Simplification results in different answer

Question

I want to find $$|1 + \frac{1}{2}z| \leq 1$$ so I use wolfram alpha and get the geometric area.

If I instead simplify it by setting $z = a + bi$ and obtaining $$|1 + \frac{1}{2}a + \frac{1}{2}bi| \leq 1$$ $$\sqrt{(1+\frac{1}{2}a)^2 + (\frac{1}{2}bi)^2}\leq 1$$

$$1 + a + a^2/4 + b^2/4 \leq 1$$ I get a different area, why?

Answer 1

The first plot from Wolfram Alpha assumes that $z$ is real -- it's not smart enough to know that when there's a $z$ and no $x$ or $y$, you probably mean that it's a complex number.

The plot has $z$ on the horizontal axis and $|1+\frac12z|$ on the vertical axis. The shaded area doesn't really mean anything, but if you were to zoom in enough to see the ticks you'd see that the real solutions are the interval $[-4,0]$.

This is also the real part of the solution set you plot in the second graph. Nothing has actually changed; they're just different plots.

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Instead of blindly plotting, it's more instructive to multiply each side of your inequality by $2$, giving $$ |2+z| \le 2 $$ which you can rewrite as $$ |z-(-2)| \le 2 $$ So you're looking for the points whose distance to $-2$ is less than or equal to $2$ -- or in other words for a closed disk with center $-2$ and radius $2$.

Answer 2

The modulus requires taking the Complex Conjugate, not merely squaring and square rooting as you have done.

Answer 3

In the first figure, Wolfram Alpha has assumed $z$ to be real. Therefore$$-2\le {z\over 2}\le 0$$or$$-4\le {z}\le 0$$which is reasonable.