All numbers from the sum of complex numbers? [Solved!]

X

Is it true, that all numbers can be made, as the sum of complex numbers, but only the ones with the argument of 45^"o" 135^"o" and 270^"o"?

Relevant page

<a href="https://www.tankonyvtar.hu/en/tartalom/tamop412A/2011-0038_25_juhasz_diszkret_matematika/ch02s03.html">
            
            
                Diszkr&#233;t matematika | 
            
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What I've done so far

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X

@BuBu: You haven't indicated any working so that I can get a sense of where you are having trouble.

To start, are you able to form the 11 integers `-5,-4,...4,5` as the sum of complex numbers, but only the ones with the argument of 45^"o" 135^"o" and 270^"o"?

X

I'll try.

The `45^"o"` ones will be like `1+i`.

The `135^"o"` ones will be like `-1+j`.

The `270^"o"` ones will be like `-j`

So we can form:

`-5 = 5*((-1+j) + (-j))`

`-4 =4*((-1+j) + (-j))`

`-3 = 3*((-1+j) + (-j))`

`-2 = 2*((-1+j) + (-j))`

`-1 = (-1+j) + (-j)`

`0 = (-1+j)+(1+j)+(-j)+(-j)`

`1 = (1+j) + (-j)`

`2 = 2*((1+j) + (-j))`

`3 = 3*((1+j) + (-j))`

`4 = 4*((1+j) + (-j))`

`5 = 5*((1+j) + (-j))`

So it works for the integers `-5,-4,-3,...5.`

We could do similar things with the decimals, so I'm inclined to think this would be possible for all real numbers.

X

Looks good to me. So yes, as long as we are talking about the reals, I think the claim is reasonable.

X

I encountered this problem again. It's good that it was solved here.