How?

You do operations like you have learnt in your primary school (or you call it elementary school it's the same) addition subtraction multiplication division.......

Imaginary number is just in the form a times i, the imaginary unit. Something called complex numbers are what you will actually duel with. They are in the form a + bi.

Complex number addition example:

\((3+2i)+(1+4i)\\ = 3+1+2i+4i\\ = 4 + 6i\)

Yeah just like how you duel with variables! Grouping terms!!

Complex number subtraction example:

\((52+6i)-(32 + 14i)\\ = 52 - 32 + 6i - 14i\\ = 20 - 8i\)

Yeah it's just grouping terms again.......

Complex number multiplication example:

\((i+2)(5i-1)\\ =(i)(5i)+(2)(5i)-(1)(i)-(2)(1)\\ =5i^2 + 10i - i - 2\\ =5i^2 + 9i - 2\\ = 9i - 2 - 5\leftarrow\text{Note that }i^2 = -1\\ =9i-7\)

Just what you do with polynomials and adding a rule that i^2 = -1...... Nothing special

Up to this point, there are nothing that feels hard. But the difficulties are all in complex number division.......

Complex number division example:

\(\dfrac{5-2i}{3-4i}\\ =\dfrac{5-2i}{3-4i} \times \dfrac{3+4i}{3+4i}\\ =\dfrac{(5-2i)(3-4i)}{(3-4i)(3+4i)}\\ =\dfrac{7-26i}{9+16}\\ \boxed{\text{Numerator:Just as how I did multilpication above.}\\\text{Denominator: }(a+bi)(a-bi)\equiv a^2+b^2}\\ =\dfrac{7-26i}{25}\\ =\dfrac{7}{25}-\dfrac{26i}{25}\)

Q: How do I know what to multiply in the 1st step?

A: You need a idea called 'conjugates'. Just times the original fraction with a fraction of the original denominator's conjugate over itself(because of the rule of x/x = 1)

Q: How do I think of a conjugate to multiply?

A: Just make sure that after some steps you can use the rule (a+bi)(a-bi) = a^2 + b^2 at the denominator.

Let the original fraction be (6+7i)/(2-5i), Then you will multiply (2+5i)/(2+5i) to it, because after further steps you can use that rule I mentioned at the denominator and that will become a real number(non-imaginary number) at the next step because as you can see, there is no 'i's on the right hand side of the rule (a+bi)(a-bi)=a^2 + b^2.

Thanks Max,

I have continues this question on a new thread - because this one is quite old already.

Max, I don't think that you really answered the intended question. :/

http://web2.0calc.com/questions/how-are-imaginary-numbers-really-useful

thx for answering!