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How would I plot plot $\{z\in\mathbb{C}:|z|^2\ge z +\bar z\}$?

So far I did: \begin{align*} |z|^2 & \ge z +\bar z \\ |a+ib|^2 & \ge a+ib+a-ib \\ a^2 + b^2 & \ge 2a\\ a(a-2)+b^2 & \ge 0 \end{align*}

But I cant seem to get the inequality into a circle or simplify it further so I can see how to plot it.

Also, how do I solve:$ \{z \in \mathbb{C}:Im\left(\frac{z+i}{2i}\right)\le 0\}$? Im not sure how to start this question.

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For the "Also, how do I solve $\dots$" the expression is incomplete. Presumably you want the imaginary part to be something or other specific. – André Nicolas May 23 '12 at 23:33
Whoops, sorry. Edited it. – Derrick May 23 '12 at 23:34
Something useful for future reference: region a^2+b^2 >= 2a (remove the word "region" for more general results) – amr May 23 '12 at 23:49
Thanks, didnt know that function on wolfram! – Derrick May 23 '12 at 23:56
up vote 1 down vote accepted

You are doing fine up to $a^2-2a+b^2 \ge 0$. Then complete the square to get $(a-1)^2+b^2 \ge 1$. These are the points on or outside the circle with centre $(1,0)$, radius $1$.

For the problem about $\Im\left(\frac{z+i}{2i}\right)\le 0$, one way to start is to let $z=a+bi$ and compute. We have $z+i=a+(b+1)i$. Divide by $2i$. We get $\frac{b+1}{2}-\frac{a}{2}i$. The imaginary part of this is $\le 0$ precisely if $a\ge 0$. We needn't have bothered adding the $i$, since we are dividing by $2i$, so the imaginary part will not be affected. And we needn't have bothered with $b$ either.

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Thanks @AndreNicolas , By the way, I've just edited my original post with the other set I wanted to add, how would I solve the second one? – Derrick May 23 '12 at 23:30
Thanks @AndreNicolas, most appreciated, understood! – Derrick May 23 '12 at 23:48

For the second one: first of all, notice that the expression $Im(\frac{z+i}{2i})$ is just $Im(\frac{z}{2i}+\frac{1}{2})$, and the latter part - being purely real - doesn't contribute anything to the imaginary part of the expression, so you're really just looking at the set $Im(\frac{z}{2i})\leq 0$; since $1/i = -i$, this is the same as $Im(-iz/2)\leq 0$. But you can multiply by the real constant -2 (since $Im(r\cdot x)=r\cdot Im(x)$ for any real $r$) to get $Im(iz)\geq 0$. Now, $Im(iz) = Re(z)$, so this is just $Re(z)\geq 0$.

The other way is just to plough through the algebra. Suppose $z=a+bi$, with $a$ and $b$ real; then $z+i$=$a+(b+1)i$, $(z+i)/2i$ = $(b+1)/2-ai/2$, $Im(z+i)/2i=-a/2$, and $\left(Im(z+i)/2i \leq 0\right) \equiv \left(-a/2\leq 0\right)\equiv \left(a\geq 0\right)$.

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Thanks Steven! Good point you made about being able to 'ignore' the real part of the expression. – Derrick May 23 '12 at 23:49

$$a(a-2) + b^2 \geq 0\\a^2 - 2a + 1 + b^2 \geq 1\\ (a-1)^2 + b^2 \geq 1$$ This is the region outside a unit circle centered at $(1,0)$.

For the second one, $\text{Im} \left( \dfrac{x+iy + i}{2i} \right) = \text{Im} \left( \dfrac{y+1}{2} -i \dfrac{x}{2} \right) = -\dfrac{x}{2} \leq 0$. This means that $x \geq 0$. Hence, the region is the half plan to the right of the imaginary axis i.e. complex of numbers whose real part is non-negative.

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