Graphing (1 Viewer)

astj · Jan 19, 2024

I don't get how to graph the left one and find the region for this q.
Thanks in advance

Average Boreduser · Jan 19, 2024

astj said:
I don't get how to graph the left one and find the region for this q.
Thanks in advance

shouldnt there also be another restriction on ur first region qn?

astj · Jan 19, 2024

Average Boreduser said:
shouldnt there also be another restriction on ur first region qn?

um I don't think the q. says anything?

Average Boreduser · Jan 19, 2024

Oh crap nevermind I thought there was a z on the denom. Have u tried cis form and then played aroud w the graph (pref also simplifying inside of arg)?

astj · Jan 19, 2024

Average Boreduser said:
Oh crap nevermind I thought there was a z on the denom. Have u tried cis form and then played aroud w the graph (pref also simplifying inside of arg)?

I ended up w/ this but I'm not sure if it's right and then part iii was to determine the min value of |z+i| which I don't get as well

Luukas.2 · Jan 19, 2024

$\begin{align*} \arg{\left(\frac{z + 2}{\sqrt{3} + i}\right)} &\ge \frac{\pi}{12} \\ \arg{(z + 2)} - \arg{\left(\sqrt{3} + i\right)} &\ge \frac{\pi}{12} \\ \arg{(z + 2)} - \frac{\pi}{6} &\ge \frac{\pi}{12} \qquad \text{as $\arg{\left(\sqrt{3} + i\right)} = \tan^{-1}{\left(\frac{1}{\sqrt{3}}\right)} = \frac{\pi}{6}$} \\ \arg{(z + 2)} &\ge \frac{\pi}{12} + \frac{\pi}{6} = \frac{3\pi}{12} = \frac{\pi}{4} \end{align*}$

So, the direction of the vector from $-2$ to $z$ is (implicitly) being restricted to

$\frac{\pi}{4} \le \text{Arg}(z + 2) \le \pi$

Combined with the restriction that $|z| \le 2$ , the region is the minor segment of the circle centred at the origin and of radius 2 bordered by the interval from $z = -2$ to $z = 2i$ . All borders are inside the locus except for the point $z = -2$ itself (as it leads to the argument of zero, which is unefined).

Luukas.2 · Jan 19, 2024

astj said:
I ended up w/ this but I'm not sure if it's right and then part iii was to determine the min value of |z+i| which I don't get as well

Correct but a little careless, in that the circle and line cross at exactly $z = 2i$ .

Average Boreduser · Jan 19, 2024

astj said:
I ended up w/ this but I'm not sure if it's right and then part iii was to determine the min value of |z+i| which I don't get as well

im just blurting out crap that might help (gonna be honest, haven't touched loci since I finished complex

, so please take anything I say with a grain of salt ) for part iii can you consider z+i as z-(-i), then observe something useful from that? provided, that you've been given |z|=2 (using vectors)?

astj · Jan 19, 2024

Average Boreduser said:
im just blurting out crap that might help (gonna be honest, haven't touched loci since I finished complex , so please take anything I say with a grain of salt ) for part iii can you consider z+i as z-(-i), then observe something useful from that provided, that you've been given |z|=2 (using vectors)?

Yeah I like started from -i and then I drew a line from it to -2 to find √5 (which looks sorta off) and then for max value I did -i to 2i = 3 but that seems insanely wrong

Luukas.2 · Jan 19, 2024

Average Boreduser said:
im just blurting out crap that might help (gonna be honest, haven't touched loci since I finished complex , so please take anything I say with a grain of salt ) for part iii can you consider z+i as z-(-i), then observe something useful from that? provided, that you've been given |z|=2 (using vectors)?

So, @astj, the third part is asking you to find the point in the locus that is furthest from $-i$ , as $|z + i| = |z - (-i)|$ , which refers to the length of the vector from $-i$ to $z$ .

This point is $z = 2i$ , so the answer is $\text{max}(|z+i|) = |2i + i| = 3$ .

