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**MarceloFantini** I'm afraid I didn't follow. What did you mean with 2CIS(40) * 2CIS(22) = 1CIS(30) + 2CIS(33)? Let's take some specific examples: if $\displaystyle z_1 = i$ and $\displaystyle z_2 = \frac{\sqrt{3}}{2} + i \frac{1}{2}$, we have that:

$\displaystyle z_1 \cdot z_2 = i \cdot \left( \frac{\sqrt{3}}{2} + i \frac{1}{2} \right) = \frac{-1}{2} + i \frac{\sqrt{3}}{2}$.

We see that $\displaystyle \arg z_1 \cdot z_2 = \frac{2 \pi}{3}$ or 120 degrees. Notice that $\displaystyle \arg z_1 = \frac{\pi}{2}$ and $\displaystyle \arg z_2 = \frac{\pi}{6}$, and so $\displaystyle \arg z_1 + \arg z_2 = \frac{\pi}{2} + \frac{\pi}{6} = \frac{2 \pi}{3} = \arg z_1 \cdot z_2$.

The underlying thought is this: a complex number can be represented by a vector starting at the origin, and thus he has an angle within $\displaystyle [0, 2 \pi) $. When you multiply two complex numbers, the result will always sum the angles and their magnitudes will be multiplied.

In our case, if instead of $\displaystyle z_1 = i$ and $\displaystyle z_2 = \frac{\sqrt{3}}{2} + i \frac{1}{2}$ we had $\displaystyle z_1 = 2i$ and $\displaystyle z_2 = \sqrt{3} + i$, the result would be $\displaystyle z_1 \cdot z_2 = -2 + 2i \sqrt{3}$, showing that magnitudes multiplied, however the sum is unchanged.

I hope this clarified it further.