Complex arithmetic may sound like a real difficult exercise to master, but after getting use to imaginary numbers it can be as simple as 2+2.
The number known as i is equal to the square root of 1. If you put that in your calculator you will find that the square root of 1 gives you an unreal answer. i is the basis for the complex number system.
i^1=i
i^2=-1
i^3=-i
i^4=1
All the exponents to follow these four will fall along this pattern.
Example:
i^5=i
i^6=-1
i^7=-i
i^8=1
and so on
So what would i^56 equal?
First you would take 56/4, which equals 14. Since 56 is divisible perfectly by 4 it would equal 1 just like i^4.
From there you could go on to figure out what i^54 and i^57 are by just having the starting point from i^56.
COMPLEX NUMBERS
Complex numbers are written in the form of a+bi, so with numbers it would look something like 4+6i or 2-3i.
When adding and subtracting complex numbers you only add and subtract the whole numbers with the whole numbers and the imaginary numbers with the imaginary numbers.
Example:
(4+3i)+(6-2i)
First add 4+6, then add 3i + -2i.
So the answer would be 10+1i.
Lets try subtracting one.
(12-5i)-(3+8i)
Again subtract 3 from 12 first then subtract 8i from 5i.
Making the answer 9-13i.
Adding and subtracting complex numbers is just like normal arithmetic. DONT LET THE is CONFUSE YOU!
MULTIPLYING COMPLEX NUMBERS
Multiplying complex numbers may be a little bit more difficult then adding and subtracting them but you still use a basic principle you learned back in middle school.
When multiplying complex numbers use the FOIL system as you used to make the general quadratic function.
First
Outer
Inner
Last
Example:
(2+3i) (5-i) = 2(5) + 2(-i) + 3i(5) + 3i(-i)
=10 2i + 15i 3i^2
=10 + 13i 3i^2
3i^2=-3 since i^2=-1
So 10 + 13i 3i^2 = 13 +13i.
(9-2i) (2+7i) = 18 + 63i 4i 14i^2
=18 + 59i 14i^2
=32 + 59i