Question

Solve for z.

`z {}^(4) - 65z {}^(2) + 64 = 0`

Answer 1

`\bf z^4-65z^2+64=0\implies (z^2)^2-65(z^2)+64=0 \\\\\\ (z^2-64)(z^2-1)=0\implies \begin{cases} z^2-64=0\\ z^2=64\\ z=\pm√(64)\\ z=\pm 8\\ ------\\ z^2-1=0\\ z^2=1\\ z=\pm√(1)\\ z=\pm 1 \end{cases}`