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**HallsofIvy** 1+ i can be written, in polar form, as $\displaystyle \sqrt{2}(cos(\theta_1)+ i sin(\theta_1)$ where $\displaystyle tan(\theta_1)= 1/1= 1$ (and, of course, $\displaystyle \theta_1= \pi/4$). 2+ i can be written as $\displaystyle \sqrt{5}(cos(\theta_2)+ i sin(\theta_2))$ where $\displaystyle tan(\theta_2)= 1/2$. 3+ i can be written as $\displaystyle \sqrt{10}(cos(\theta_3)+ i sin(\theta_3)$ where $\displaystyle tan(\theta_3)= 1/3$.

By DeMoivre's formula, their product is $\displaystyle (\sqrt{2})(\sqrt{5})\sqrt{10}(cos(\theta_1+ \theta_2+ \theta_3)+ i sin(\theta_ 3+ \theta_2+ \theta_3))$.

The fact that the product is 10 i tells you that $\displaystyle cos(\theta_1+ \theta_2+ \theta_3)= 0$ and $\displaystyle cos(\theta_1+ \theta_2+ \theta_3)= 1$ or that $\displaystyle \theta_1+ \theta_2+ \theta_3= \pi/2$.

$\displaystyle 1- 2i= \sqrt{5}(cos(\theta+ i sin(\theta)$ where $\displaystyle tan(\theta_1)= 2$ and, again by DeMoivre's formula, that is equal to $\displaystyle 25(cos(4\theta)+ i sin(4\theta))$. -7+ 24i can be written as $\displaystyle 25 (cos(\phi)+ i sin(\phi))$ where $\displaystyle tan(\phi)= 24/7$.