Principal argument of $z=\left(\left(\sqrt{3}+i)/(\sqrt{2}-i\right)\right)^{50}$

Question

I am trying to compute the principal argument (which belongs to $(-\pi,\pi]$) of the complex number $$ z=\left(\dfrac{\sqrt{3}+i}{\sqrt{2}-i}\right)^{50}$$

---

My attempt:

Let $a=\sqrt{3}+i$ and $b=\sqrt{2}-i$. Then $\operatorname{Arg} (a)=\frac{\pi}{6}$ and $\operatorname{Arg}(b)=-\tan^{-1}\left(\frac{1}{\sqrt{2}}\right)$, so $\operatorname{arg}(z)=50\operatorname{arg}\left(\frac{\sqrt{3}+i}{\sqrt{2}-i}\right)=50\operatorname{arg}(a)-50\operatorname{arg}(b)=\frac{50\pi}{3}-50\tan^{-1}(\frac{1}{\sqrt{2}})=\frac{2\pi}{3}-50\tan^{-1}\left(\frac{1}{\sqrt{2}}\right)$, modulo $2\pi$, where $\operatorname{Arg}$ and $\operatorname{arg}$ denote the principal argument and argument, respectively.

Now it suffices to add a suitable integer multiple of $2\pi$ to $\frac{2\pi}{3}-50\tan^{-1}\left(\frac{1}{\sqrt{2}}\right)$, but I got stuck here. Any hints?

Answer 1

$$\frac{\sqrt{3}+i}{\sqrt{2}-i} = \frac{2\exp(\frac{\pi}{6}i)}{\sqrt{3}\exp(-\arctan(\frac{1}{\sqrt{2}})i)} = \frac{2}{\sqrt{3}}\exp\left(\left[\frac{\pi}{6}+\arctan\left(\frac{1}{\sqrt{2}}\right)\right]i\right)$$

So

$$z = \left(\frac{\sqrt{3}+i}{\sqrt{2}-i}\right)^{50} = \left(\frac{2}{\sqrt{3}}\right)^{50}\exp\left(50\left[\frac{\pi}{6}+\arctan\left(\frac{1}{\sqrt{2}}\right)\right]i\right)$$

So the (non-principal) argument is $50 \left[\frac{\pi}{6}+\arctan\left(\frac{1}{\sqrt{2}}\right)\right]$, and the principal argument happens to be

$$\theta =50 \left[\frac{\pi}{6}+\arctan\left(\frac{1}{\sqrt{2}}\right)\right] - 18\pi \approx 0.4053 \,\,\text{radians} \approx 23.22^\circ $$

I don't know if there is a simple way to figure out how many multiples of $2\pi$ to subtract. I simply used a computational tool to calculate the numerical value of the (non-principal) argument.

Answer 2

Converting to trigonometry form, we have: $$z=\frac{2(\cos(\pi/6)+i\sin(\pi/6))}{\sqrt3(\cos(\theta)+i\sin(\theta))}$$ where $\theta=\arctan\left(-\frac{1}{\sqrt2}\right)+2\pi\approx 2\pi-0.615\approx 5.25$.

Now, using De-Moivre's rule, we have: $$z^{50}=\left(\frac{2}{\sqrt3}\right)^{50}=(\cos(50(\pi/6-\theta))+i\sin(50(\pi/6-\theta)))$$ So, the argument is: $$\arg(z^{50})=50\left(\frac{\pi}{6}-\theta\right)=50\cdot\left(\frac{\pi}{6}-\arctan\left(-\frac{1}{\sqrt2}\right)-2\pi\right)=-\frac{550}{6}\pi-50\cdot\arctan\left(-\frac{1}{\sqrt2}\right)$$ Here, we can compute $\arg(z^{50}) \mod 2\pi$, and so: $$\arg(z^{50})=\left(\frac{1}{3}\pi-50\cdot\arctan\left(-\frac{1}{\sqrt2}\right)\right)\mod 2\pi$$