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As the video on
It says that: $1\cdot(-1)=-1$ can represent as $1$ rotated $180^{\circ}$ around the Origin to get $-1$

$$1\cdot(-1)\mapsto 1\quad\text{Rotate}(180 ^{\circ})$$ so it must be:

$$1\cdot\sqrt{-1}\mapsto 1\quad\text{Rotate}(90 ^{\circ})$$

but how to map: $$(-1\rightarrow\sqrt{-1}\ )\mapsto (180 ^{\circ}\rightarrow90^{\circ}) $$ Is there any principles to map the two transforms, or it's just a basic define as $1+1=2$.

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What happens if you multiply $-1$ by $-i$? – J. M. Aug 10 '12 at 7:14
@J.M. it rotate $(-90^{\circ})$ ,you mean it is just a define? – NFDream Aug 10 '12 at 7:23
@NFDream It's not an arbitrary definition. It's the result of complex number multiplication! – rschwieb Aug 10 '12 at 12:34

I'm a little fuzzy on the question, but let's just review using complex numbers to rotate the complex plane.

Multiplication by a fixed complex number creates a transformation of the plane. Explicitly we're talking about the map $x\mapsto cx$ for some fixed $c\in \mathbb{C}$. The transformation will stretch and rotate the plane. If $|c|=1$ there will be no stretching, so we'll restrict our attention to this case.

If you experiment using $c=i$, then you'll find that
$1\mapsto i$,
$i\mapsto -1$,
$-1\mapsto -i$ and
$-i\mapsto 1$.

You can play a bit and show that this is truly a 90 degree counterclockwise rotation of the complex plane.

In general, the rotation caused by multiplication with a complex number can be seen in the polar form of the complex number. Complex numbers with modulus 1 have the form $e^{i\theta}$, and $\theta$ is going to give the angle of rotation achieved. In the case of $i$, $i=e^{i \pi/2}$, and in the case of $-1$, $-1=e^{i\pi}$

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This is related to your last question about the matrix representation of complex numbers. Recall that a 2d rotation can be represented by the matrix

$$R_\theta = \pmatrix{\cos \theta&\sin \theta\\-\sin\theta&\cos\theta}$$

Notice that $R_0 = I$ and

$$R_{\pi/2} = J = \pmatrix{0&1\\-1&0} \cong i\\ R_{\pi} = \pmatrix{-1&0\\0&-1} \cong -1$$

The complex number $z$ "given" by a rotation matrix will always have $|z|=1$.

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