Complex Sum to Zero

When I first saw this problem, I was like wow, roots of unity ! Well, we will get back to this idea later. Instinct wise, clearly, only $b$ and $d$ can possibly have an irrational imaginary part since their exponent is an odd number. It is also clear that $a = ±1$ . Next, the last equation, $a+b+c+d = 0$ , is very interesting, since it reminds me of conjugate pairs . Aha! If $b$ and $d$ indeed have an irrational imaginary part, then we need them to be complex conjugates of each other, which then implies that under this condition, $c \ \text{must be}\ ± 1$ .

So, for simplicity sake, we shall split into cases:

(i) $a = ±1, c = ∓1$ , where the signs of $a,c$ are opposite.

Basically, this implies that $b+d = 0$ . At this point it is probably helpful to know something about roots of unity . However, we can also determine $b^3 = 1$ by factoring out from $b^3 - 1 = 0$ the non-principal root $+ 1$ , to solve $(b-1)(b^2+b+1) = 0$ . Eitherways, we should be able to get:

$b = 1, -\frac{1}{2} + i(\frac{\sqrt{3}}{2}) \ \text{and} \ -\frac{1}{2} - i(\frac{\sqrt{3}}{2})$

With some knowledge of primitive roots of unity , we can conclude that the two primitive sixth roots of unity are the negatives of the two primitive third roots of unity. Which basically means that we have $2$ pairs of $(b,d)$ that satisfy the conditions, giving a total of $2 \times 2 = 4$ quadruples for this case.

(ii) $b, d$ have no imaginary part.

Clearly, $a = ±1, b = 1, c = ±1, d = ±1$ , so by pure listing we can get $3$ quadruples:

$(a,b,c,d) = (1,1,-1,-1), (-1,1,-1,1), (-1,1,1,-1)$

Since we've exhausted all cases, the final answer is $4 + 3 =$ 7 .

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Now, suppose we want to generalise the problem, say there are more variables. A case bashing will not do. Perhaps it's time to interpret roots of unity another way, such as noticing that in general, the roots of unity form a regular polygon with $n$ sides , and each vertex lies on the unit circle. This reinterpretation allows us to say that, suppose we consider a $4$ sided-gon, with 4 equal sides, then there exists two pair of sides that will translate into each other .

Now if we consider back the problem, then $a+b+c+d = 0$ will imply that $a+c = 0$ , $a+b = 0$ or $a + d = 0$ . We can then proceed as above with the case checking. However, it is simple now to generalise .

Draw the complex circle. all numbers can be represented as $(\cos x + i\sin x)$ let the angles for $a,b,c,d$ be $x_a, x_b, x_c, x_d$ respectively.

$x_a = 0$ or $\pi$

$x_b = 0, 2\pi/3,$ or $4\pi/3$

$x_c = 0, \pi/2, \pi,$ or $3\pi/2$

$x_d = 0, \pi/3, 2pi/3,\ pi,\ 4pi/3,$ or $5\pi/3$

Let $M$ be the number of $x_a, x_b, x_c, x_d = 0$ , and $N$ be the number of $x_a, x_b, x_c, x_d = \pi$

$(M,N) = (2,0), (0,2), (2,1), (1,2)$ are all easily checked and rejected

Since $M+N$ has to be at least $1$ due to $x_a,$ we reject $(0,0)$ and split into $4$ cases: $(M,N) = (2,2), (1,1), (1,0), (0,1)$ .

$(2,2)$ is easily verified to have $3$ possibilities. If $(M,N) = (1,0)$ or $(0,1)$ , $x_b, x_c, x_d$ are all not $0$ or $\pi$ . Since $x_c$ either $\pi/2$ or $3\pi/2,$ the imaginary values of $b$ and $d$ must sum to $i$ or $-i,$ which is impossible. If $(M,N) = (1,1)$ we have $5$ cases:

Case 1: $x_a = 0, x_c = \pi$

Case 2: $x_a = 0, x_d = \pi$

Case 3: $x_a = \pi, x_b = 0$

Case 4: $x_a = \pi, x_c = 0$

Case 5: $x_a = \pi, x_d = 0$

by considering the imaginary component of $c$ , in cases 2, 3 and 5, it is clear that a sum of $0$ is impossible since the imaginary component of c has magnitude $1,$ which the other of $(b,d)$ cannot cover.

