## anonymous 5 years ago add or subtract: 1.) (8-2i) + (6+4i) 2.) (9+5i) - (1+8i) + (9+10i)

1. anonymous

kathyworld, do you know how to do this? If you were paying attention to the past problems you should know how to do these. These are far easier than the other questions you posted and to solve those problems you needed to know how to add these. You could not have solved the other problems without knowing how to do this. Are you just writing down what people are saying the answers are? I just want you to do good in math and this is not how to do it.

2. anonymous

1.) (8-2i) + (6+4i)=(8+6)+(-2+4)i=14+2i 2.) (9+5i) - (1+8i) + (9+10i) = (9-1+9)+(5-8+10)i=17+7i. kathy, i know where your problem is. you're not very familiar with the addition, subtract logarithm of imaginaries.

3. anonymous

i really don't get the whole "i" process

4. anonymous

yes , i get really confuse of how the "i" used

5. anonymous

are you supid?

6. anonymous

Ok. That's a good place to start :) $i = \sqrt{-1}$ Which is obviously not a "real number" because you cannot take the square root of a negative number. So instead, it uses i to mean the square root of negative one which is called an imaginary number. So you can treat it like x like you usually would in algebra.

7. anonymous

lol !! oh yes very stupid

8. anonymous

i is just an imaginary number. i know it;s hard for you to change your, maybe deap-seated, mind that i has to be a real number...

9. anonymous

So solve the problems you posted as if it were x instead of i.

10. anonymous

you can regard i as simply an inseparable thing there.

11. anonymous

The only difference is if you multiply i by i, which becomes -1.

12. anonymous

as in just another variable ?

13. anonymous

WOW im not surprised some weirdo bully who takes joy in puttin others down is on this site being rude >:O not surprised!!!!

14. anonymous

right, so you have to classify the given numbers into reals and imaginaries

15. anonymous

No bullying happening here... I'm trying to help. Writing down the answers without knowing why is a bad idea.

16. anonymous

i think John's purpose is good in helping kathy to solve the problem autonomously, but a little rude, at least in my opinion...

17. anonymous

she didnt mean to you john , she meant to the "someguy"

18. anonymous

ah, i made a mistake it's someguy//... i'm sorryyyy

19. anonymous

yes i did mean the someguy! HAHA sowee if u thought i meant u john! ^^ i just do NOT like bullies at ALL ..pet peeve i guess :P lol

20. anonymous

agreed/ back to the problem itself, kathy, do u still have something unclear?

21. anonymous

oh jeez. Sorry. Thats a funny username. Sorry if I came off as rude. Do you understand how to do the problem now? When you have the standard form a + bi of a complex number ( a complex number is a number with an imaginary part), a is the real part and bi is the imaginary part. When working with complex numbers you can just act as if i is a variable like x except when it is i^2 it "magically" becomes real. Just think of what i means. The square root of -1 times the square root of -1 is just -1!

22. anonymous

yesss i understand now , sorry it was just very confusing to me in the beginning

23. anonymous

except for the fact that i^2=-1