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Math Help - Complex numbers

  1. #1
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    Complex numbers

    simplify 1+2j/1+j

    I got 1+2j/1+j=(1+2j)(1-j)/(1+j)(1-j)=3+j/2
    is this right?

    and
    z^2+4z+13=0
    i have no idea how to do this.... can someone tell me how to start?
    thanks.
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  2. #2
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    Quote Originally Posted by MK47 View Post
    simplify 1+2j/1+j

    I got 1+2j/1+j=(1+2j)(1-j)/(1+j)(1-j)=3+j/2
    is this right?
    It is!


    Quote Originally Posted by MK47 View Post
    and
    z^2+4z+13=0
    i have no idea how to do this.... can someone tell me how to start?
    thanks.

    Have you tried the quadratic formula?
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  3. #3
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    Have you tried the quadratic formula?[/QUOTE]
    i got -2-3j and -2+3j

    another question that i am kind of stuck on
    show that 3+2j is a root of the quation z^3+7z^2+19z-13=0
    and find the other two roots.

    this is what i thought i should do but it's wrong....
    z=(3+2j)
    z^2=(3+2j)(3+2j)
    =5+12j
    z^3=(5+12j)(3+2j)
    =-9+46j

    (-9+46j)+7(5+12j)+19(3+2j)-13=0
    66+148j=0
    but it doesn't......
    i know the other root would be 3-2j but what did i do wrong?
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  4. #4
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    You aren't doing anything wrong unless you copied the problem wrong. 3+ 2j is NOT a root of z^3+7z^2+19z-13=0.
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  5. #5
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    well i just checked and it is 3+2j and i put the equation into my calculator ,it gave me 0.56,-3.78+2.99i and -3.78-2.99j..... so i guess my teacher wrote the question wrong.
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