I was doing an exercise on complex numbers, most of which was about finding the moduli and the arguments of complex numbers of the form $\displaystyle \dfrac{\left(x_{1}+iy_{1}\right)^m}{\left(x_{2}+iy _{2}\right)^n}$ and $\displaystyle \left(x_{1}+iy_{1}\right)^m\left(x_{2}+iy_{2}\righ t)^n$ (m and n integers). I had to first convert them into the polar form, use De Moivre's theorem, and then use the fact that the quotient of two complex numbers is their moduli divided and arguments subtracted, and that the product of two complex numbers is their moduli multiplied and their arguments added. After finishing, I thought that someone kind of a formula that gives the moduli and the argument straight away would be quite nice.

I have found that if $\displaystyle z = \dfrac{\left(x_{1}+iy_{1}\right)^m}{\left(x_{2}+iy _{2}\right)^n} $, ($\displaystyle m$ and $\displaystyle n$ integers), then $\displaystyle |z| = \dfrac{\sqrt{(x_{1}^2+y_{1}^2)^m}}{\sqrt{(x_{2}^2+ y_{2}^2)^n}}$, and $\displaystyle \arg{z} = m\theta_{1}-{n}\theta_{2} $, where $\displaystyle \theta_{1} = \arctan\left\{\dfrac{y_{1}}{x_{1}}\right\} $ and $\displaystyle \theta_{2} = \arctan\left\{\dfrac{y_{2}}{x_{2}}\right\} $.

Because:

$\displaystyle \dfrac{\bigg(x_{1}+iy_{1}\bigg)^m}{\bigg(x_{2}+iy_ {2}\bigg)^n}$ $\displaystyle =\dfrac{\left(\sqrt{x_{1}^2+y_{1}^2}\right)^m\bigg \{\cos(\theta_{1})+i\sin(\theta_{1})\bigg\}^m}{\le ft(\sqrt{x_{2}^2+y_{2}^2}\right)^n\bigg\{\cos(\the ta_{2})+i\sin(\theta_{2})\bigg\}^n}$ = $\displaystyle \dfrac{\sqrt{\left(x_{1}^2+y_{1}^2\right)^m}\bigg\ {\cos(m\theta_{1})+i\sin(m\theta_{1})\bigg\}}{\sqr t{\left(x_{2}^2+y_{2}^2\right)^n}\bigg\{\cos(n\the ta_{2})+i\sin(n\theta_{2})\bigg\}}$ $\displaystyle = \dfrac{\sqrt{(x_{1}^2+y_{1}^2)^m}}{\sqrt{(x_{2}^2+ y_{2}^2)^n}}\bigg\{\cos\bigg(m\theta_{1}-{n}\theta_{2}\bigg)+i\sin\bigg(m\theta_{1}-n\theta_{2}\bigg)\bigg\}.$

Similarly, if $\displaystyle z = \bigg(x_{1}+iy_{1}\bigg)^m\bigg(x_{2}+iy_{2}\bigg) ^n$, then $\displaystyle |z| = \sqrt{\bigg(x_{1}^2+y_{1}^2\bigg)^m\bigg(x_{2}^2+y _{2}^2\bigg)^n}$ and $\displaystyle \arg{z} = m\theta_{1}+{n}\theta_{2} $, where again $\displaystyle \theta_{1} = \arctan\left\{\dfrac{y_{1}}{x_{1}}\right\} $ and $\displaystyle \theta_{2} = \arctan\left\{\dfrac{y_{2}}{x_{2}}\right\} $.

Because:

$\displaystyle \bigg(x_{1}+iy_{1}\bigg)^m\bigg(x_{2}+iy_{2}\bigg) ^n$ $\displaystyle = \left(\sqrt{x_{1}^2+y_{1}^2}\right)^m\bigg\{\cos(\ theta_{1})+i\sin(\theta_{1})\bigg\}^m\left(\sqrt{x _{2}^2+y_{2}^2}\right)^n\bigg\{\cos(\theta_{2})+i\ sin(\theta_{2})\bigg\}^n$ = $\displaystyle \sqrt{\left(x_{1}^2+y_{1}^2\right)^m}\sqrt{\left(x _{2}^2+y_{2}^2\right)^n}\bigg\{\cos({m}\theta_{1}) +i\sin({m}\theta_{1})\bigg\}\bigg\{\cos({n}\theta_ {2})+i\sin(n\theta_{2})\bigg\}$

$\displaystyle = \sqrt{\left(x_{1}^2+y_{1}^2\right)^m\left(x_{2}^2+ y_{2}^2\right)^n}\bigg\{\cos({m}\theta_{1}+{n}\the ta_{2})+i\sin({m}\theta_{1}+{n}\theta_{2})\bigg\}$.

Is what I have done correct?