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Math Help - Plotting solutions on Argand diagram

  1. #1
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    Plotting solutions on Argand diagram

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

    Attempt:

    I found 2 solutions:

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

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

    but I am not sure what I am suppose to do next
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  2. #2
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    Quote Originally Posted by SyNtHeSiS View Post
    Plot the solutions of z^2 = -2 + \sqrt{12}i on an Argand diagram

    Attempt:

    I found 2 solutions:

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

    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.
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  3. #3
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    I got:

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

    z^2 = 4(cos\frac{2\pi}{3} + isin\frac{2\pi}{3})
    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
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  4. #4
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    Quote Originally Posted by SyNtHeSiS View Post
    Plot the solutions of z^2 = -2 + \sqrt{12}i on an Argand diagram

    Attempt:

    I found 2 solutions:

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

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

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

    However, it does have 2 square roots.

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

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

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


    Alternatively,

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

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

    Therefore the length of the square root is 2.

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

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

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

    These angles are 2\theta\Rightarrow\ \theta=\frac{{\pi}}{3},\;\;{\pi}+\frac{{\pi}}{3}
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  5. #5
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    Hello, SyNtHeSiS!

    There is an algebraic solution . . .


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

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

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

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


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


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

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

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

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


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

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


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

    Now plot those two points . . .
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  6. #6
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    Quote Originally Posted by Archie Meade View Post
    You will have to use the 2 values of \beta,...... \frac{2{\pi}}{3},\;\;2{\pi}+\frac{2{\pi}}{3}
    Why does B have 2 values why not only \frac{2{\pi}}{3}?
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  7. #7
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    Quote Originally Posted by SyNtHeSiS View Post
    Why does B have 2 values why not only \frac{2{\pi}}{3}?
    From an algebraic perspective, z^2 has two square roots, z and -z

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

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

    These complex numbers differ by {\pi} radians.

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

    \frac{2{\pi}}{3} and 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 {\pi} and 2{\pi}
    when squared has an angle between 2{\pi} and 4{\pi}.
    Last edited by Archie Meade; October 25th 2010 at 02:12 AM. Reason: typo
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