$Re(z)$ and $Im(z)$ in the complex plane "work as" $x$ and $y$, respectively, in the Cartesian plane, so you have something like
\begin{align*}
|x|+2|y|\leq1 &\Leftrightarrow \:|y|\leq\frac{-|x|+1}{2}\\[1em]
&\Leftrightarrow \:y\leq\frac{-|x|+1}{2} \: \wedge \: y\geq-\frac{-|x|+1}{2}\\[1em]
&\Leftrightarrow \:y\leq-|\frac{x}{2}|+\frac{1}{2} \: \wedge \: y\geq|\frac{x}{2}|-\frac{1}{2}
\end{align*}
If it is still not obvious, start by sketching $y=x/2$, then $y=\pm|x/2|$, and finally, $y=\pm|x/2|\mp1/2$. Then look at the conditions above, and it should be straighforward.
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Bumbble Comm
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denote $z=a+ib$.
Then $Re(z) = a$ and $Im(z) = b$
Plug into your equation and solve.
$Re(z)$ and $Im(z)$ in the complex plane "work as" $x$ and $y$, respectively, in the Cartesian plane, so you have something like \begin{align*} |x|+2|y|\leq1 &\Leftrightarrow \:|y|\leq\frac{-|x|+1}{2}\\[1em] &\Leftrightarrow \:y\leq\frac{-|x|+1}{2} \: \wedge \: y\geq-\frac{-|x|+1}{2}\\[1em] &\Leftrightarrow \:y\leq-|\frac{x}{2}|+\frac{1}{2} \: \wedge \: y\geq|\frac{x}{2}|-\frac{1}{2} \end{align*}
If it is still not obvious, start by sketching $y=x/2$, then $y=\pm|x/2|$, and finally, $y=\pm|x/2|\mp1/2$. Then look at the conditions above, and it should be straighforward.