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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?

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    $\begingroup$ Only the phase of the complex numbers matters here; so you're just looking for three real numbers which are not multiples of $\pi$ but which sum to a multiple of $\pi$. I think you can probably manage that. $\endgroup$ – Patrick Stevens Mar 3 '18 at 15:06
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    $\begingroup$ I know that if $a$ were complex, for $abc$ to be real, $bc=a^*$ -- uh, no. That's false. Just take $b=0$, for instance. $\endgroup$ – Federico Poloni Mar 3 '18 at 16:31
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    $\begingroup$ $(1+0i)(1+0i)(1+0i)$ $\endgroup$ – imallett Mar 3 '18 at 19:18
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    $\begingroup$ I think the more interesting question would be three distinct and non-real complex numbers $\endgroup$ – Mitch Mar 3 '18 at 19:29
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    $\begingroup$ @Mitch It is not hard. Say, $(-i)(1+2i)(2+i)$ is real. As long as the "phases" or arguments of the numbers add up to something parallel to the real axis, it will work. It is in no way difficult to make the factors distinct. $\endgroup$ – Jeppe Stig Nielsen Mar 3 '18 at 19:57

15 Answers 15

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

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

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    $\begingroup$ This is the right answer, but it doesn't explain why such a z has to exist, and the rectangular form doesn't really clear that up either. Pointing out that this value is simply a third of a circle in polar coordinates might be more useful. $\endgroup$ – Draconis Mar 3 '18 at 19:01
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    $\begingroup$ Don’t you find it pretentious to say “id est” when i.e. was adopted into English as an abbreviation for “that is”? $\endgroup$ – Chase Ryan Taylor Mar 4 '18 at 0:15
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    $\begingroup$ @ChaseRyanTaylor "Id est" is simply the non-abbreviated form of i.e. Of course, I didn't even know that just now, but his usage made me look it up and now I know. I don't see a problem with using i.e., so I don't see a problem with using "Id est". $\endgroup$ – Nelson Mar 4 '18 at 0:51
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    $\begingroup$ @Nelson So if I don't see a problem with using e.g., it would follow that using exempli gratia can not sound a bit pretentious? $\endgroup$ – oerkelens Mar 4 '18 at 14:22
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    $\begingroup$ I think more importantly (by which I mean more pedantically) eg would be more correct than ie anyway. There are two solutions for the equation and since only one is given we are only giving an example of what is meant by the first line, not the full solution. "ie" is just giving a different form of the exact same information, eg is giving one example of a set of items as is being done here. Also I would tend to prefer "ie" because that is such a commonly used abbreviation that most people probably don't know that "id est" is the same as ie. $\endgroup$ – Chris Mar 5 '18 at 9:28
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I'm not sure I understood your question, but I suppose that the equality$$i\times(1+i)\times(1+i)=-2$$answers it.

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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.

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    $\begingroup$ This is really the best explanation. $\endgroup$ – gnasher729 Mar 5 '18 at 0:00
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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 $\overline{ab}$.

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  • $\begingroup$ I posted very similar, which I've deleted - just to note that introducing a scalar factor into $c$ makes it possible to choose the product as any desired real number (the terms of the question implicitly exclude zero). $\endgroup$ – Mark Bennet Mar 3 '18 at 15:11
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I think the easiest example to come up with is $e^{2i\pi/3}$,

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

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Polar coordinates.

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

Then $abc = e^0=1.$

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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)

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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 $\pi$.

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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.

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    $\begingroup$ Here's a MathJax tutorial. $\endgroup$ – g.kov Mar 4 '18 at 4:00
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    $\begingroup$ Brilliant! Also: $t_1+t_2+t_3=k\pi$, $k\in\mathbb Z$. $\endgroup$ – MattAllegro Mar 4 '18 at 9:55
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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.

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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.

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  • $\begingroup$ If there's at least one trio, it should be pretty clear there are infinitely many trios. You can multiply any of those 3 numbers by any real, the product will stay real. $\endgroup$ – Eric Duminil Mar 4 '18 at 14:49
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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.

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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 $2\pi k$.

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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.

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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.

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protected by Daniel Fischer Mar 4 '18 at 15:17

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