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May 23, 2019, 07:35:23 am

### AuthorTopic: Locus of Arg(z1) - Arg(z2) = 0 EXPLAIN PLS  (Read 312 times) Tweet Share

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#### borismone

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• Respect: 0 ##### Locus of Arg(z1) - Arg(z2) = 0 EXPLAIN PLS
« on: December 05, 2018, 04:30:36 pm »
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Hi there im really struggling to understand why the locus of Arg(z- 2i) - Arg(z+3) = 0 doesn't include that bit between 2i and -3.
The same goes for the other case where it = 180 degreees
PLEASSEEE EXPLAINUR BOY GOT AN EXAM ON FRIDAY AND IS GONNA DIEEE THANK U VERY MUCH
Diagram of answer attached:

#### VanillaRice

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• Respect: +266 ##### Re: Locus of Arg(z1) - Arg(z2) = 0 EXPLAIN PLS
« Reply #1 on: December 05, 2018, 05:44:06 pm »
+2
Hey there,

This is one of the graph types which is best solved logically (rather than algebraically).
You can rearrange the equation as
Arg(z - 2i) = Arg(z - (-3))
That is, we want the locus of points such that if we take the angle between any point on that locus and z = -3, and then take the angle between the same point and z = 2i, the angle will be the same. Try this for yourself - pick any point along the solid lines (in your attached diagram), and you will see that the angle between that point, and z = -3 and z = 2i are the same. Now, try pick a point along the dotted line. You will see that the angle between that point and z = -3 will be exactly 180 degrees different to the angle between that point and z = 2i. As a matter of fact, this point satisfies the case Arg(z- 2i) - Arg(z+3) = 180 degrees!

Hope that helps - post if you're still stuck « Last Edit: December 05, 2018, 05:45:57 pm by VanillaRice »
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#### borismone

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« Reply #2 on: December 05, 2018, 06:39:44 pm »
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I tried the case for the solid line and it worked but i also tried the case for the dotted line and it did too??
The line is 2/3x + 2 so i subbed in -1 + 4/3i to both arguments and inversed tan and i got 33 degrees for both of them?
Probably did some algebra wrong but anyways, why does this work? I dont see how when u take the angle from the line in the first quadrant and then subtract the angle with the line in the 3rd quadrant you get 0? Wouldnt you get a small positive angle - a bigger negative angle = -ve angle?

#### RuiAce

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• Respect: +2229 ##### Re: Locus of Arg(z1) - Arg(z2) = 0 EXPLAIN PLS
« Reply #3 on: December 05, 2018, 06:49:53 pm »
+5
Normally I'd put more depth into my answer but I'm extremely busy so I'm only going to provide a sketch solution.

Firstly, ensure you understand the locus of $\arg \left( \frac{z-z_1}{z-z_2} \right) = \theta$ when $\theta$ is NOT equal to $0$ or $\pi$. You can use this video by Eddie Woo to start thinking about it.

After you've understood it, you'll note that $\theta$ is the angle that the two line segments make on the arc of the circle. What happens as $\theta$ approaches $\pi$? The answer is that the arc of the circle gradually flattens out. We know that when $\theta=\frac\pi2$ we have a semi-circle, and when $\frac\pi2 < \theta < \pi$ we have a minor-arc. That minor arc has to be stretched further and further out if we keep increasing the angle.

And then what happens when $\theta = \pi$, i.e. it actually reaches $\pi$? Well the two line segments merge into one line segment to form the straight angle. The only way that can happen (in an intuitive sense) is if the arc flattens out into becoming a straight line segment as well.

On the contrary, when $0 < \theta < \frac\pi2$ we know that we have a major arc. What happens as $\theta$ constantly decreases, and also when $\theta\to 0$? That angle being made is becoming smaller and smaller, and that sorta implies that the two line segments are getting closer and closer together. The only way this is possible is if that major arc is being inflated ever so large so that we can actually incorporate such a small angle.

So what do we 'intuitively' expect when $\theta = 0$. That major arc is going to be inflated so much that its limiting behaviour is that essentially it's been popped open. Leaving you with two diametrically opposed rays.

(Note: Why should they be diametrically opposed and not pointing in random angles from each other? This is harder to describe intuitively and relies on you understanding what the limiting gradient-of-the-tangent of the arc at the two endpoints should be.)
« Last Edit: December 05, 2018, 06:56:23 pm by RuiAce »

#### borismone

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« Reply #4 on: December 05, 2018, 07:06:55 pm »
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thank u so much !!! IREEALLLY APPRECIATE THIS