Results 1 to 7 of 7

Thread: Plotting solutions on Argand diagram

  1. #1
    Member
    Joined
    Apr 2010
    Posts
    156

    Plotting solutions on Argand diagram

    Plot the solutions of $\displaystyle z^2 = -2 + \sqrt{12}i$ on an Argand diagram

    Attempt:

    I found 2 solutions:

    $\displaystyle z = \sqrt{2(-1 + i\sqrt{3})}$

    $\displaystyle z = -\sqrt{2(-1 +i\sqrt{3})}$

    but I am not sure what I am suppose to do next
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by SyNtHeSiS View Post
    Plot the solutions of $\displaystyle z^2 = -2 + \sqrt{12}i$ on an Argand diagram

    Attempt:

    I found 2 solutions:

    $\displaystyle z = \sqrt{2(-1 + i\sqrt{3})}$

    $\displaystyle z = -\sqrt{2(-1 +i\sqrt{3})}$

    but I am not sure what I am suppose to do next
    You should find the solutions in polar form by using deMoivre's Theorem.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Apr 2010
    Posts
    156
    I got:

    r = 4
    $\displaystyle \theta = \frac{2\pi}{3} =\pi - \frac{\pi}{3}$ (since this angle lies in the 2nd quadrant)

    $\displaystyle z^2 = 4(cos\frac{2\pi}{3} + isin\frac{2\pi}{3})$
    $\displaystyle z^2 = 4(-cos\frac{2\pi}{3} + isin\frac{2\pi}{3})$ (since cos is negative in 2nd quadrant)

    I am not sure what else to do
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4
    Quote Originally Posted by SyNtHeSiS View Post
    Plot the solutions of $\displaystyle z^2 = -2 + \sqrt{12}i$ on an Argand diagram

    Attempt:

    I found 2 solutions:

    $\displaystyle z = \sqrt{2(-1 + i\sqrt{3})}$

    $\displaystyle z = -\sqrt{2(-1 +i\sqrt{3})}$

    but I am not sure what I am suppose to do next
    De Moivre's theorem is most likely the method of solution expected.

    You have $\displaystyle z^2=4\left(Cos\frac{2{\pi}}{3}+iSin\frac{2{\pi}}{3 }\right)$

    Your second expression is incorrect and unnecessary, since $\displaystyle z^2$ is a single point on the complex plane.

    However, it does have 2 square roots.

    De Moivre's Theorem gives $\displaystyle z=|z|^{\frac{1}{2}}\left(Cos\beta+iSin\beta\right) ^{\frac{1}{2}}$

    $\displaystyle =\sqrt{4}\left(Cos\frac{\beta}{2}+iSin\frac{\beta} {2}\right)$

    You will have to use the 2 values of $\displaystyle \beta$,...... $\displaystyle \frac{2{\pi}}{3},\;\;2{\pi}+\frac{2{\pi}}{3}$


    Alternatively,

    $\displaystyle z=|z|e^{i\theta}\Rightarrow\ z^2=|z|^2e^{i2\theta}$

    Using Pythagoras' theorem, the length of $\displaystyle z^2$ is $\displaystyle 4$

    Therefore the length of the square root is 2.

    The argument of $\displaystyle z^2$ is $\displaystyle arctan\left(-\frac{\sqrt{12}}{\sqrt{4}}\right)=arctan(-\sqrt{3})={\pi}-arctan{\sqrt{3}}=\frac{2{\pi}}{3}$

    To find solutions for $\displaystyle z$ from $\displaystyle 0^o\rightarrow\ 2{\pi}$, realise that in squaring $\displaystyle z$, the double angle of $\displaystyle z^2$ is $\displaystyle <4{\pi}$

    We therefore also take a 2nd angle for $\displaystyle z$, the angle $\displaystyle 2{\pi}+\frac{2{\pi}}{3}$

    These angles are $\displaystyle 2\theta\Rightarrow\ \theta=\frac{{\pi}}{3},\;\;{\pi}+\frac{{\pi}}{3}$
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    849
    Hello, SyNtHeSiS!

