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I have some problems to understand the following situation:

Let $A=\{z\in\mathbb C : |z|\ge 1$ and $|\Re(z)|\le\frac{1}{2}\}$, then the inequality $|cz+d|\le 1$ with $c,d\in\mathbb Z$ doesn't have solutions in $A$ if $|c|\ge 2$.

I have tried to write $z=x+iy$ so $|cz+d|^2=c^2(x^2+y^2)+d^2+2cdx\ge 5+2cdx...$

but probably this is the wrong way because there is the obnoxious term $2cdx$.

1 Answers 1

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Draw a sketch of $A$ in the complex plane and note that the imaginary part of $z\in A$ must be (absolutely) at least $\frac{\sqrt 3}2$. Therefore, if $|c|>2$ the imaginary part of $cz+d$ is at least $\sqrt3$, which is more than $1$.