find all z in C that verify simultaneously:
|z|² + |z| - 2 = 0
and
z^8 + z^6 + z^4 + z^2 = 0
Of the first equation I know that 1 is a root.
And I think i is a solution for the second one.
But I cant find a convincing solution.
Hello, kezman!
Find all $\displaystyle z \in C$ that satisfy simultaneously: .$\displaystyle \begin{array}{cc}|z|^2 + |z| - 2\:=\:0\\z^8 + z^6 + z^4 + z^2 \:= \:0\end{array}$
The first equation factors: .$\displaystyle (|z| -1)(|z| + 2)\:=\:0$
And we get: .$\displaystyle |z| = 1$
. . (A magnitude cannot be negative.)
The second equation factors: .$\displaystyle z^2(z^6 + z^4 + z^2 + 1) \:= \:0$
Factor: .$\displaystyle z^2\left[(z^4(z^2 + 1) + (z^2 + 1)\right] \:= \:0$
Factor: .$\displaystyle z^2(z^2 + 1)(z^4 + 1)\:=\:0$
And we have: .$\displaystyle z^2 = 0\quad\Rightarrow\quad z = 0 $ . . . which does not satisfy the first equation
But these do: $\displaystyle z^2 = -1\quad\Rightarrow\quad z = \pm i$
And so do these: $\displaystyle z^4 = -1\quad\Rightarrow\quad z \,= \,\pm\sqrt{i} \,= \,\pm\frac{1 + i}{\sqrt{2}}$