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

### My question

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

Diszkrét matematika | Digital Textbook Library

### What I've done so far

.

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 |

Digital Textbook Library
</a>

What I've done so far

.

## Re: All numbers from the sum of complex numbers?

@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

@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"?

## Re: All numbers from the sum of complex numbers?

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

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.

## Re: All numbers from the sum of complex numbers?

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

X

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

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