Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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

Answer 4

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

Answer 5

are you supid?

Answer 6

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.

Answer 7

lol !! oh yes very stupid

Answer 8

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...

Answer 9

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

Answer 10

you can regard i as simply an inseparable thing there.

Answer 11

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

Answer 12

as in just another variable ?

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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!

Answer 22

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

Answer 23

except for the fact that i^2=-1