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

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?

• HINT: Rationalize the fraction. Commented Apr 15, 2020 at 9:02
• @Hussain-Alqatari Doesn't it make the expression more complicated? Commented Apr 15, 2020 at 9:08
• To get the answer algebraically, use a CAS. From Maple: $$\arctan \left( -{\frac {717897987691852588770249\,\sqrt {3}}{ 311576564512748849025623}}+{\frac {973758949636344298055992\,\sqrt {2} }{311576564512748849025623}} \right)$$ But is that of any use? Commented Apr 15, 2020 at 9:40

## 2 Answers

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

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.

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: $$\arg(z^{50})=\left(\frac{1}{3}\pi-50\cdot\arctan\left(-\frac{1}{\sqrt2}\right)\right)\mod 2\pi$$

• But I should compute it algebraically, without using approximations Commented Apr 15, 2020 at 9:25
• @Troaro: please see my edits. Commented Apr 15, 2020 at 9:34
• @Troaro: if this answer has been useful, maybe a $+1$ or a favourite answer? Commented Apr 15, 2020 at 19:13