Hello all,

I was hoping for some help on turning complex number to polar form. I'm working on a nonhomogenous difference equation, and while solving I got the roots $\displaystyle \frac{1.5\pm\sqrt{-1.75}}{2}$.

Ok, so far so good. The modulus, $\displaystyle \rho=\sqrt{(\frac{3}{4})^2+(\frac{\sqrt{1,75}}{2}) ^2}$, which gives me $\displaystyle \rho=1$. Then the argument, $\displaystyle \cos \theta = \frac{3/4}{1}=\frac{3}{4}\mbox{, } \sin \theta = \frac{\sqrt{1,75}/2}{1}=\frac{\sqrt{1,75}}{2}$. This gives me $\displaystyle \cos^{-1} \theta =\sin^{-1} \theta = 0.7227342478$. Is this correct?

The root I got, $\displaystyle \frac{1.5\pm\sqrt{-1.75}}{2}$, when I want to find $\displaystyle \sin \theta = \frac{b}{\rho}$, what value of b do I use (because of the $\displaystyle \pm$)? The negative or positive value of $\displaystyle \frac{\sqrt{-1.75}}{2}$?

Is there any way to make the argument a "pretty" $\displaystyle \pi$-fraction (if it exists)?

Thank you for any help!