Question

z4 = 16One solution is z = 2, another is -2, and there are 2 more distinct complex solutions.What are those solutions?Choose 2 answers:Аz = 2iB2= -42 = -2i

Answer 1

The given equation is

$z^4=16$

Take square root for both sides

$\begin{gathered} \sqrt[]{z^4}=\pm\sqrt[]{16} \\ z^2=\pm4 \end{gathered}$

Now we have 2 values of z^2, we will find the square roots for each

$\begin{gathered} z^2=4 \\ \sqrt[]{z^2}=\pm\sqrt[]{4} \\ z=\pm2 \\ z=2,-2 \end{gathered}$$\begin{gathered} z^2=-4 \\ \sqrt[]{z^2}=\pm\sqrt[]{-4} \\ z=\pm\sqrt[]{-1}.\sqrt[]{4} \end{gathered}$

Replace square root -1 by i

$\begin{gathered} z=\pm i\sqrt[]{4} \\ z=\pm2i \\ z=2i,-2i \end{gathered}$

The other two roots area 2i, -2i

The answers are A, C