Complex number from a region After sketching this region$$-4\sqrt{2}\le\operatorname{Re} z\le0, $$ $$\operatorname{Im} z\ge0,$$ $$|z|\ge8$$ I need the polar form and the rectangular form of the complex number that is in the region which has the smallest possible imaginary part.
I have sketched the figure of the region but what do I do now?
 A: Let's look step-by-step at this region.
First, you constrain your region to $-4\sqrt{2} \le \text{Re}\ z \le 0$. This is a vertical band between $x = -4\sqrt{2}$ and $0$.
Then, $\text{Im}\ z \ge 0$ further restricts this band to being only in the upper half-plane.
So now we have a half-infinite band sitting on the real axis.
$|z| \ge 8$ says that any such number must be outside the open disc of radius 8 centered at the origin. So this disc chops off a bit of the rectangular region described above.
We know that along the arc $|z| = 8$ in the upper-left quadrant, $\text{Im}\ z$ attains its maximum at $\text{Re}\ z = 0$, and is monotonically decreasing as $\text{Re} z$ becomes "more negative". So, it should be clear that the number with minimal imaginary part will like on the line $\text{Re}\ z = -4\sqrt{2}$.
Let $z = -4\sqrt{2}+i y$. Since we know that the number will be on the arc $|z| = 8$, then we have $64 = |z|^2 = z\overline{z} = (-4\sqrt{2})^2 + y^2 = 32+y^2$, so $y^2 = 32$, and $y = \pm 4\sqrt{2}$.
But since the number is in the upper half-plane, we can exclude $y = -4\sqrt{2}$. Thus, $y = 4\sqrt{2}$, so $z = 4\sqrt{2}(-1+i)$, and $\text{Arg}\ z = \text{Arg}\ (-1+i)$. This should be more than enough work to reach the conclusion.
