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Thread: Roots of an equation (complex numbers)

  1. #1
    MHF Contributor arbolis's Avatar
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    Roots of an equation (complex numbers)

    I must find the roots of the following equation : $\displaystyle z^2-2i=0$ and graph them.
    My attempt : $\displaystyle z^2-2i=0 \Leftrightarrow r^2e^{2i\theta}=2i \Rightarrow r^2=2 $ and $\displaystyle e^{2i\theta}=i$ (I don't trust the implication here) and I don't reach anything.

    Another attempt : $\displaystyle z^2-2i=0 \Leftrightarrow a^2-b^2+i(2ab)=2i \Rightarrow a=b=1 \Rightarrow z=1+i$ (easy to graph).
    I notice that $\displaystyle |z|=\sqrt 2$ in my both attempts.

    So what are the roots? I didn't know an equation could have roots!
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  2. #2
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    Quote Originally Posted by arbolis View Post
    I must find the roots of the following equation : $\displaystyle z^2-2i=0$ and graph them.
    My attempt : $\displaystyle z^2-2i=0 \Leftrightarrow r^2e^{2i\theta}=2i \Rightarrow r^2=2 $ and $\displaystyle e^{2i\theta}=i$ (I don't trust the implication here) and I don't reach anything.

    Another attempt : $\displaystyle z^2-2i=0 \Leftrightarrow a^2-b^2+i(2ab)=2i \Rightarrow a=b=1 \Rightarrow z=1+i$ (easy to graph).
    I notice that $\displaystyle |z|=\sqrt 2$ in my both attempts.

    So what are the roots? I didn't know an equation could have roots!
    My attempt would be $\displaystyle z^2-2i=0 \Leftrightarrow z^2-(\sqrt{2}\times -1)^2=0 \Leftrightarrow (z-(\sqrt{2}\times -1))(z+(\sqrt{2}\times -1)) $$\displaystyle \Leftrightarrow (z+\sqrt{2})(z-\sqrt{2}) \Leftrightarrow z=\pm\sqrt{2}$
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  3. #3
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    Quote Originally Posted by arbolis View Post
    I must find the roots of the following equation : $\displaystyle z^2-2i=0$ and graph them.
    My attempt : $\displaystyle z^2-2i=0 \Leftrightarrow r^2e^{2i\theta}=2i \Rightarrow r^2=2 $ and $\displaystyle e^{2i\theta}=i$ (I don't trust the implication here) and I don't reach anything.

    Another attempt : $\displaystyle z^2-2i=0 \Leftrightarrow a^2-b^2+i(2ab)=2i \Rightarrow a=b=1 \Rightarrow z=1+i$ (easy to graph).
    I notice that $\displaystyle |z|=\sqrt 2$ in my both attempts.

    So what are the roots? I didn't know an equation could have roots!
    The other solution to your cartesian approach is a = b = -1.

    So the roots are $\displaystyle z = \pm (1 + i)$.

    As a check, note that $\displaystyle (1 + i)^2 = 1 + 2i - 1 = 2i$.


    If you take the polar approach, then you should note that

    $\displaystyle e^{2 i \theta} = i \Rightarrow e^{2 i \theta} = e^{\pi/2 + 2 n \pi}$ where n is an integer

    $\displaystyle \Rightarrow 2 \theta = \frac{\pi}{2} + 2 n \pi \Rightarrow \theta = \frac{\pi}{4} + n \pi$.

    n = 0: $\displaystyle \theta = \frac{\pi}{4}$.

    n = 1: $\displaystyle \theta = \frac{5 \pi}{4}$.

    Therefore the two roots are:

    $\displaystyle z = \sqrt{2} e^{i \pi/4} = \sqrt{2} \left( \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) = 1 + i$

    $\displaystyle z = \sqrt{2} e^{i 5 \pi/4} = \sqrt{2} \left( - \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) = -1 - i$

    Quote Originally Posted by pickslides View Post
    My attempt would be $\displaystyle z^2-2i=0 \Leftrightarrow z^2-(\sqrt{2}\times -1)^2=0 \Leftrightarrow (z-(\sqrt{2}\times -1))(z+(\sqrt{2}\times -1)) $$\displaystyle \Leftrightarrow (z+\sqrt{2})(z-\sqrt{2}) \Leftrightarrow z=\pm\sqrt{2}$
    Unfortunately this is very wrong, as expanding shows. You've found the roots of $\displaystyle z^2 -2 = 0$.
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