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Thread: All solutions - Complex Number

  1. #1
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    All solutions - Complex Number

    $\displaystyle z^4+16=0$. Solve and graph all solutions.
    Hint: Then use the solutions to factor $\displaystyle z^4+16=0$ into quadratic factors with real coefficients.

    It gave me a hint, but I don't understand this hint.

    For me, now I know only $\displaystyle z = -2i$. And after that, I am not sure how to make it in $\displaystyle z=re^{i\theta}$

    Would you please tell me the direction.
    Thank you.
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  2. #2
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    Look here.
    In my last reply, there is a general way of finding roots.
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  3. #3
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    Quote Originally Posted by noppawit View Post
    $\displaystyle z^4+16=0$. Solve and graph all solutions.
    Hint: Then use the solutions to factor $\displaystyle z^4+16=0$ into quadratic factors with real coefficients.

    It gave me a hint, but I don't understand this hint.

    For me, now I know only $\displaystyle z = -2i$. And after that, I am not sure how to make it in $\displaystyle z=re^{i\theta}$

    Would you please tell me the direction.
    Thank you.
    Hello : Pose : Z2 = t
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  4. #4
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    Quote Originally Posted by noppawit View Post
    $\displaystyle z^4+16=0$. Solve and graph all solutions.
    Hint: Then use the solutions to factor $\displaystyle z^4+16=0$ into quadratic factors with real coefficients.

    It gave me a hint, but I don't understand this hint.

    For me, now I know only $\displaystyle z = -2i$. And after that, I am not sure how to make it in $\displaystyle z=re^{i\theta}$

    Would you please tell me the direction.
    Thank you.
    $\displaystyle z = -2i$ isn't one of the solutions. $\displaystyle z^4 = (-2i)^4 = 16$.

    This is what I get:

    $\displaystyle \begin{aligned}
    z^4 + 16 &= 0 \\
    z^4 &= -16
    \end{aligned}$

    $\displaystyle -16 = 16e^{i\pi}$

    If $\displaystyle z = re^{i\theta}$, then the nth roots of a complex number would be in the form of

    $\displaystyle \sqrt[n]{z} = \sqrt[n]{r}\cdot e^{\frac{\theta + 2\pi k}{n}i}$ ,
    where k = 0, 1, 2, ... n - 1.

    In our case, n = 4, r = 16, and k will go from 0 to 3. The angle simplifies to
    $\displaystyle \frac{\theta + 2\pi k}{n} = \frac{\pi + 2\pi k}{4} = \frac{\pi}{4} + \frac{\pi k}{2}$

    The four fourth roots of $\displaystyle -16 = 16e^{i\pi}$ are:

    $\displaystyle \begin{aligned}
    z_1 &= \sqrt[4]{16}\cdot e^{\left(\frac{\pi}{4} + \frac{0\pi}{2}\right)i} \\
    &= 2e^{\frac{\pi}{4}i} \\
    &= 2\left[\cos\left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right] \\
    &= 2\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i\right) \\
    &= \sqrt{2} + \sqrt{2}i
    \end{aligned}$


    $\displaystyle \begin{aligned}
    z_2 &= \sqrt[4]{16}\cdot e^{\left(\frac{\pi}{4} + \frac{\pi}{2}\right)i} \\
    &= 2e^{\frac{3\pi}{4}i} \\
    &= 2\left[\cos\left(\frac{3\pi}{4}\right) + i \sin \left(\frac{3\pi}{4}\right)\right] \\
    &= 2\left(-\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i\right) \\
    &= -\sqrt{2} + \sqrt{2}i
    \end{aligned}$


    $\displaystyle \begin{aligned}
    z_3 &= \sqrt[4]{16}\cdot e^{\left(\frac{\pi}{4} + \frac{2\pi}{2}\right)i} \\
    &= 2e^{\frac{5\pi}{4}i} \\
    &= 2\left[\cos\left(\frac{5\pi}{4}\right) + i \sin \left(\frac{5\pi}{4}\right)\right] \\
    &= 2\left(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\right) \\
    &= -\sqrt{2} - \sqrt{2}i
    \end{aligned}$


    $\displaystyle \begin{aligned}
    z_4 &= \sqrt[4]{16}\cdot e^{\left(\frac{\pi}{4} + \frac{3\pi}{2}\right)i} \\
    &= 2e^{\frac{7\pi}{4}i} \\
    &= 2\left[\cos\left(\frac{7\pi}{4}\right) + i \sin \left(\frac{7\pi}{4}\right)\right] \\
    &= 2\left(\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\right) \\
    &= \sqrt{2} - \sqrt{2}i
    \end{aligned}$


    Furthermore, I don't think $\displaystyle z^4+16=0$ can be factored into quadratic factors with real coefficients. $\displaystyle z^4+16 = (z^2 + 4i)(z^2 - 4i)$, but that doesn't work.

    Are you sure the problem isn't $\displaystyle z^4 {\color{red}-} 16=0$ (with a minus sign)?


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  5. #5
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    Thank you for your reply. Anyway, $\displaystyle z^4+16=0$ is a correct one. This problem came from "Advance Engineering Mathematics 9th edition", from Wiley publisher.

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  6. #6
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by yeongil View Post

    Furthermore, I don't think $\displaystyle z^4+16=0$ can be factored into quadratic factors with real coefficients. $\displaystyle z^4+16 = (z^2 + 4i)(z^2 - 4i)$, but that doesn't work.
    Your intuition was right; you can factor it like you did, and it actually makes things quite easy. Your four roots become $\displaystyle \pm\sqrt{4i}, \pm i\sqrt{4i}$. Notice that you have

    $\displaystyle \sqrt{i} = \exp(\pi i/4) = \frac{1+i}{\sqrt 2}$
    Last edited by Bruno J.; Jun 18th 2009 at 12:39 AM.
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  7. #7
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    Quote Originally Posted by yeongil View Post
    [snip]
    Furthermore, I don't think $\displaystyle z^4+16=0$ can be factored into quadratic factors with real coefficients. $\displaystyle z^4+16 = (z^2 + 4i)(z^2 - 4i)$, but that doesn't work.

    [snip]
    Not so:

    $\displaystyle z^4 + 16 = z^4 + 8z^2 - 8z^2 + 16 = (z^4 + 8z^2 + 16) - 8z^2 = (z^2 + 4)^2 - 8z^2$

    $\displaystyle = (z^2 + 4 - 2 \sqrt{2} z) (z^2 + 4 + 2 \sqrt{2} z)$

    $\displaystyle = (z^2 - 2 \sqrt{2} z + 4) (z^2 + 2 \sqrt{2} z + 4)$.
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  8. #8
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    ^ Quite clever! Would have never thought of that!


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  9. #9
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    Quote Originally Posted by yeongil View Post
    In our case, n = 4, r = 16, and k will go from 0 to 3. The angle simplifies to
    $\displaystyle \frac{\theta + 2\pi k}{n} = \frac{\pi + 2\pi k}{4} = \frac{\pi}{4} + \frac{\pi k}{2}$
    From this, why $\displaystyle \theta = \pi$?
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  10. #10
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    Quote Originally Posted by noppawit View Post
    From this, why $\displaystyle \theta = \pi$?
    Hi noppawit,

    Note that we have $\displaystyle z^4=-16$ therefore we are trying to find the 4th roots of $\displaystyle -16$ and so $\displaystyle \theta = \arg(-16) = \pi$

    Hope this helps.
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