We are left with cases 1 and 4. In both, $b = -d.$ It is then clear that there are 2 possibilities: $b = 2\pi/3, d = 5\pi/3$ or $b = 4\pi/3, d = \pi/3$

Thus the number of solutions for $a,b,c,d$ is just $2*2 + 3$

$2*2$ comes from the case $(M,N) = (1,1)$ with $2$ from each subcase; $3$ from $(M,N) = (2,2)$

First, let’s find all the possible values of $a$, $b$, $c$, and $d$, according to the first four equations. These can be found a number of different ways. One of the easiest and most straight forward ways to do this is expanding $(x+yi)^n=1$ for $n=$2, 3, 4, 6 and comparing the complex and imaginary parts of the RHS and the LHS and solving for all the different $x$ and $y$. I, however, will use DeMoivre’s theorem, which states that if $z=r(cos\theta +isin\theta)$ is a complex number and $n$ is a natural number, then $z^n=r^n(cosn\theta +isinn\theta)$, where $r$ is the absolute value of $z$ and $\theta$ is the angle from the positive real axis to the ray containing z in the complex plain. We can show that $r$ must be 1 in all four of our variables and thus need only find the values of $\theta$ for which $cosn\theta =1$.

To prove the first conjecture, it suffices to remember that the product of the Norms is the Norm of the product. Let N(z) be the Norm of z. Thus, $1=N(z^n)=N(r^n(cosn\theta +isinn\theta))=N(r^n) \cdot N(cosn\theta +isinn\theta) = N(r^n) \cdot (cos^2 n\theta + sin^2 n\theta)) = N(r^n)$ Because r is a positive real number (it is the absolute value of z) and $r^n=1$, $r$ must be 1 for all $n$.

Now we can work on the different values of $n$. For $n=2$, $a^2=1 \Rightarrow cos2\theta = 1$, $sin2\theta = 0$, both of which have solutions of $\theta = 0^\circ$, $180^\circ$ in one revolution. Thus $a$ can equal either 1 or -1.

The others can be done similarly. For $n=3$, we have $cos3\theta = 1$, with solutions in the form $\theta = 0^\circ$, $120^\circ$, $240^\circ$ in one revolution. Thus $b$ can equal 1 or $\frac{-1}{2} \pm \frac{\sqrt{3}}{2}$.

Similarly c can equal $\pm 1$ or $\pm i$

And d can equal $\pm 1$, $\frac{1}{2} \pm \frac{\sqrt{3}}{2}$, or $\frac{-1}{2} \pm \frac{\sqrt{3}}{2}$

Now we can start working with the fifth condition. So when we add all four variables together, both the real and imaginary parts must cancel to zero. So, focusing first on the imaginary parts, one can see that if $c=\pm i$, it will not be able to cancel. Thus $c=\pm 1$. One can also see that the imaginary part of $b$ must be the opposite of $d$, and everything will cancel, as far as imaginaries go. As for the reals, it seems best to first focus on $b$ and $d$, and then adjust $a$ and $c$ likewise. Note that if $b$ and $d$ have equal real parts of $\frac{-1}{2}$, they will sum to -1, and $a$ and $c$ will not properly cancel that value. Thus $b$ and $d$ must have opposite real parts of magnitude $\frac{1}{2}$ (or the same or opposite real parts of magnitude one, as shown in the casework below) so that they cancel. Casework on $d$ will give us our answer.

So, if $d=\frac{1}{2} + \frac{\sqrt{3}}{2}$, b must be $\frac{-1}{2} - \frac{\sqrt{3}}{2}$, causing $a+c=0$. This gives us two solutions.

If $d=\frac{1}{2} - \frac{\sqrt{3}}{2}$, b must be the opposite again, causing $a+c=0$. This gives us another two solutions.

If $d=1$, then $b=1$, $a=-1$, $c=-1$. This is one solution.

If $d=-1$, then $b=1$, and $a+c=0$. This gives us two solutions.

We have in total 7 solutions.

By roots of unity, we have $a=\pm1$ , $b=1,-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}$ , $c=\pm1,\pm i$ , and $d=\pm1,\pm\frac{1}{2}\pm i\frac{\sqrt{3}}{2}$ . So we shall split it into two cases, based on $b$ .

