# |z⁴| + 1 - 2(z*)²= 0

Solutions are ±1. How do I get there ?

Note that $|z|$ is a real number, so if the equation $|z⁴| + 1 - 2(z^*)²= 0$ is satisfied, $(z^*)^2$ should also be a real number. This means that $z^*$ must be either real or purely imaginary, i.e. $z=x$ or $z=ix$ for some real number $x$.

I) If $z=x$. Then the equation becomes
$x^4-2x^2+1=0 \Rightarrow (x^2-1)^2=0 \Rightarrow x=\pm 1.$

I) If $z=xi$. Then the equation becomes
$x^4+2x^2+1=0 \Rightarrow (x^2+1)^2=0 \Rightarrow \text{no solution}.$
So $x=x=\pm 1$ are the only solutions.

You can use this expression $a^2 - b^2 = (a+b)(a-b)$  by taking either $a = +z^2$ or $a=-z^2$ and $b=z$.

For $a = +z^2$ you will end up with $(z^2+z)(z^2-z) + 1 - z^2 = 0$ and solving this will give $z=-1$

Repeat for $a=-z^2$ and it will give $z=1$

Hope it helps :)

• Gubiv
+2

I think it would work if it was z⁴ -2z² +1 = 0. But |z⁴| -2(z*)² + 1 = 0 is different

• Agus8
+1

You can take ±$z^2$ so it the same as an absolute value

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