Luukas.2 · Jan 19, 2024

Luukas.2 said:
So, @astj, the third part is asking you to find the point in the locus that is furthest from $-i$ , as $|z + i| = |z - (-i)|$ , which refers to the length of the vector from $-i$ to $z$ .

This point is $z = 2i$ , so the answer is $\text{max}(|z+i|) = |2i + i| = 3$ .

Oops, you wanted min... So, you have:

$\sqrt{5} < |z + i| \le 3$

Average Boreduser · Jan 19, 2024

Luukas.2 said:
Oops, you wanted min... So, you have:

$\sqrt{5} < |z + i| \le 3$

Isn't the lenght from -i to -2i smaller though for z? sorry if i sound stupid
ie

Luukas.2 · Jan 19, 2024

Luukas.2 said:
Oops, you wanted min... So, you have:

$\sqrt{5} < |z + i| \le 3$

Oops x 2, that's wrong too.

The closest point in the locus will be on the interval from $z = -2$ to $z = 2i$ where the line to $z = -i$ is perpendicular to the interval. I think this gives

$\frac{3\sqrt{2}}{2} \le |z + i| \le 3$

Luukas.2 · Jan 19, 2024

Average Boreduser said:
Isn't the lenght from -i to -2i smaller though for z? sorry if i sound stupid
ie
View attachment 42195

I'm assuming that the $z$ must lie in the region established earlier in the question.

If it was to the region where $|z| \le 2$ , the minimum value of $|z + i|$ would be zero, occurring when $z = -i$ .

Average Boreduser · Jan 19, 2024

Luukas.2 said:
I'm assuming that the $z$ must lie in the region established earlier in the question.

If it was to the region where $|z| \le 2$ , the minimum value of $|z + i|$ would be zero, occurring when $z = -i$ .

Oh right lmfao mbmb. so then its point-line distance formula for perp distance from origin?

Luukas.2 · Jan 19, 2024

Average Boreduser said:
Oh right lmfao mbmb. so then its point-line distance formula for perp distance from origin?

Perpendicular distance from $z = -i$ to the interval... I get it as $3\sqrt{2} / 2 = \sqrt{4.5}$ .

Luukas.2 · Jan 21, 2024

@ISAM77... have you noticed this question / thread, as it touches on what we were discussing. Taking the question as written, the locus of $z$ where

$\arg{\left(\frac{z + 2}{\sqrt{3} + i}\right)} \ge \frac{\pi}{12} \qquad \qquad \text{and} \qquad \qquad |z| \le 2$

is actually not the answer given above, because that answer (which is the one intended, I am sure), is actually to the question

$\text{Arg}\left(\frac{z + 2}{\sqrt{3} + i}\right) \ge \frac{\pi}{12} \qquad \qquad \text{and} \qquad \qquad |z| \le 2$

Some teachers will explain this by saying that you need to take $\arg{z}$ to mean / imply the principal argument when not including the restriction leads to an absurd aoutcome, but I see that as an excuse for imprecise question writing.

---

For everyone, as an illustration of how the proof topic can be used in a variety of ways, consider this extension.

Suppose that $z$ is any complex number that lies in the locus

$\text{Arg}\left(\frac{z + 2}{\sqrt{3} + i}\right) \ge \frac{\pi}{12} \qquad \qquad \text{and} \qquad \qquad |z| \le 2$

It has been shown above that

$\frac{3\sqrt{2}}{2} \le |z + i| \le 3$

Prove that there are infinitely many solutions of the equation $|z + i| = k$ when

$\frac{3\sqrt{2}}{2} < k < 3$

and exactly one solution for the equation $|z + i| = k$ when

$k = \frac{3\sqrt{2}}{2} \qquad \text{or} \qquad k = 3.$

There are a variety of approaches that can be taken here, so I am interested in people exploring what ways might be used and how much needs to be said to "prove" these results as opposed to "explain" / "justify" / "show".

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