Results 1 to 3 of 3

Math Help - Determine the square roots of...

  1. #1
    Newbie
    Joined
    Mar 2012
    From
    Hong Kong
    Posts
    11

    Determine the square roots of...

    DETERMINE THE SQUARE ROOTS OF 1-(√3)j

    Help will be much appreciated.
    Thanks in advance
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,735
    Thanks
    642

    Re: Determine the square roots of...

    Hello, Scoplex!

    Here's an algebraic approach . . .


    Determine the square roots of . 1 - i\sqrt{3}

    \text{Let }\,a + bi \:=\:\sqrt{1 - i\sqrt{3}}\;\text{ where }a\text{ and }b\text{ are real numbers.}

    \text{Square: }\:(a + bi)^2 \:=\:1 - i\sqrt{3} \quad\Rightarrow\quad (a^2 - b^2) + 2abi \:=\:1 - i\sqrt{3}

    \text{Equate real and imagiary components: }\:\begin{Bmatrix} a^2-b^2 \:=\:1 & [1] \\ 2ab \:=\:\text{-}\sqrt{3} & [2] \end{Bmatrix}

    \text{From [2]: }\:b \:=\:\text{-}\frac{\sqrt{3}}{2a}\;\;[3]

    \text{Substitute into [1]: }\:a^2 - \left(\text{-}\frac{\sqrt{3}}{2a}\right)^2 \:=\:1 \quad\Rightarrow\quad a^2 - \frac{3}{4a^2} \:=\:1

    . . . . 4a^4 - 4a^2 - 3 \:=\:0 \quad\Rightarrow\quad (2a^2 + 1)(2a^2 - 3) \:=\:0

    2a^2 + 1 \:=\:0 \quad\Rightarrow\quad a^2 \:=\:-\frac{1}{2} \quad \text{ but }a\text{ must be real.}

    2a^2 - 3 \:=\:0 \quad\Rightarrow\quad a^2 \:=\:\frac{3}{2} \quad\Rightarrow\quad a \:=\:\pm\frac{\sqrt{3}}{\sqrt{2}}

    \text{Substitute into [3]: }\:b \:=\:-\frac{\sqrt{3}}{2\left(\pm\frac{\sqrt{3}}{\sqrt{2}  }\right)} \quad\Rightarrow\quad b \:=\:\mp\frac{1}{\sqrt{2}}


    \text{Therefore: }\:a + bi \;=\;\pm\frac{\sqrt{3}}{\sqrt{2}} \mp\frac{1}{\sqrt{2}}i \;=\; \pm\left(\frac{\sqrt{3} - i}{\sqrt{2}}\right)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2012
    From
    Unknown
    Posts
    29
    Thanks
    3

    Re: Determine the square roots of...

    1-\sqrt{3}i = 2 \bigg(\cos{\frac{\pi}{3}}-i\sin{\frac{\pi}{3}}\bigg), thus by De Moivre

    \pm \sqrt{1-\sqrt{3}i} = \pm \sqrt{2}\left(\cos{\frac{\pi}{6}}-\sin\frac{\pi}{6}\right)  = \pm \frac{1}{\sqrt{2}}(\sqrt{3}-i).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Determine Roots of Complex Polynomial
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: September 30th 2011, 08:03 AM
  2. Replies: 3
    Last Post: February 21st 2009, 08:47 AM
  3. square roots
    Posted in the Algebra Forum
    Replies: 5
    Last Post: April 20th 2008, 11:07 PM
  4. square roots
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 14th 2007, 12:52 PM
  5. Algebra determine possible roots
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 1st 2007, 01:42 PM

Search Tags


/mathhelpforum @mathhelpforum