Answer
$\left\{ -3, 3, -3i, 3i \right\}$
Work Step by Step
Using the properties of equality, the given equation, $
z^4=81
,$ is equivalent to
\begin{array}{l}\require{cancel}
z^4-81=0
.\end{array}
Using $a^2-b^2=(a+b)(a-b)$, then the factored form of the expression, $
z^4-81=0
,$ is
\begin{array}{l}\require{cancel}
(z^2+9)(z^2-9)=0
.\end{array}
Equating each factor to zero, then,
\begin{array}{l}\require{cancel}
z^2+9=0
\\\\
z^2=-9
\\\\
z=\pm\sqrt{-9}
\\\\
z=\pm\sqrt{-1}\sqrt{9}
\\\\
z=\pm i\sqrt{(3)^2}
\\\\
z=\pm 3i
,\\\\\text{OR}\\\\
z^2-9=0
\\\\
z^2=9
\\\\
z=\pm\sqrt{9}
\\\\
z=\pm\sqrt{(3)^2}
\\\\
z=\pm3
.\end{array}
Hence, the solutions are $
\left\{ -3, 3, -3i, 3i \right\}
$.
