Can the product of three complex numbers ever be real?

Question

Say I have three numbers, $a,b,c\in\mathbb C$. I know that if $a$ were complex, for $abc$ to be real, $bc=\overline a$. Is it possible for $b,c$ to both be complex, or is it only possible for one to be, the other being a scalar?

Answer 1

For example $z^3=1$, where $z\neq1.$

Id est, $$a=b=c=-\frac{1}{2}+\frac{\sqrt3}{2}i.$$

Answer 2

Another approach: suppose $a, b$ are complex and not real and $ab$ isn't real. Then let $c=\overline{ab}$.

Note that in a precise sense this is universal: if $abc$ is real (and each is nonzero), then $c$ is a real multiple of $\overline{ab}$.

Answer 3

It is definitely, entirely possible. $$a=i\qquad b=c=1+i\qquad abc=-2$$ This example demonstrates that $bc$ doesn't even have to be $a^*$, merely that the sum of their arguments is a multiple of $\pi$.

Answer 4

I'm not sure I understood your question, but I suppose that the equality$$i\times(1+i)\times(1+i)=-2$$answers it.

Answer 5

Polar coordinates.

$a=e^{i\alpha}$, $b= e^{i\beta}$, $c=e^{-i(\alpha+\beta)}$.

Then $abc = e^0=1.$

Answer 6

I think the easiest example to come up with is $e^{2i\pi/3}$,

$$e^{2i\pi/3}\cdot e^{2i\pi/3}\cdot e^{2i\pi/3} = e^{2i\pi}=1.$$

Answer 7

Algebraic approach:

Let the three complex numbers by $z_1=a+bi,z_2=c+di,z_3=e+fi$. Then $$\begin{align}(a+bi)(c+di)(e+fi)&=((ac-bd)+(ad+bc)i)(e+fi)\\&=[e(ac-bd)-f(ad+bc)]+[e(ad+bc)+f(ac-bd)]i\end{align}$$ so for the product to be real we have $$e(ad+bc)+f(ac-bd)=0\implies ac-bd=-\frac{e(ad+bc)}f$$ giving $$(a+bi)(c+di)(e+fi)=-\frac{e^2(ad+bc)}f-f(ad+bc)=-\frac{ad+bc}f(e^2+f^2)$$ So any three complex numbers $z_1,z_2,z_3$ satisfying $$\Re(z_3)(\Re(z_1)\Im(z_2)+\Im(z_1)\Re(z_2))+\Im(z_3)(\Re(z_1)\Re(z_2)-\Im(z_1)\Im(z_2))=0$$ will do. Hence there are infinitely many trios whose product is real.

Answer 8

If you represent a complex number using polar coordinates (angle and a distance from zero), it is known that multiplying the numbers in this trigonometric form is way easier than in the algebraic form - you simply multiply the distance and add the angles:

$$z_1=r_1(\cos(ϕ_1)+i\sin(ϕ_1))$$ $$z_2=r_2(\cos(ϕ_2)+i\sin(ϕ_2))$$ $$z_1z_2=r_1r_2(\cos(ϕ_1+ϕ_2)+i\sin(ϕ_1+ϕ_2))$$

Once you are accustomed to this, the rest is simple. If $ϕ$ is parallel with the x axis (0 or 180°, $\sin ϕ=0$), the number is real, and so your only task is to find three angles that add up to 0 (mod 180°). There is an infinite number of them.

Answer 9

Let the numbers be of the form $r_1 \exp(i t_1)$. Then the product is $r_1 r_2 r_3 \exp (i(t_1+t_2+t_3))$. The imaginary part of this is $\sin(t_1+t_2+t_3)$. So any set of $t$'s where this is zero will do the trick. For example, $t_1+t_2+t_3 = 0$. You can generalize this proof to any number of complex numbers, not just three.

Answer 10

Your problem is presumably that you are approaching the problem as:

• Think of non-real values for $a$, $b$, and $c$. Hope that $abc$ is real.

A much easier way to deal with the problem is

• Think of non-real values for $a$, $b$, and think of a real value for $abc$. Hope that $c$ is non-real.