Case 1. $b=-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}$ : Clearly $d=\frac{1}{2}\mp i\frac{\sqrt{3}}{2}$ since none of the possible values of $a$ or $c$ allow the $\frac{1}{2}$ or $i\frac{\sqrt{3}}{2}$ terms to cancel out. And there are only $2$ ways for $a+c=0$ , giving us $2\times2=\boxed{4}$ results. Note that we multiplied by two because of the $\pm$ in the case statement.

* Case 2. * $b=1$ : Clearly with this criterion, $a=\pm1$ , $c=\pm1$ , and $d=\pm1$ , so the generating function (a Laurent polynomial) is

$(x^{-1}+x)^3=x^{-3}+3x^{-1}+3x+x^3$

Since the $x^{-1}$ coefficient is $3$ , this yields $\boxed{3}$ results.

In total, there are $4+3=\boxed{7}$ solutions.

$a$ has roots $1,-1$ , $b$ has roots $1,w,w^{2}$

$c$ has roots $1,-1,i,-i$ and $c$ has roots $1,-1,w,-w,w^{2},-w^{2}$

Now sum of $a,b,c,d$ is equal to $0$

Hence possible sets are $1 or -1,\boxed{1},1 or -1,1 or -1$ which has $3$ solutions, $1 or -1, \boxed{w} ,1 or -1, \boxed{-w}$ which has $2$ solutions and lastly 1 or -1, \boxedw^{2} ,1 or -1, \boxed{-w^{2} which has $2$ solutions.

Total solutions $=3+2+2= \boxed{7}$

a= -1, 1 b= 1, w, w2 c= 1, -1, i, -i d= 1, -1, w, w2, -w, -w2

a+b+c+d=0 --> routes of C i & -i cant be solution coz due to balance problem --> if w is there for B the only -w can balance from D and same for w2 --> so available 7 solution solutions (1, 1, -1, -1) (1, w, -1, -w) (1, w2, -1, -w2) (-1, 1, 1, -1) (-1, 1, -1, 1) (-1, w, 1, -w) (-1, w2, 1, -w2)

"If b=1 then one of a, c, and d must be 1 and the other two have to be -1" Not clear why this s true.

for a^2=1 we have a=-1,1; for b^3=1 we have b=1,-0.5-0.866i(let this be x),-0.5+0.866i(let this be y) where i^2=-1 ie i is complex no.; for c^4=1 we have c=1,-1,i,-i; for d^6=1 we have d=1,-1,-0.5-0.866i,-0.5+0.866i,0.5-0.866i,0.5+0.866i; Therefore we can have 7 quadruplets of a,b,c,d given by (1,1,-1,-1),(-1,1,-1,1),(-1,1,1,-1),(1,x,-1,-x),(1,y,-1,-y),(-1,x,1,-x),(-1,y,1,-y). Hence solved.

The roots of $a^2 = 1$ are $-1,1$ and these are also the roots of $c^4 = 1$ with the addition of $i,-i$ .

Let $p = e^{2\pi i/3}$ . The roots of $b^3 = 1$ are $1, p, p^2$ and these are also the roots of $d^6 = 1$ with the addition of their negatives form (-1, -p,-p^2). Both $p,p^2$ are the roots of $x^2+x+1=0$ .

If $b = p$ , then $a+c+d = -p = p^2+1$ . The latter is not achievable since while we can set $d = p^2$ , but $a+c = 1$ is not achievable since $a = 1,-1$ implies $c = 0,2$ respectively.Hence $c = -p$ and $a+d = 0$ gives the possible pairs $(a,d) = (1,-1),(-1,1)$

If $b = p^2$ , then $a+c+d = -p^2 = p+1$ . The latter is not achievable since while we can set $d = p$ , but $a+c = 1$ is not achievable since $a = 1,-1$ implies $c = 0,2$ respectively. Hence $c = -p^2$ and $a+d = 0$ gives the possible pairs $(a,d) = (1,-1),(-1,1)$

If $b = 1$ , then $a+c+d = 0$ . We can't have $c = i.-i$ as they can only be cancelled by their negative forms. Hence $(a,c,d)$ must be permutations of $(-1,-1,1)$ , there are 3 of them.

