Complex equation: $z^8 = (1+z^2)^4$ What's up with this complex equation?
$ z^8 = (1+z^2)^4 $
To start with, there seems to be a problem when we try to apply root of four to both sides of the equation:
$ z^8 = (1+z^2)^4 $
$ z^2 = 1 + z^2 $
which very clearly doesn't have any solutions, but we know there are solutions: the problem is from an exam, and, besides, wolphram alpha gladily gives them to us.
We've tried to solve it using the trigonomectric form, but the sum inside of the parenthesis is killing all of our attempts.
Any help? Ideas?
 A: When you take fourth roots in the complex plane, you get four solutions that differ by factors of $i$.  This is similar to the fact from the reals that a square root gives you two solutions.  Besides the fourth root you show, you also have the possibilities $$z^2=i(1+z^2)\\z^2=-(1+z^2)\\z^2=-i(1+z^2)$$
A: HINT: we have $z^8-(1+z^2)^4=- \left( 2\,{z}^{2}+1 \right)  \left( 2\,{z}^{4}+2\,{z}^{2}+1 \right) $
A: You have the same phenomenon over the real numbers and with much simpler equations. Consider $x^2 = (x + 1)^2$. If you "take square root", you get $x = x + 1$ which doesn't have any solutions while $x = -\frac{1}{2}$ is clearly a solution. The reason is that a non-zero number has two possible square roots which you need to consider. If $x^2 = (x+1)^2$ then $x = (x+1)$ or $x = -(x + 1)$. The first equation doesn't have any solutions while the second equation has one solution.
In your case, just like Ross Millikan noted, the equation $(z^2)^4 = (1 + z^2)^4$ leads to four possible equations $z^2 = \pm (1 + z^2), z^2 = \pm i (1 + z^2)$ that you should analyze.
A: The solutions of $z^8 = (1+z^2)^4$, and the problem you see, are clearly seen from a factorization process:
$$\begin{align} 
z^8 - (1+z^2)^4 &= 0 \,, \\
[z^4 + (1+z^2)^2] [z^4 - (1+z^2)^2] &= 0 \,, \\
[z^2 + i(1+z^2)] [z^2 - i (1+z^2)] [z^2 - (1+z^2)] [z^2 + (1+z^2)] &= 0 \,.
\end{align}$$
From the last line, it is clear that you have only considered one case (the third term) over four other cases. This case, in particular, has no solutions, 
$$\begin{align} 
[z^2 + i(1+z^2)] [z^2 - i (1+z^2)] [z^2 - (1+z^2)] [z^2 + (1+z^2)] &= 0 \,, \\
[(1+i)z^2 + i] [(1-i)z^2 - i ] [ - 1 ] [2z^2 + 1] &= 0 \,, \\
\end{align}$$
I believe this clarify a bit the answers posted by Millikan and Levap.
A: $$z^8=(z^2+1)^4\Longleftrightarrow$$
$$z^4=(z^2+1)^2\Longleftrightarrow\space\space\vee\space\space z^4=-(z^2+1)^2\Longleftrightarrow$$
$$z^2=z^2+1\Longleftrightarrow\space\space\vee\space\space z^2=-1-z^2\Longleftrightarrow\space\space\vee\space\space z^4=-(z^2+1)^2\Longleftrightarrow$$
$$0\ne1\space\space\vee\space\space z^2=-1-z^2\space\space\vee\space\space\Longleftrightarrow z^4=-(z^2+1)^2\Longleftrightarrow$$
$$z^2=-1-z^2\Longleftrightarrow\space\space\vee\space\space z^4=-(z^2+1)^2\Longleftrightarrow$$
$$2z^2=-1\Longleftrightarrow\space\space\vee\space\space z^4=-(z^2+1)^2\Longleftrightarrow$$
$$z^2=-\frac{1}{2}\Longleftrightarrow\space\space\vee\space\space z^4=-(z^2+1)^2\Longleftrightarrow$$
$$z=\pm\frac{i}{\sqrt{2}}\Longleftrightarrow\space\space\vee\space\space z^4=-(z^2+1)^2\Longleftrightarrow$$
$$z=\pm\frac{i}{\sqrt{2}}\Longleftrightarrow\space\space\vee\space\space z^4=-z^4-2z^2-1\Longleftrightarrow$$
$$z=\pm\frac{i}{\sqrt{2}}\Longleftrightarrow\space\space\vee\space\space 2z^4+2z^2+1=0\Longleftrightarrow$$

Substitute $x=z^2$:

$$z=\pm\frac{i}{\sqrt{2}}\Longleftrightarrow\space\space\vee\space\space 2x^2+2x+1=0\Longleftrightarrow$$
$$z=\pm\frac{i}{\sqrt{2}}\Longleftrightarrow\space\space\vee\space\space x=\frac{-2\pm\sqrt{-4}}{4}\Longleftrightarrow$$
$$z=\pm\frac{i}{\sqrt{2}}\Longleftrightarrow\space\space\vee\space\space x=\frac{-2\pm 2i}{4}\Longleftrightarrow$$
$$z=\pm\frac{i}{\sqrt{2}}\Longleftrightarrow\space\space\vee\space\space z^2=\frac{-2\pm 2i}{4}\Longleftrightarrow$$
$$z=\pm\frac{i}{\sqrt{2}}\space\space\vee\space\space z=\pm\sqrt{\frac{-2\pm 2i}{4}}$$
A: $$\displaystyle z^8 = (1+z^2)^4 \Rightarrow 1=(1+\frac{1}{z^2})^4$$
Now find 4-th root of unity.
Thus $\displaystyle (1+\frac{1}{z^2})=1,i,-1,-i$
Now solve for $z^2$.
