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**Leroy****Guest**

How do i solve these

5 mod 2+2i

4+5i mod 3

3+i mod 1+i

i mod 3

56 mod 5i

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,771

Hi Leroy;

For 5 mod 2+2i;

So the answer is 1.

For 4+5i mod 3;

So the answer is 1 - i.

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**Leroy****Guest**

I didn't really understood the first method and it seems to me the second one could be (4+5i)-3(1+i)=1+2i

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,771

Hi;

That is not the correct expression. Mathematica confirms what I did in post #3.

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**Ronald****Guest**

Ok,but could you explain the method to me?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,771

Hi;

I am just plugging into a formula:

For a mod n

I have been interpreting the int as the greatest integer.

I would say to be careful with the work done above. I can find no standardization for that formula for complex numbers. As a matter of fact some say floor or int are not defined for complex numbers. Others define them differently than Mathematica does.

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**zetafunc.****Guest**

Interesting, I never thought about modular arithmetic with complex numbers.

**Leroy****Guest**

I think,as complex number is also a number,so it should have all properties of normal number-mod,factorial,floor,ceil,ratio,integer part,...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,771

Not necessarily, there are differences. For one thing the mod function is different.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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