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?
 A: 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.
A: One easy solution is to use polar coordinates. Write $z=re^{i\theta}$. Then $r^4 e^{i4\theta}=z^4 =-16 = 16 e^{i\pi}$ implies $r=2$ and $4\theta \equiv \pi \bmod 2\pi$. This gives $\theta = \pi/4 + k \pi/2$, which is the picture you posted.
A: Let's concentrate on the geometry.  We look at $w^4+1=0$, and multiply by $2$ later.
Instead of looking at $w^4+1=0$, look at 
$$(w^4+1)(w^4-1)=0,\qquad\qquad(\ast)$$
that is, 
$$w^8=1.$$
The solutions of this are the vertices of a regular octagon inscribed in the unit circle. The "first" vertex is $1$, the second is $1$ rotated counterclockwise by $45$ degrees, and so on. Draw the circle. Put in, in pencil, the vertices of this octagon.  
Every solution of $w^4+1=0$ is among these vertices. But also among these vertices are, by $(\ast)$, the solutions of $w^4-1=0$. These are the $4$ vertices of the square inscribed in the unit circle, with first vertex at $1$, the next at $1$ rotated $90$ degrees, and so on. Erase these $4$ vertices.  
What's left is the $4$ solutions of $w^4+1=0$.  The "first" one is $1$ rotated through $45$ degrees, that is, $\cos(\pi/4)+i\sin(\pi/4)$.
