anonymous
  • anonymous
add or subtract: 1.) (8-2i) + (6+4i) 2.) (9+5i) - (1+8i) + (9+10i)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • 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.
anonymous
  • 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.
anonymous
  • anonymous
i really don't get the whole "i" process

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anonymous
  • anonymous
yes , i get really confuse of how the "i" used
anonymous
  • anonymous
are you supid?
anonymous
  • 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.
anonymous
  • anonymous
lol !! oh yes very stupid
anonymous
  • 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...
anonymous
  • anonymous
So solve the problems you posted as if it were x instead of i.
anonymous
  • anonymous
you can regard i as simply an inseparable thing there.
anonymous
  • anonymous
The only difference is if you multiply i by i, which becomes -1.
anonymous
  • anonymous
as in just another variable ?
anonymous
  • anonymous
WOW im not surprised some weirdo bully who takes joy in puttin others down is on this site being rude >:O not surprised!!!!
anonymous
  • anonymous
right, so you have to classify the given numbers into reals and imaginaries
anonymous
  • anonymous
No bullying happening here... I'm trying to help. Writing down the answers without knowing why is a bad idea.
anonymous
  • 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...
anonymous
  • anonymous
she didnt mean to you john , she meant to the "someguy"
anonymous
  • anonymous
ah, i made a mistake it's someguy//... i'm sorryyyy
anonymous
  • 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
anonymous
  • anonymous
agreed/ back to the problem itself, kathy, do u still have something unclear?
anonymous
  • 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!
anonymous
  • anonymous
yesss i understand now , sorry it was just very confusing to me in the beginning
anonymous
  • anonymous
except for the fact that i^2=-1

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