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I am currently working on a complex analysis problem and I am struggling with one of the questions. The problem is as follows:

Let $z=(\sqrt{3}+i)^n$. Determine all positive integer values of $n$ for which it holds that $\operatorname{Re}(z)=0$.

I have attempted to solve the problem by using the fact that $\operatorname{Re}(z)=\frac{z+\bar{z}}{2}$, where $\bar{z}$ is the complex conjugate of $z$. So, I first found the value of $z$:

$z=(\sqrt{3}+i)^n=e^{n\operatorname{Arg}(\sqrt{3}+i)}=e^{n\operatorname{arctan}(\frac{1}{\sqrt{3}})+2k\pi i}$, where $k$ is an integer.

Then, I used the fact that $e^{i\theta}=\cos(\theta)+i\sin(\theta)$ to write:

$z=\cos(n\operatorname{arctan}(\frac{1}{\sqrt{3}})+2k\pi)+i\sin(n\operatorname{arctan}(\frac{1}{\sqrt{3}})+2k\pi)$

And, I found that $\operatorname{Re}(z)=\cos(n\operatorname{arctan}(\frac{1}{\sqrt{3}})+2k\pi)$. However, I am not sure where to go from here.

Any help or hints would be greatly appreciated!

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  • $\begingroup$ As the answer below indicates, your next step is to actually calculate $\arctan 1/\sqrt3$. $\endgroup$
    – Brian Tung
    May 5, 2023 at 4:28

2 Answers 2

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Think geometrically. The argument is $\pi/6$. What integer multiples of this are also odd integer multiples of $\pi/2$?

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  • $\begingroup$ @user2661923 Thanks for catching the mistake, I misread the question. $\endgroup$
    – DanDan面
    May 5, 2023 at 10:18
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$$z=(\sqrt{3}+i)^n=2^n\cdot\left(\frac{\sqrt3}{2}+\frac{1}2i\right)^n=2^n\cdot e^{n\frac{\pi}{6}i}=2^n\cdot \left(\cos\frac{n\pi}6+i \sin\frac{n\pi}6 \right)$$

If the real part is $0$, then we get:

$$\cos\frac{n\pi}6=0\Rightarrow \frac{n\pi}6=\frac{\pi}2+k\pi,~~~k\in \mathbb{Z}$$ Therefore,

$$n=3+6k, ~~~k\in\mathbb{Z}$$

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