# Determining values of $n$ for which the real part of a complex number is $0$

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!

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

Think geometrically. The argument is $$\pi/6$$. What integer multiples of this are also odd integer multiples of $$\pi/2$$?
$$z=(\sqrt{3}+i)^n=2^n\cdot\left(\frac{\sqrt3}{2}+\frac{1}2i\right)^n=2^n\cdot e^{n\frac{\pi}{6}i}=2^n\cdot \left(\cos\frac{n\pi}6+i \sin\frac{n\pi}6 \right)$$
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,
$$n=3+6k, ~~~k\in\mathbb{Z}$$