Proving the square root of $z\in \mathbb{C}$ geometrically

Question

I hope somebody could help me to prove that the squareroot exists for all numbers in $\mathbb{C}$. A complex number is Always a stretch and a Rotation at the same time. For example (3,2) Projects (1,0) to (3,2), if we trat (3,2) like we did treat (3,2), we do the same stratch and Rotation again. Visually it would look like this:

Can one derive an explicit Formula with that idea or with other words is it possible to argue that every Point on a plane can be described as the composition of two equal streteches and Rotation?

Many Thanks for your Input and time.

Here is also another idea of mine to find such a Point but not sure whether it is Right and how to prove it

score 2 · Accepted Answer · answered Dec 02 '18 at 12:53

2

From that point of view, consider half of the rotation and the square root of the strecht. If you multiply this by itself, you will get the original number.

answered Dec 02 '18 at 12:53

José Carlos Santos

427,504

For example if I want to find the squareroot of (2,3) it would be $(\sqrt(2),\frac{3}{2})$? – RM777 Dec 02 '18 at 13:12
1

The complex number $2+3i$ corresponds to a strecht of $\sqrt{13}$ and a rotation of $\arccos\left(\frac2{\sqrt{13}}\right)$. – José Carlos Santos Dec 02 '18 at 13:18

score 0 · Answer 2 · answered Dec 02 '18 at 12:53

0

I am not sure I understood your question correctly, but I think this answer will have all you are looking for:

How do I get the square root of a complex number?

answered Dec 02 '18 at 12:53

Jeffery

278

Proving the square root of $z\in \mathbb{C}$ geometrically

2 Answers2