Skip to main content

IntMath forum | Complex Numbers

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° 135° and 270°?

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° 135° and 270°?
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

.

Continues below

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° 135° and 270°?

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° 135° and 270°?

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.

Reply

You need to be logged in to reply.

Related Complex Numbers questions

Complex Numbers lessons on IntMath

top

Search IntMath, blog and Forum

Search IntMath