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Math Help - Complex number

  1. #1
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    Complex number

    Under the section of complex number, i faced 2 questions which i couldn't answer... Here they go...

    -Show that x=1-2i is a root of the equation x3-3x2+7x-5=0. Hence, find all the roots of the equation.

    -Express Complex number-msp180019e92f47ie4d5afg0000585fbhf5c1f804b0.gif in the form a+ib, where a>0

    Note: for both question, 'i' represents imaginary number... Please show step-by-step working if possible... Thanks...
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  2. #2
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    Q1 First note that \displaystyle x^3 - 3x^2 + 7x - 5 = 0 when \displaystyle x=1 , so \displaystyle (x-1) is a factor...

    Long dividing gives the quadratic factor \displaystyle x^3 - 3x^2 + 7x - 5 = (x - 1)(x^2 - 2x + 5).

    Now factorise the quadratic factor to get the required result.



    Q2: Express \displaystyle 1 + i\sqrt{3} in polar form, then use DeMoivre's Theorem.
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  3. #3
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    Quote Originally Posted by MichaelLight View Post
    -Show that x=1-2i is a root of the equation x^3-3x^2+7x-5 = 0. Hence, find all the roots of the equation.
    • Show that x^3-3x^2+7x-5 = 0 yields zero at x = 1-2i.
    • Hint for finding the rest of the roots: note that 1+2i is also a root.
    • For the second question, use De Moivre's theorem, as Prove It suggested.


    Please show step-by-step working if possible... Thanks..
    Notes like this usually serve as a reminder for me not to spoon-feed detailed solutions!
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  4. #4
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    Quote Originally Posted by MichaelLight View Post
    Under the section of complex number, i faced 2 questions which i couldn't answer... Here they go...

    -Show that x=1-2i is a root of the equation x3-3x2+7x-5=0. Hence, find all the roots of the equation.

    -Express Click image for larger version. 

Name:	MSP180019e92f47ie4d5afg0000585fbhf5c1f804b0.gif 
Views:	34 
Size:	1.2 KB 
ID:	20803 in the form a+ib, where a>0

    Note: for both question, 'i' represents imaginary number... Please show step-by-step working if possible... Thanks...
    Since the polynomial coefficients are real...

    x=1-2i is a root

    \Rightarrow\ x=1+2i is also a root

    (x-1+2i)(x-1-2i)(x-\alpha)=x^3-3x^2+7x-5

    \left(x^2-x-2ix-x+1+2i+2ix-2i-4i^2\right)(x-\alpha)=x^3-3x^2+7x-5

    \left(x^2-2x+5\right)(x-\alpha)=x^3-3x^2+7x-5

    which gives \alpha=1 from the product of the last 2 terms


    (a+ib)^2=1+i\sqrt{3}\Rightarrow\ a^2-b^2=1

    and 2ab=\sqrt{3}

    Hence solve the simultaneous equations

    a=\frac{\sqrt{3}}{2b}

    \frac{3}{4b^2}-b^2=1\Rightarrow\ 3-4b^4=4b^2\Rightarrow\ 4b^4+4b^2-3=0

    \left(2b^2-1\right)\left(2b^2+3\right)=0

    and since "b" is real

    2b^2=1\Rightarrow\ b^2=\frac{1}{2}\Rightarrow\ b=\pm\frac{1}{\sqrt{2}}

    \Rightarrow\ a=\pm\sqrt{\frac{3}{2}}

    while "a" and "b" must have same signs as 2ab=\sqrt{3}
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by MichaelLight View Post
    -Express \sqrt{1+i\sqrt{3}} in the form a+ib, where a>0
    Another way (without using polar form):

    \sqrt{1+i\sqrt{3}}=a+bi\Leftrightarrow (a+bi)^2=1+i\sqrt{3}\Leftrightarrow a^2-b^2+2abi=1+i\sqrt{3}\Leftrightarrow


    \begin{Bmatrix} a^2-b^2=1\\2ab=\sqrt{3}\end{matrix}\Leftrightarrow\;\l  dots


    Fernando Revilla


    Edited: Sorry, now I see that the problem was already solved this way.
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