Find all complex numbers $z$ such as $z$ and $2/z$ have both real and imaginary part integers I am really struggling to solve this one. I feel like I am missing the key part of the solution, so I would like to see how it's done.
Find all complex numbers $z=x+yi$ such as $z$ and $2z^{-1}$ have both real and imaginary part integers
This is what I thought:
$$2z^{-1} =\frac{2}{z} = \frac{2}{x+yi}= \frac{2}{x+yi}\cdot \frac{x-yi}{x-yi} = \frac{2x-2yi}{x^2+y^2}.$$
In order to $2z^{-1}$ have its imaginary part $\in \mathbb Z$, we should equal $2y$ to $0$
$$2y=0 \Rightarrow y=0$$ $yi=0$ is an integer.
$x$ must also be an integer. We simply assume $x \in \mathbb Z$ (no matter what value $x$ has, as long as it's an integer, we are good).
We do the same for $z$ and find out the same values $yi=0$ and $x \in \mathbb Z$.
Therefore, the set $A=\{(x,y)\in \mathbb C|x\in \mathbb Z\text{ and }y=0\}$ is the set of all complex numbers whose real and imaginary part are integers.
 A: Recall that if a complex number $w$ has integral real and complex parts, then $|w|^2$ must be an integer.
Let's start by discussing the possible values of $|z|^2$. Note that $|z|^2$ is an integer, as is $|2z^{-1}|^2 = 4 / |z|^2$. Therefore, we see that $|z|^2$ is an integer factor of 4. This gives us 3 cases for $|z|^2$; it can either be 1, 2, or 4.
Suppose that $|z|^2 = 1$. The only possibilities are $z = \pm 1$ and $z = \pm i$, all of which work.
Now suppose that $|z|^2 = 2$. The only possibilities here are $z = \pm 1 \pm i$ (where the two $\pm$s are independent). A quick check shows that all of these possibilities work.
Finally, the last possibility is that $|z|^2 = 4$. The only possibilities here are $\pm 2$ and $\pm 2i$; again, a quick check shows that all of these work.
These are all the possibilities.
A: If $z$ and $2/z$ both have real and imaginary parts that are integers, say $z=a+bi$ and $2/z=c+di$, then also
$$(a^2+b^2)(c^2+d^2)=|z|^2\cdot\left|\frac{2}{z}\right|^2=4.$$
This leaves very few options for $a$, $b$, $c$ and $d$.
A: Starting where you left off we have  $$\frac{2x-2yi}{x^2+y^2}=\frac{2x}{x^2+y^2}-\frac{2y}{x^2+y^2}i$$ which implies that $\frac{2x}{x^2+y^2}$ is an integer and so by symmetry $-\frac{2y}{x^2+y^2}$ will be an integer as well. Since the numerator grows linearly and the denominators grows quadratically there can only be finitely many such $x$ and $y$ and indeed we see that with $y=0$ and $x=3$ that $x$ has grown too large so $x,y \leq 2$. Now we also see that $2$ divides the numerator, and so the denominator must be a multiple of $2$ as well. This gives us that $x=\pm 1$ or $x=\pm2$. From here we see that the only possible values are $\pm 1 \pm i, \pm 1, \pm i, \pm 2 $ and $\pm 2i$,
A: If $|\frac {2}{z}| < 1$ then either $Re(\frac 2z)$ and $Im(\frac 2z)$ are either zero or non-integer.
But we are not interested in the non-integer cases.
$|z| \le 2$
But if the components of $z$ are integers, there are only a handful complex numbers to work with.
