Question 1026094
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Furthermore,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \text{Arg}\left(\frac{z_1}{z_2}\right)\ =\ \text{Arg}(z_1)\ -\ \text{Arg}(z_2)]


So for your example:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \text{Arg}\left(\frac{2\ -\ 5i}{1\ -\ 3i}\right)\ =\ \arctan\left(\frac{-5}{2}\right)\ -\ \arctan\left(-3\right)]


"arctan" may be designated as *[tex \Large \tan^{-1}] or *[tex \Large atan] on your scientific calculator.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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