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I know the result and how to solve it using trigonometry and De Moivre.

However, given that the complex number $z$ can be rewritten as $a+bi$, how can I solve it algebraically?

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Wouldn't you just write $$ z^4+1=0\\ \Updownarrow\\ w^2=-1,z^2=w $$ And conclude that $z^2=w=\pm i$. Then with $z=a+bi$ the equation $z^2=i$ becomes $$ (a+bi)^2=i\\ \Updownarrow\\ a^2-b^2+2abi=i $$ leads to $a^2-b^2=0$ and thus $a=\pm b$ and then $2abi=i$ gives us $ab=\frac{1}{2}$. So either $a=b=\pm\frac{\sqrt{2}}{2}$ or $a=-b$ which then yields non-real $a,b$ which is absurd.

Then you can repeat the strategy from above for the case $z^2=-i$ to get the other two solutions.

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Inserting $z = a + bi$ we get $$ a^4 - 6a^2b^2 + b^4 + 1 + ab(a - b)(a+b)i = 0 $$ which becomes two real equations $$ a^4 - 6a^2b^2 + b^4 + 1 = 0 \quad \wedge\quad ab(a-b)(a+b) = 0 $$ The second one is already factored for you, giving you four specific cases to work with, two without real solutions. The first one can be seen as a quadratic equation in $a^2$ or $b^2$, depending on which case you choose from the second equation.

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    $\begingroup$ I corrected two critical errata. The coefficient on $a^2b^2$ is $-6$, not $-1$. I hope that you don't mind my doing so. $\endgroup$
    – Mark Viola
    Mar 26 '15 at 15:59
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Using polar form is the standard method, but there is another trick. Some creative rewriting gives:

$$z^4 + 1 = (z^2+1)^2 - 2z^2 = (z^2 + z\sqrt 2 + 1)(z^2 - z\sqrt 2 +1)$$

and you can reduce the problem to solving two quadratic equations.

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$$ z^4 + 1 = 0 \Rightarrow z^4 + 2z^2 + 1 - 2z^2 = 0 $$

$$ \Rightarrow (z^2 + 1)^2 -(\sqrt{2}z)^2 = 0 $$

$$ \Rightarrow (z^2 - \sqrt{2} z + 1)( z^2 + \sqrt{2}z + 1 ) = 0 $$

use $$ ax^2 + bx + c = 0 \Rightarrow x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

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Hints:

  • $z^4+1=(z^2+1)^2-2z^2=(z^2+\sqrt2z+1)(z^2-\sqrt2z+1)$
  • $z^2\pm\sqrt2z+1=(z\pm\tfrac12\sqrt2)^2+\tfrac12$
  • $u^2+a^2=(u+\mathrm ia)(u-\mathrm ia)$
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Hint: $z^4+1=(z^2+i)(z^2-i)$ and $i=(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)^2$ and $-i=(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i)^2$ therefore: $$ z^4+1=\left(z-(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)\right)\left(z+(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)\right)\left(z-(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i)\right)\left(z+(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i)\right) $$

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  • $\begingroup$ That's not really a hint, is it? :) +1 thought. $\endgroup$
    – JMCF125
    May 10 '14 at 17:19
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With total cheat, use the change of variable $w=\dfrac{1+i}{\sqrt2}z$.

The equation becomes

$$w^4-1=0$$

with the obvious solutions $$w=\pm1,\pm i.$$

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