Math question

[BRUTAL]

Can someone explain to me how the imaginary number 'i' works in equations?
For example (5-2i)-(6+i), the answer is -1-3i, but i have no idea how to answer it properly.
Edit; i misread the answer thinking it was -1, but now it makes a bit more sense.

 

[Xienoph]

i works like any other variables in that equation. If you're still confused, try replacing the i's with x's.

The only exception comes when you have powers of i. i^2 is -1.

 

[monoVertex]

And

i^3 = i^2 * i = -i
i^4 = i^3 * i = 1

Then they get repetitive.
Except that, it's like Xienoph said. You can treat it like a variable (a so called known variable, in this case). So your equation would be like this:

(5-2i)-(6+i) = 5-2i-6-i = -1-3i

 

[ElderKingpin]

you can make it simpler

i^0 = 1
i^1 = i
i^2 = -1
i^3 = -i

And then it cycles.
After that, when you get a large number
i^72

Divide by 4, and your remainder is your exponent.

Your remainder is 0, so that is 1