(Note: I assume the form of the problem above is what the OP actually intends to ask)

Answer 11

If you represent complex numbers as vectors on the complex plane, then multiplication of two complex numbers produces a result whose angle is the sum of the two multiplicand angles.

It's trivial to note that for one vector with any arbitrary angle, multiplication by another vector with the negative of that angle will produce a result on the real line.

Answer 12

First, $(x + iy) \times (x - iy) = x^2 - i^2 y^2 = x^2 + y^2$ is real.

So you can take any complex numbers $a, b$ where the product $ab$ is not real, write $ab = (x + iy)$, and choose $c = (x - iy)$, or choose $c = t(x - iy)$ for any real $t ≠ 0$, and the product $abc$ is real.

Answer 13

Break down the argument into two parts:

1. Find two complex numbers which when multiplied give you a complex number
2. Find a third complex number which when multiplied with result from (1) gives a real.

It will be easy to digest.

Answer 14

There are lots of what I would call trivial or near trivial answers, those that include only real numbers or include duplicates.

The intention of the problem seems to me that what is sought are distinct and having non-zero complex part.

For motivation, note that multiplication by a complex on the unit circle is like rotating by or adding the angle of it.

For the first part, distinct, any full set of roots of unity will work. For 3,

$$e^{0} \cdot e^{i 2\pi/3} \cdot e^{i 4\pi/3} = e^{i(0+2+4))\pi/3} = e^{i 6\pi/3} = e^{i 2\pi} = 1$$

Or in explicit complex notation,

$$1 \cdot (-1 + i \sqrt{3})/2 \cdot (-1 - i \sqrt{3})/2 = 1 + i 0$$

(this works for all angles plus $2\pi k$).

Again, because multiplication by a complex (on the unit circle) is like rotating or adding by an angle, we can take any distinct triple (like the one above), and adjust a bit so that none are on the real line. Rotating the first by $\pi/6$ and the last back by the same, we get:

$$e^{i \pi/6} \cdot e^{i 2\pi/3} \cdot e^{i 7pi/6} = e^{i(1+4+7)\pi/6} = e^{i 12\pi/6} = e^{i 2\pi} = 1$$

or

$$(\sqrt{3} + i)/2 \cdot (- 1 + i\sqrt{3})/2 \cdot (- \sqrt{3} - i)/2 = 1$$

This is only one example. The entire space of solutions is, for any number of multiplicands on the unit circle, the set of angles that sum to $2\pi k$.

Answer 15

It's exceedingly easy to generate any number of complex numbers that satisfy this property. By starting with an arbitrary complex number and generating a new one by swapping the magnitudes of the real and imaginary parts, we can generate pairs of complex numbers that when multiplied together will always produce an imaginary number, either positive or negative. $$(x+iy)(y+ix)=xy+x^2i+y^2i+xyi^2=(x^2+y^2)i$$ $$(x+iy)(-y-ix)=-xy-x^2-y^2i+xyi^2=(-x^2-y^2)i$$

We can find two such pairs: $$(1+i)(-1-i)=-1-i-i-i^2=-2i$$ $$(1+2i)(2+i)=2+i+4i+2i^2=5i$$

When multiplied together, they will give $-2i\times5i=10$. This result will always be real because both numbers are imaginary with no real part. Now take one complex number from each of the original pairs and multiply them together: $$(1+i)(1+2i)=1+2i+i+2i^2=-1+3i$$

Using this result with the remaining numbers gives: $$(-1+3i)(-1-i)(2+i)$$ $$=(1+i-3i-3i^2)(2+i)$$ $$=(4-2i)(2+i)$$ $$=8+4i-4i-2i^2$$ $$=10$$

I have now demonstrated that three seemingly unrelated complex numbers: $-1+3i$; $-1-i$; & $2+i$, all with non-zero real and imaginary parts (i.e. not counting imaginary numbers as complex numbers), and all with integer magnitudes, can be multiplied together to produce a real number. You can repeat this process with any starting numbers you feel like to generate more triples that satisfy this condition.