Find all $|z|=1$ such that $|z^4+4| = \sqrt{5}.$
I've tried doing $$|z^4+4|^2 = 5 \implies (z^4+4)(\overline{z^4}+4) = 5 \implies |z|^8 + 4(z^4+\overline{z^4}) +11=0,$$ but i'm not sure how to solve that.
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In your last equation it should be $|z|^8$, which is then $1$. Observe that if $z=e^{it}$, then $z^4+\overline{z^4}=2\cos(4t)$. From it should get directly that $\cos(4t)=-3/2$, where $t$ is the argument of $z$. – plop Jun 14 '21 at 17:34
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The last equation is $z^4+\dfrac{1}{z^4}+3=0$ which is a quadratic in $z^4$. – dxiv Jun 14 '21 at 18:11
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This problem can be solved by simple geometry.
Let $z^4=w$
Using $|z^4|=|z|^4$, as $|z|=1$ then $|w|=1$.
Now $w$ satisfies two properties
$|w+4|=\sqrt{5}$ and $|w|=1$, plotting in Argand plane we get that former is a circle with centre at $(-4,0)$ and radius $\sqrt{5}$
while the latter is a circle with centre at $(0,0)$ and radius $1$.
Now the solutions exist on the Argand plane where these two circles intersect and it can be clearly observed that they do not intersect at all . Hence no solution exists for the complex equations.
plop
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Hint: try defining a new variable $w = z^4$ so that the equation in terms of $w$ reads (correcting the $z^8$ term to $|z|^8$)
$$ |w|^2 + 4(w+\bar{w}) + 11 = 0.$$
There are probably multiple ways to solve this, but I would break $w$ into real and imaginary parts and solve, noting that since $|z|=1$, $|w|=1$. I imagine there's a more elegant approach. Then after solving for $w$, you can solve for $z$.
Cameron Williams
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1But this gives $8\Re(w)+12=0\implies\Re(w)=-3/2$ which is impossible if $|w|=1$. – saulspatz Jun 14 '21 at 18:13