We have a total of $2+2+3 = 7$ solutions.

a={1,-1} b={w,w^2,1} c={1,-1,i,-i} d={w,w^2,1,-w,-w^2,-1} where i is iota and w is omega

note:-it is important to choose the negation of a chosen number as magnitude of all the numbers is 1 i.e if we choose 1 we have to choose -1

so if a=1 c= -1 and for 3 values of b we get corresponding 3 values of d if a=-1 c=1 and for 3 values of b we get corresponding 3 values of d and 1 more for a=-1 and c=-1

so total 7 quadruples

An important possibility is not addressed: the imaginary part being balanced not by one other root's , but by the sum of two or three.

We consider a graphical representation of all the solutions on the Argand Plane. These should be relatively simple since they are simply equally spaced on the unit circle with different spacings.

(pardon the bad drawing; O represents the origin, each of the slanted lines should be 60deg to the horizontal real number line)

                 (b,d)     (c)      (d)

                     ^         ^        ^

                       \       |       /

                         \     |     /

                           \   |   /

                             \ | /

(a,c,d) <-------------- O --------------> (a,b,c,d)

                             / | \

                           /   |   \

                         /     |     \

                       /       |       \

                     v        v        v

                 (b,d)     (c)      (d)

Since all of these complex numbers have magnitude 1, in order for a + b + c + d to be 0, the vector sum of these 4 values ought to be 0, and by the diagram above, each vector needs to have the correct opposing vector to cancel it out. We consider the following pairs (bear in mind that the final a+b+c+d will have to have 2 of these pairs):

1. Left/Right: (acd, abcd)
2. Up/Down: (c, c)
3. Upleft/Downright: (bd, d)
4. Upright/Downleft: (d, bd)

For number 2, we note that both vectors can only be represented by c, so these are not usable for the question.

For number 3 and 4, it is clear that d has to be used for 1 vector , so the other vector , by elimination, has to be represented by b, i.e. 3 and 4 both use b and d once.

With these points in mind, we can consider the cases for a+b+c+d having 2 of the above pairs in the following situations (remember that 2 cannot be used):

1. Pair 1 , Pair 1 - we may choose the lefts first: (a,c), (a,d), and (c,d); the rights become uniquely determined
2. Pair 1 , Pair 3 - 3 uses (b,d) in a unique order, 1 has 2 permutations of (a,c)
3. Pair 1 , Pair 4 - 4 uses (b,d) in a unique order, 1 has 2 permutations of (a,c)
4. Pair 3 , Pair 3 - requires (b,d) and (b,d), hence not possible
5. Pair 3 , Pair 4 - requires (b,d) and (b,d), hence not possible
6. Pair 4 , Pair 4 - requires (b,d) and (b,d), hence not possible

Therefore, the total number of ordered quadruples is simply 3 (from case 1) + 2 (from case 2) + 2 (from case 3) = 7.

$a=1,-1; b=1, \omega, \omega^{2}; c=1,-1,i,-i; d=1, \omega, \omega^{2}, -1, -\omega, -\omega^{2}$ .

as $a+b+c+d=0$ , the solutions are $(-1,1,1,-1), (-1,1,-1,1), (-1,\omega,1,-\omega), (-1,\omega^{2},1,-\omega^{2}),(1,1,-1,-1), (1,\omega,-1,-\omega), (-1,\omega^{2},-1,-\omega^{2})$ .

A python solution:

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from numpy import roots, absolute
from itertools import product
def solve():
  A = roots([1,0,-1])
  B = roots([1,0,0,-1])
  C = roots([1,0,0,0,-1])
  D = roots([1,0,0,0,0,0,-1])
  return len([(a,b,c,d) for a, b, c, d in product(A, B, C, D) if absolute(a + b + c + d) < 1e-6])

We can take a=1,-1 b=1,-1±√3i/2 c=1,i,-1 d=1,-1,-1±√3i/2,1±√3i/2. following the constraints given in the question , we can get 7 ordered quadruples as follows;- (-1,1,1,-1) (-1,1,-1,1) (1,1,-1,-1) (1,(-1+√3i)/2,-1,(1-√3i)/2 (1,(-1-√3)/2,-1,(1+√3)/2) (-1,(-1+√3)/2,1,(1-√3)/2) (-1,(1-√3)/2,1,(-1+√3)/2)