Drawing $z^4 +16 = 0$ I need to draw $z^4  +16 = 0$ on the complex numbers plane.
By solving $z^4  +16 = 0$ I get:
$z = 2 (-1)^{3/4}$ or $z = -2 (-1)^{3/4}$ or $z = -2 (-1)^{1/4}$ or $z = 2 (-1)^{1/4}$
However, the suggested solution by my teacher is:

Where you can see that the solutions that he found are:
$z = \sqrt{2} + i\sqrt{2}$ or $z = \sqrt{2} - i\sqrt{2}$ or $z = - \sqrt{2} + i\sqrt{2}$ or $z = - \sqrt{2} - i\sqrt{2}$
How did my teacher find these solutions?
Am I missing something here? Maybe I don't need to solve the equation in order to draw it on the plane of complex numbers?

The solutions that your teacher found are the $4$ solutions to $z^4=16$. Here we will find these roots algebraically.
As you point out, the solutions should be of the form $z=2\cdot (-1)^{1/4}$, where "$(-1)^{1/4}$" stands for any fourth root of $-1$. Thus, we need to find the solutions to 
$$z^4=-1,$$
or equivalently, the solutions to $z^4+1=0$. Over the complex numbers, $-1$ is a square, namely $i^2=-1$, so
$$z^4+1 = z^4-(-1) = z^4-i^2$$
and therefore $z^4+1$ factors as $(z^2-i)(z^2+i)$. In order to factor this expression completely, you need to find the square roots of $i$ and the square roots of $-i$. 
Let us find the square roots of $i$, i.e., we are looking for a complex number $a+bi$ such that $(a+bi)^2=i$. Since $(a+bi)^2=a^2-b^2+2abi$, we conclude that $a^2-b^2=0$ (so $a=\pm b$) and $1=2ab=\pm 2b^2$. Thus, $a=b=\pm \frac{\sqrt{2}}{2}$. In other words, the square roots of $i$ are
$$z_1=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i,\quad \text{and} \quad z_2=-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i.$$
You can similarly find that the square roots of $-i$ are given by
$$z_3=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i,\quad \text{and} \quad z_4=-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i.$$
Therefore, it follows from our previous discussion that the roots of $z^4+16=0$ are $2z_1$, $2z_2$, $2z_3$ and $2z_4$, which are the solutions your teacher found.
