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- Thread starter astj
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Average Boreduser
Rising Renewal
shouldnt there also be another restriction on ur first region qn?
um I don't think the q. says anything?
Average Boreduser
Rising Renewal
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
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So, the direction of the vector from
Combined with the restriction that
Correct but a little careless, in that the circle and line cross at exactly
Average Boreduser
Rising Renewal
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)?
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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 wrongim just blurting out crap that might help (gonna be honest, haven't touched loci since I finished complex
So, @astj, the third part is asking you to find the point in the locus that is furthest fromim just blurting out crap that might help (gonna be honest, haven't touched loci since I finished complex
This point is
Oops, you wanted min... So, you have:So, @astj, the third part is asking you to find the point in the locus that is furthest from, as
, which refers to the length of the vector from
.
This point is, so the answer is
.
Average Boreduser
Rising Renewal
Isn't the lenght from -i to -2i smaller though for z? sorry if i sound stupidOops, you wanted min... So, you have:
ie
Oops x 2, that's wrong too.Oops, you wanted min... So, you have:
The closest point in the locus will be on the interval from
I'm assuming that the
If it was to the region where
Average Boreduser
Rising Renewal
Oh right lmfao mbmb. so then its point-line distance formula for perp distance from origin?I'm assuming that the
If it was to the region where, the minimum value of
would be zero, occurring when
.
@ISAM77... have you noticed this question / thread, as it touches on what we were discussing. Taking the question as written, the locus of where
} \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
 \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
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 is any complex number that lies in the locus
It has been shown above that
Prove that there are infinitely many solutions of the equation
when
and exactly one solution for the equation when
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".
is actually not the answer given above, because that answer (which is the one intended, I am sure), is actually to the question
Some teachers will explain this by saying that you need to take
---
For everyone, as an illustration of how the proof topic can be used in a variety of ways, consider this extension.
Suppose that
It has been shown above that
Prove that there are infinitely many solutions of the equation
and exactly one solution for the equation