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Math Help - more complex numbers

  1. #1
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    more complex numbers

    hey guys! i have to solve this equation but i have tried diferent things n i cant work it out. please help?

    z^2+2z+iz=0


    i have tired taking a common factor out but that doesnt really get me anywhrere...

    any ideas?

    thank you.
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  2. #2
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    First thing to note is that it is a quadratic, so you will get 2 solutions (roots) to the equation.

    You are correct to factorize the z which is common in each term i.e.

    z(z+2+i) = 0

    This implies that z = 0 is the first solution.

    Then you have z+2+i=0 which when re-arranged gives

    z = -2-i

    You can check that you are correct by substituting in these values for z into the original quadratic equation - you will find the left hand side equates to 0 for both z's as required.
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  3. #3
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    Thumbs up

    thank you so much! i'm so tired , didnt even realise the answer was right infront of me! thank you!
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by boomshine57th View Post
    hey guys! i have to solve this equation but i have tried diferent things n i cant work it out. please help?

    z^2+2z+iz=0


    i have tired taking a common factor out but that doesnt really get me anywhrere...

    any ideas?

    thank you.
    Also you could use the quadratic formula

    z=\frac{(-2+i)-\sqrt{(2+i)^2-0}}{2}=-2-i or z=0
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  5. #5
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    Hello, boomshine57th!

    Solve: . z^2+2z+iz\:=\:0

    i have tried taking a common factor out,
    but that doesnt really get me anywhrere. . . . . really?
    You know the drill ... factor, then set each factor equal to zero, etc. . . .

    Factor: . z(z + 2 + i) \:=\:0

    Then: . \begin{array}{ccccc}\boxed{z \:=\:0} \\ z+2+i \:=\:0 & \Rightarrow & \boxed{z \:=\:\text{-}2-i} \end{array}

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