I get by learning these things?" So Euclid said "Give him threepence, since he must make a gain out of what he learns.

When I encounter the "why do I need to learn this" sentiment of this question in the classroom, I turn the focus to a discussion of growth mindset and delayed gratification.

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    $\begingroup$ My opinion is by providing few hints about relationships between the issues students can be encouraged to find connections between the issues and it would create interest in the subject. When you come to contextual teaching and learning process making sense of what they are doing is very important. $\endgroup$ Commented Nov 8, 2023 at 16:33

Students who are interested in geometry might find this proof of Ptolemy's theorem to be a compelling application of complex numbers.

$$ AC\cdot BD = AB\cdot CD+BC\cdot AD $$

  • $\begingroup$ Yes you have lot more such as proving Napoleon theorem, Van aubel's theorem. Those proofs may be much simpler if you use complex numbers to represent vertices. $\endgroup$ Commented Nov 10, 2023 at 14:02

Complex numbers don't have an ordering like real numbers, i.e. $a + i b\ngtr c + id$. This is both subtle and profound. To explain, assign north as imaginary and east as real and then talk about locations in a classroom as complex numbers. Perform distance calculations between those locations using complex numbers to get real-valued distances. If you're there, figure out the angle between two lines from their point of intersection. Will the answers to either of these questions change if we move or rotate the origin? Does the origin matter for complex numbers?


My favorite goes approximately like this:

The geometric series $\sum\limits_{i=0}^{\infty}\frac{1}{x^n}$ converges to a function $f =\frac{1}{1-x}$ only on $-1\lt x \lt 1$. On the other hand, $f$ is well defined way beyond the interval of convergence. Is it a coincidence, a freak of nature, or what?

It is more or less easy to demonstrate how it can be continued towards $-\infty$. However, there is no way - as long as we stay in a domain of reals - to continue it into $x \gt 1$. We cannot jump over the singularity.

In the complex domain we don't need to jump over the singularity. We can walk around it, and lo and behold, we have a continuation.


Indeed, a general cubic equations $ax^3+bx^2+cx+d=0$ can be transformed to a depressed cubic equation $x^3 - 3px - 2q = 0$ and it can be stated as $x = \root3\of{q+ \sqrt{q^2 - p^3}}+ \root3\of{q-\sqrt{q^2 - p^3}}$, however, how was such formula be derived? or is it only a shot in the dark?

Of course, there are many ways to derive cubic equation roots, below is just one.

  1. first, we start with the depressed cubic equation $x^3 - 3px - 2q = 0$ (1)
  2. (1) can be transformed to equation $y^3 - 3y + s = 0$ (2)
  3. then, use substitution $y:=z+\frac{1}{z}$, we can transform it to $z^3+\frac{1}{z^3}+s=0$ (3)
  4. further, with substitution $w:=z^3$ we get $w+\frac{1}{w}+s=0$, or, $w^2 +sw +1 =0$ (4), which is a quadratic equation.
  5. solve the equation (4) with the quadratic equation formula, we got $w$, notice we always have $w_1 w_2=1$
  6. moving back, we got $z=\sqrt[3]{w}$, notice we always have $z_1 z_2=1$
  7. continue, we got $y=z+\frac{1}{z}$, notice we always have $y_1 = y_2$
  8. fold up, we got one root of $x$
  9. by factoring, we can get the other two roots of $x$


  1. first, we start with the depressed cubic equation $x^3-6x-9=0$ (1)
  2. (1) can be transformed to equation $y^3 - 3y -\frac{9}{2\sqrt{2}} = 0$ (2)
  3. then, use substitution $y:=z+\frac{1}{z}$, we can transform it to $z^3+\frac{1}{z^3}-\frac{9}{2\sqrt{2}}=0$ (3)
  4. further, with substitution $w:=z^3$ we get $w+\frac{1}{w}-\frac{9}{2\sqrt{2}}=0$, or, $w^2 -\frac{9}{2\sqrt{2}}w +1 =0$ (4), which is a quadratic equation
  5. solve the equation (4) with the quadratic equation formula, we got $w_1=2\sqrt{2}$, $w_2=\frac{1}{2\sqrt{2}}$, notice we indeed have $w_1 w_2=1$
  6. moving back, we got $z_1=\sqrt[3]{w_1}=\sqrt{2}$ and $z_2=\frac{1}{\sqrt{2}}$
  7. fold up, we got $y_1=w_1+\frac{1}{w_1}=\frac{1}{w_2}+w_2=y_2$
  8. ...

However, at step 6, actually for either $w_1$ or $w_2$, in complex number we can find 3 roots. With such result we will end up with 3 roots of $x$ at step 8, need not to use step 9.

Note: The above example has 1 real number root and 2 complex number roots. You might want to change accordingly to $x^3-15x-4=0$.


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