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Math Help - All solutions - Complex Number

  1. #1
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    All solutions - Complex Number

    z^4+16=0. Solve and graph all solutions.
    Hint: Then use the solutions to factor 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 z = -2i. And after that, I am not sure how to make it in 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
    Super Member dhiab's Avatar
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    Quote Originally Posted by noppawit View Post
    z^4+16=0. Solve and graph all solutions.
    Hint: Then use the solutions to factor 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 z = -2i. And after that, I am not sure how to make it in 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
    z^4+16=0. Solve and graph all solutions.
    Hint: Then use the solutions to factor 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 z = -2i. And after that, I am not sure how to make it in z=re^{i\theta}

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

    This is what I get:

    \begin{aligned}<br />
z^4 + 16 &= 0 \\<br />
z^4 &= -16<br />
\end{aligned}

    -16 = 16e^{i\pi}

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

    \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
    \frac{\theta + 2\pi k}{n} = \frac{\pi + 2\pi k}{4} = \frac{\pi}{4} + \frac{\pi k}{2}

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

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


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


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


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


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

    Are you sure the problem isn't 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, 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 z^4+16=0 can be factored into quadratic factors with real coefficients. 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 \pm\sqrt{4i}, \pm i\sqrt{4i}. Notice that you have

    \sqrt{i} = \exp(\pi i/4) = \frac{1+i}{\sqrt 2}
    Last edited by Bruno J.; June 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 z^4+16=0 can be factored into quadratic factors with real coefficients. z^4+16 = (z^2 + 4i)(z^2 - 4i), but that doesn't work.

    [snip]
    Not so:

    z^4 + 16 = z^4 + 8z^2 - 8z^2 + 16 = (z^4 + 8z^2 + 16) - 8z^2 = (z^2 + 4)^2 - 8z^2

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

    = (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
    \frac{\theta + 2\pi k}{n} = \frac{\pi + 2\pi k}{4} = \frac{\pi}{4} + \frac{\pi k}{2}
    From this, why \theta = \pi?
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  10. #10
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    Quote Originally Posted by noppawit View Post
    From this, why \theta = \pi?
    Hi noppawit,

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

    Hope this helps.
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