    There is an algebraic solution . . .


    $\displaystyle \text{Plot the solutions of }\,z^2 \:=\: -2 + \sqrt{12}i\,\text{ on an Argand diagram.}$

    $\displaystyle \text{Let: }\:z \:=\: a + bi\:\text{ where }a\text{ and }b\text{ are real number.}$

    $\displaystyle \text{We have: }\:(a+bi)^2 \:=\:-2 + 2\sqrt{3}\,i$

    . . .$\displaystyle (a^2 - b^2) + 2abi \:=\:-2 + 2\sqrt{3}\,i$


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


    From [2]: .$\displaystyle 2ab \:=\:2\sqrt{3} \quad\Rightarrow\quad b \:=\:\frac{\sqrt{3}}{a}$ .[3]

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

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

    . . $\displaystyle \begin{array}{cccccccccc}a^2 - 1 \:=\:0 & \Rightarrow & a^2 \:=\:1 & \Rightarrow & a \:=\:\pm1 \\
    a^2+3 \:=\:0 & \Rightarrow& a^2 \:=\:\text{-}3 & \Rightarrow & a \:=\:\pm\sqrt{3}\,i \end{array}$


    Since $\displaystyle \,a$ is real, $\displaystyle a \:=\:\pm1$

    Substitute into [3]: .$\displaystyle b \:=\:\frac{\sqrt{3}}{\pm1} \quad\Rightarrow\quad b \:=\:\pm\sqrt{3} $


    Therefore: .$\displaystyle z \;=\;\pm1 \pm\sqrt{3}\,i \;=\;\pm(1 + \sqrt{3}\,i)$

    Now plot those two points . . .
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Apr 2010
    Posts
    156
    Quote Originally Posted by Archie Meade View Post
    You will have to use the 2 values of $\displaystyle \beta$,...... $\displaystyle \frac{2{\pi}}{3},\;\;2{\pi}+\frac{2{\pi}}{3}$
    Why does B have 2 values why not only $\displaystyle \frac{2{\pi}}{3}$?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4
    Quote Originally Posted by SyNtHeSiS View Post
    Why does B have 2 values why not only $\displaystyle \frac{2{\pi}}{3}$?
    From an algebraic perspective, $\displaystyle z^2$ has two square roots, $\displaystyle z$ and $\displaystyle -z$

    so the square root of $\displaystyle z^2$ at angle $\displaystyle \frac{{\pi}}{3}$ is $\displaystyle 1+\sqrt{3}i$

    The negative square root is $\displaystyle -1-\sqrt{3}i$

    These complex numbers differ by $\displaystyle {\pi}$ radians.

    The reason we take 2 angles $\displaystyle \beta$ is because in the complex world, the angles

    $\displaystyle \frac{2{\pi}}{3}$ and $\displaystyle 2{\pi}+\frac{2{\pi}}{3}$ are the same, as they reference the exact same position.

    When a complex number is squared, it's angle doubles,
    so a complex number with an angle between $\displaystyle {\pi}$ and $\displaystyle 2{\pi}$
    when squared has an angle between $\displaystyle 2{\pi}$ and $\displaystyle 4{\pi}$.
    Last edited by Archie Meade; Oct 25th 2010 at 02:12 AM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Argand diagram
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: Sep 16th 2010, 04:28 AM
  2. Help me with Argand diagram
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Jun 19th 2008, 01:52 AM
  3. Argand Diagram
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Mar 30th 2008, 09:13 AM
  4. Argand Diagram
    Posted in the Math Topics Forum
    Replies: 6
    Last Post: Jan 8th 2008, 07:59 AM
  5. Plotting Complex Variables on Argand Diagrams
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Aug 8th 2006, 11:04 AM

Search Tags


/mathhelpforum @mathhelpforum