# A sum and product from first principles

A question was put before Iron Man and Miss Tress. They were allowed to choose from the set $\mathbb{C}$ of complex numbers and asked to find a subset satisfying the conditions below:

• sum is zero
• product is one

Can you provide two suitably different solutions for Iron Man and Miss Tress?

• This question is unclear. What are Iron Man's and Miss Tress's respective aims, over & above making the sum $0$ and product $1$ as stated? (Presumably those aims are different, for different solutions for them to be required.) – Rosie F Jun 14 '16 at 9:03
• Good point. Their aim is to make sum 0 and product 1 as stated but hopefully also to choose quite different answers which have some relevance to their names - the title is a hint. – Tom Jun 14 '16 at 9:19
• You accepted an answer that does not answer the puzzle part? – Jonathan Allan Jun 14 '16 at 13:56
• @Jonathan Allan - you have a point. I have unaccepted for now. – Tom Jun 14 '16 at 14:05
• The accepted answer certainly satisfies the mathematical conditions. But I think the wordplay part of the puzzle needs more work: its wording as it stands, IMO, doesn't motivate clearly enough the wordplay that turned out to be involved. – Rosie F Jun 16 '16 at 14:29

My guess is that the answer you are after is from first principles
Iron Man

$\{0+i, 0-i\}$
Iron Man $\rightarrow$ IM $\rightarrow \Bbb{Im}$

...or any extension thereof, such as $\{0+ai,0-ai,0+\frac{i}a,0-\frac{i}a\}\forall a \in A \subseteq \Bbb{R}$

Miss Tress

$\emptyset$
Miss Tress $\rightarrow$ MT $\rightarrow$ "em tee" $\rightarrow$ empty

• This resolves the clues and title hint of the puzzle as hoped for - the initials and then Imaginary and Empty . Miss Tress's inspiration was from the empty product in e.g. mathoverflow.net/questions/45951/sexy-vacuity/45997#45997 – Tom Jun 14 '16 at 15:03
• Yes, or simply that $1$ is the identity element of the group under the operation of multiplication. – Jonathan Allan Jun 14 '16 at 15:06

{-i,i} where i+(-i) = 0 and i*(-i) = 1
{-i/2,i/2,-2,-1,1,2}

But any solution as:

{-i/k, i/k, -k,-1,1,k} k>0

would do.

• Thanks @Sechiro for this very general and correct solution to make sum 0 and product 1 and you have provided (an infinite number of) different solutions. I'm sorry I added the wordplay tag after your answer. The wordplay is tiny but it helps to distinguish Iron Man's and Miss Tress' answers which I hope will be different in nature. – Tom Jun 14 '16 at 10:02
• Ho, ok. then I think { 26 ,- 26, 1/26 , - 1/26 } is a better answer for Iron Man – Sechiro Jun 14 '16 at 10:11
• This has certainly addressed the different solutions part! I think the wordplay tag which wasn't initially included is needed now. – Tom Jun 14 '16 at 10:38
• I just got your Iron atomic number 26 solution - though it wasn't what I thought it does work. Still holding on to my initial thoughts including the title and Miss Tress – Tom Jun 14 '16 at 11:16
• Your very first solution was the one intended for Iron Man - also thanks for your further very good ones. – Tom Jun 14 '16 at 13:44

The empty set is the most obvious solution. There's also an easy-to-find two-element solution: the sum requirement means that one of the numbers is the negative of the other, and the product requirement thus means that the square of that number is $-1$, i.e. the numbers are $i$ and $-i$. I don't know who is going to choose what. “Iron Man” sounds like a jock name so I guess that Miss Tress will be first to pick the easy solution and Iron Man will be left scrambling because he doesn't know what a complex number is.

What if we wanted more solutions? Let's characterize them all.

Let $A$ be any finite subset of $\mathbb{C}$ not containing $0$, $S$ their sum and $P$ their product. $A \cup \{x,y\}$ is a solution iff $x+y+S=0$ and $x\,y\,P=1$, i.e. iff $y=-S-x$ and $P\,x^2 + PSx + 1 = 0$. We know that $P \ne 0$ so this quadratic equation has two distinct solutions if $(PS)^2-4P \ne 0$ i.e. $PS^2 \ne 4$. The sum of the solutions is $-S$, so solutions to the original puzzle are obtained by taking the two roots $x$ and $y$. If neither $x$ nor $y$ is in the original set $A$ then $A \cup \{x,y\}$ is a solution to the puzzle.

Conversely, any solution to the puzzle with a set of at least two elements can be broken down as above (let $A$ be any subset containing all but two elements; $0$ can't be in $A$ since it would cause the product to be $0$).

In particular, for $A = \varnothing$, we get the single two-element solution: $x+y=0$ and $x\,y=1$, i.e. $-x^2=1$ i.e. $\{x,y\}=\{i,-i\}$.

Finally, there is no one-element solution (the number would have to be both equal to $0$ and $1$).

Even with the generalized solution, an association between names and solutions doesn't stand out. You can draw all kinds of pictures with the cloud of points $A$.

• Sorry for a stupid question. Why is the sum of the elements of the empty set 0 and the product 1? How can the product be well-defined? – Floris Jun 14 '16 at 20:47
• @Floris That's the general definition of the $n$-ary generalization of a binary operation with a neutral element: when you apply it to $0$ items, the result is the neutral element. – Gilles 'SO- stop being evil' Jun 14 '16 at 20:53

For Iron Man:

$\zeta^k, k\in\{0,\dots,n-1\}$ and $\zeta=e^{i2\pi/n}$, where $i^2=-1$ and $n$ odd. This gives:
$\dfrac{\zeta^n-1}{\zeta-1}=\sum_\limits{k=0}^{n-1} \zeta^k=0$ because $\zeta^n=1$ and $\prod_\limits{k=0}^{n-1} \zeta^k=1$, provable using the property of $e$ that $e^a\cdot e^b=e^{a+b}$.

For Miss Tress:

The rectangle over $\mathbb{C}$ given by $\pm a \pm ib$ such that $a^2+b^2=1$, because $(a+ib)(a-ib)=a^2+b^2$, and $(-a+ib)(-a-ib)=a^2+b^2$ and $a,b\in\mathbb{R}$

• I am missing the obvious - his defensive energy shield that can be extended up to 360 degrees? – Jonathan Allan Jun 14 '16 at 11:19
• OK, I have no idea of the reasoning you've used. – Jonathan Allan Jun 14 '16 at 11:25