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According to the fundamental theorem of algebra, any polynomial: $ax^w + bx^v + cx^u \ldots$ has exactly $n$ zeroes, $n$ being the highest power of any term.

How can $y=x^4+1$ have 4 zeroes? There must be something I'm missing. My solving works as follows:

$$0=x^4+1 \\ -1 = x^4 \\ x = \pm \sqrt[4]{-1}$$

Isn't this just two solutions: $x= 0.707106781 + 0.707106781 i$ and $x= -0.707106781 - 0.707106781 i$? Where do the other two come from?

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    Hint: treat $x^4$ as $(x^2)^2$. – 2017-01-20
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    Regardless, some of the zeroes could have been repeated. For example $y=(x-1)^4$ has four zeroes, all of which are $1$. The fundamental theorem of algebra does not state that the zeroes must all be distinct. – 2017-01-20
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    $\sqrt2+\sqrt2i,\ \sqrt2-\sqrt2i,\ -\sqrt2+\sqrt2i,\ -\sqrt2-\sqrt2i$ – 2017-01-20
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    $\pm\sqrt i$ and $\pm\sqrt-i$ – 2017-01-20

4 Answers 4

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Noticing that $x^2 = \pm i$ which means that $x^2 \in \left \{ e^{i\pi/2}, e^{3i\pi/2} \right \}$. Noticing that $-1 = e^{i\pi}$. We have that

$$ x \in \left \{ \sqrt{i}, - \sqrt{i}, \sqrt{-i}, - \sqrt{-i} \right \} $$

which we can write in polar form as

$$ \sqrt{i} \;\; =\;\; e^{i\pi/4} \\ -\sqrt{i} \;\; =\;\; e^{i\pi} e^{i\pi/4} \;\; =\;\; e^{5i\pi/4} \\ \sqrt{-i} \;\; =\;\; \sqrt{e^{i\pi} e^{i\pi/2}} \;\; =\;\; e^{3i\pi/4} \\ -\sqrt{-i} \;\; =\;\; e^{i\pi} e^{3i\pi/4} \;\; =\;\; e^{7i\pi/4}. $$

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$x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-2x^2=(x^2+\sqrt2x+1)(x^2-\sqrt2x+1)$

and solve two quadratic equations.

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Thanks to all of you in the comments, I've come up with a fully solved relatively elementary answer.

$$0=x^4+1 \\ -1 = (x^2)^2 \\ x^2 = \pm i \\ x=\pm\sqrt{\pm i} \\ or \\x= \pm\sqrt{i}, \pm \sqrt{-i} \\ or \\ x= \sqrt{i}, \sqrt{-i}, -\sqrt{i}, -\sqrt{-i}$$

So the 4 solutions are $\left[ \sqrt{i}, \sqrt{-i}, -\sqrt{i}, -\sqrt{-i} \right]$

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    Just something extra, but the square root of $i$ isn't well defined. So assume $\sqrt{i}=a+bi$ and solve for $a,b$. – 2017-01-20
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Actually, $x=\pm\sqrt{\pm\sqrt{-1}} \Rightarrow x=\pm\sqrt{\pm i}$. So, you have the two roots of plus-or-minus $i$, which are $\pm\frac{\sqrt{2}}{2}\pm\frac{\sqrt{2}i}{2}$.