Question: After I found out that $\sqrt{2i} = i+1$, I realised that it makes perfect sense if you visualize it on the plane with triangles.

So I continued drawing triangles
\begin{align\*}
(i+1)^2 & = 2i\\
(2i)^2 & = -4\\
(-4)^2 & = -8i
\end{align\*} 
Can anyone explain to me how this makes sense or why I am wrong?

[![](https://i.stack.imgur.com/ZIE7B.jpg)](https://i.stack.imgur.com/ZIE7B.jpg)

Looking at the picture I'm starting to feel like $(2i)^2 = -4$ is not right, but I'm not sure how.

Answer: 
You are confusing the computation of successive powers of $(1+i)$, i.e. $(1+i)^k$ and what is obtained by successive squarings, i.e. $(1+i)^{2^k}$.

In the first case, you get a complex number whose magnitude doubles every time and argument increases by $45°$, hence your triangles.

In the second case, you get a complex number whose magnitude is squared (and grows very quickly like $2,4,16,256,65536\cdots$) and argument doubles. Actually, a subsample of the previous sequence.
Complex numbers can be written in two ways, as $a+bi$ with $a,\,b$ real or as $re^{i\theta}$, which is called polar form. This is where the Argand diagram is most helpful. Here $r,\,\theta$ are the usual polar coordinates in 2D. Hence $r$ is the distance from $0$, i.e. $\sqrt{a^2+b^2}$, while $\theta$ is the anticlockwise angle formed with the positive half of the real axis so $a=r\cos\theta,\,b=r\sin\theta$. (You can prove $e^{i\theta}=\cos\theta +i\sin\theta$ with well-known Taylor series.) Squaring $re^{i\theta}$ squares $r$ while doubling $\theta$. You should be able to show this implies the square of $-4$ is $16$, as you expected before you learned about complex numbers.