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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?

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What you are asking makes no sense. Complex numbers do not have "small" or "large" values - they are not ordered like real numbers. –  Ron Gordon Apr 23 '13 at 14:30
... the complex number that is in the area which has the smallest possible value. How do you define "small" for complex numbers? Small modulus? –  rschwieb Apr 23 '13 at 14:30
@rschwieb: if the OP were asking that, then he already answered his own question. No? –  Ron Gordon Apr 23 '13 at 14:35
@RonGordon It's impossible to tell from the broken English. He has certainly not provided both solution sets in polar and rectangular form, so I should say there is no answer here. –  rschwieb Apr 23 '13 at 14:37
@OP: I have edited your question. I hope the meaning is what you intended. If not, I apologize. –  Shahab Apr 23 '13 at 14:44

1 Answer 1

up vote 1 down vote accepted

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.

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How do I write the answer in polarform? –  user1838781 May 2 '13 at 13:33

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