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Math Help - Algebra of complex numbers - a few questions

  1. #1
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    Algebra of complex numbers - a few questions

    ok guys a couple more if you dont mind (this is a math help forum so im sure most you wont mind ;D)

    Directions - Simplify the following problems

    24. (3-5i) + (4+6i) = 7+i
    25. (2-2i) - (2+2i) = -4i
    26. (-6i)(-4i) = 24i^2

    I have a few more that need to be checked but they have radical signs in them and I dont know how to type it out on the keyboard =/
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    Quote Originally Posted by fallenx33 View Post
    ok guys a couple more if you dont mind (this is a math help forum so im sure most you wont mind ;D)

    Directions - Simplify the following problems

    24. (3-5i) + (4+6i) = 7+i
    25. (2-2i) - (2+2i) = -4i
    26. (-6i)(-4i) = 24i^2

    I have a few more that need to be checked but they have radical signs in them and I dont know how to type it out on the keyboard =/
    What you have posted are equalities. Are you asking how the right hand side is got from the left hand side?

    If you go to the Latex subforum you will learn how to post mathematical equations and symbols, including radical signs: http://www.mathhelpforum.com/math-help/latex-help/
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    Quote Originally Posted by fallenx33 View Post
    ok guys a couple more if you dont mind (this is a math help forum so im sure most you wont mind ;D)

    Directions - Simplify the following problems

    24. (3-5i) + (4+6i) = 7+i
    25. (2-2i) - (2+2i) = -4i
    26. (-6i)(-4i) = 24i^2

    I have a few more that need to be checked but they have radical signs in them and I dont know how to type it out on the keyboard =/
    If you're asking if your answers are correct, the first two are.

    You can simplify the third further.

    i^2 = -1, so 24i^2 = 24(-1) = -24.
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    Quote Originally Posted by fallenx33 View Post
    24. (3-5i) + (4+6i) = 7+i
    Correct.

    25. (2-2i) - (2+2i) = -4i
    Correct.

    26. (-6i)(-4i) = 24i^2
    This can be simplified further. What is i^2?

    I have a few more that need to be checked but they have radical signs in them and I dont know how to type it out on the keyboard =/
    You may write sqrt(x) for \sqrt x. This is the usual convention. "cbrt" may be used for cube roots, and for anything higher it is probably best to use rational exponents. Or, take the advice that I gave you before and look into LaTeX.

    Quote Originally Posted by mr fantastic View Post
    What you have posted are equalities. Are you asking how the right hand side is got from the left hand side?
    I think this was split from his last thread. Those are his answers on the right.
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    Quote Originally Posted by Reckoner View Post
    You may write sqrt(x) for \sqrt x. This is the usual convention. "cbrt" may be used for cube roots, and for anything higher it is probably best to use rational exponents. Or, take the advice that I gave you before and look into LaTeX.
    Actually for higher roots, the code is

    \sqrt[n]{x} for \sqrt[n]{x}.
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    Quote Originally Posted by Prove It View Post
    Actually for higher roots, the code is

    \sqrt[n]{x} for \sqrt[n]{x}.
    Sorry, I wasn't clear. I was telling him how he should write it in plain-text, not LaTeX. The latter would be preferable, of course.
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    Ok well I have some more problems that I need to be checked, but since I've posted about 10 here, and they've all been pretty much right so far, then im pretty sure I did the rest of them right as well. Thanks for the help guys
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