How to solve problem with complex number?

Question

I believe the letter i is called a compex number.... anyway, how do I go about solving this:

sq(2)*sq(-2)+sq(-3)*sq(-3)+sq(-4)-i

Please help me UNDERSTAND this! I don't want to just be given the answer! Thank you so so much!!

Answer 1

two things:

i = sq(-1)

sq(-n) = sq(n*-1) = sq(n)*sq(-1) = sq(n)*i

So now you have that, take sq(2)*sq(-2)

you can write that as sq(2)*sq(2)*i = 2i

take sq(-3)*sq(-3)

you can write that as sq(3)*i*sq(3)*i = 3*i*i

since i = sq(-1), i^2 = -1, the above expression becomes 3*-1 = -3

take sq(-4) it becomes sq(4)*i = 2i

putting it all together

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

The second problem..

[3 - i + sq(4)*i]*[2-i-sq(9)*i]

= [3 - i - 2i] * [ 2 - i - 3i]

= [3-3i]*[2-4i]

now open out the brackets and you get

3*2 - 3*4i - 3i*2 + 3i*4i

= 6 - 12i - 6i -12  (i got the -12 by 3*4*i^2 .. remember i^2 = -1)

= -6 - 18i

understood?