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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    i really don't get the whole "i" process

  4. anonymous
    • 5 years ago
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    yes , i get really confuse of how the "i" used

  5. anonymous
    • 5 years ago
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    are you supid?

  6. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    lol !! oh yes very stupid

  8. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    So solve the problems you posted as if it were x instead of i.

  10. anonymous
    • 5 years ago
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    you can regard i as simply an inseparable thing there.

  11. anonymous
    • 5 years ago
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    The only difference is if you multiply i by i, which becomes -1.

  12. anonymous
    • 5 years ago
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    as in just another variable ?

  13. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    right, so you have to classify the given numbers into reals and imaginaries

  15. anonymous
    • 5 years ago
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    No bullying happening here... I'm trying to help. Writing down the answers without knowing why is a bad idea.

  16. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    she didnt mean to you john , she meant to the "someguy"

  18. anonymous
    • 5 years ago
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    ah, i made a mistake it's someguy//... i'm sorryyyy

  19. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    agreed/ back to the problem itself, kathy, do u still have something unclear?

  21. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    yesss i understand now , sorry it was just very confusing to me in the beginning

  23. anonymous
    • 5 years ago
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    except for the fact that i^2=-1

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