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Math Help - complex number help

  1. #1
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    complex number help

    I have a couple of questions that i cant figure out to save myself =/

    1.these are expressing in the form a+ib \frac{1}{(2+i)(z+1)} = 5-3i

    i got z= -1/14 + i/14

    2. 2z^2=6iz + 3
    didnt know how to do this.

    3. (z)(zcon)+2(z-zcon)=10 +2i (where zcon is the conjugate of z)

    okay and theres:
    " 1-zi is the root of the equation
    z^2+(i+1)z+k=0

    find the other root and the value of k."

    now i assume you use the \Delta but im not sure.

    im sorry theres so many but i need the help.
    also if this isnt the correct forum also sorry as there was a similar problem listed here too
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  2. #2
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    Dear link0099,

    The first one is incorrect.

    \frac{1}{(2+i)(z+1)}=5-3i

    1=(5-3i)(2+i)(z+1))

    1=(13-i)(z+1)

    z+1=\frac{1}{(13-i)}\times\frac{13+i}{13+i}


    Hope you will be able to continue.

    The second one is a quadratic equation of z.

    2z^2=6iz + 3\Rightarrow{2z^2-6iz - 3=0}

    z^2-3iz-\frac{3}{2}=0

    (z-\frac{3i}{2})^2+\frac{9}{4}-\frac{3}{2}=0

    (z-\frac{3i}{2})^2+\frac{9}{4}-\frac{3}{2}=0

    (z-\frac{3i}{2})^2+\frac{3}{4}=0

    Can you continue from here??
    Last edited by Sudharaka; October 28th 2010 at 05:59 AM. Reason: Mistake in calculation.
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  3. #3
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    ok thanks i understand the first one

    is the second
    +/- \frac{3i}{2}+\frac{9}{4}?

    i think i got my last posted one as well.
    its a 2 not a z
    do you use the fact that the sum of the roots would be let x=second root<br />
x+(1-2i)=i+1 ?
    therfore figuring out k?
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  4. #4
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    that didnt post as expected

    should be
    "letx = the second root

    then the eqn starting with x
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  5. #5
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    is the first one z=\frac{-157}{170}+\frac{i}{170}??
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  6. #6
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    Quote Originally Posted by link0099 View Post
    is the first one z=\frac{-157}{170}+\frac{i}{170}??
    Correct.

    ok thanks i understand the first one

    is the second
    +/- \frac{3i}{2}+\frac{9}{4} ?

    Incorrect.

    (z-\frac{3i}{2})^2=-\frac{3}{4}

    (z-\frac{3i}{2})=\pm{\frac{i\sqrt{3}}{2}}

    z=\frac{3i}{2}\pm{\frac{i\sqrt{3}}{2}}

    z=(\frac{3}{2}\pm\frac{\sqrt{3}}{2})i

    3. (z)(zcon)+2(z-zcon)=10 +2i (where zcon is the conjugate of z)

    Use z=x+iy and simplify. You would end up with two equations with x and y; hence you would be able to solve them simultaneously.

    okay and theres: 1-zi" is the root of the equation z^2+(i+1)z+k=0 find the other root and the value of k."

    You could use the sum and product of roots of a quadratic equation. That is if \alpha~and~\beta are the roots of the quadratic, ax^2+bx+c=0; then, \alpha+\beta=-\frac{b}{a}~and~\alpha\beta=\frac{c}{a}
    Last edited by Sudharaka; October 28th 2010 at 06:33 AM.
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  7. #7
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    Quote Originally Posted by link0099 View Post
    1.these are expressing in the form a+ib \frac{1}{(2+i)(z+1)} = 5-3i
    i got z= -1/14 + i/14
    There is a much easier way to do this one.

    Recall that  \dfrac{1}{z} = \dfrac{{\overline z }}{{\left| z \right|^2 }}

    So  z + 1 = \dfrac{1}{{2 + i}}\dfrac{1}{{5 - 3i}} = \dfrac{{2 - i}}{5}\dfrac{{5 + 3i}}{{34}}.
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  8. #8
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    would my way of working out k in question 4 be right?
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  9. #9
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    Quote Originally Posted by link0099 View Post
    would my way of working out k in question 4 be right?

    Dear link0099,

    If the two roots of this equation are \alpha~and~\beta then, \alpha+(1-2i)=-(1+i)~and~\alpha(1-2i)=k. From these two equations you could find k.
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