Cambridge EXT 2 - Chapter 1 | Q+As (1 Viewer)

ISAM77 · Jan 15, 2024

Hi guys, I have a few questions about this chapter. Please help, if you can. And if you have questions of your own from this chapter of the cambridge book, pls post them here and I'll try help as well.

My questions:
1F Q12.a) After drawing the graph and looking at the answers, it's clear that the maximum for |z| is the horizontal distance from the origin to z=3. However, how can we know that this horizontal distance is longer than all possible distances from the origin to the circle.

1G Q12.b)ii: I solved this with long division. Is there another way? (using the fact that (x-2i)(x+2i) = x^2 + 4)

1F Q13.a)ii: After drawing the graph and comparing it to what the question asks, it's obvious that the maximum value for arg(z) is when x=1.5 which is midway between the first root, and the maximum y value. Is there a circle geometry rule about the maximum angle from a point to a circle? I doubt it. Can anyone share there proof to this question pls.

note: the original questions + my working is attached below.

Luukas.2 · Jan 15, 2024

For the first one...

From any point outside a circle, the closest and furthest points must lie on the line through the external point and the centre of the circle.

Draw a circle, centred at O, and an external point P. Join OP, which crosses the circle at C. PO produced meets the circle at F. If X is any point on the circle (other than C or F), PX must be longer than PC and shorter than PF, making C and F the closest and furthest points (respectively) from P.

Luukas.2 · Jan 15, 2024

For 12G, you could say that

$P(x) = x^3 + x^2 + 6x - 3 = \left(x^2 + 4\right)Q(x) + R(x)$

where $R(x) = Ax + B$ and solve for $A$ and $B$ using

$P(2i) = R(2i) = 2Ai + B \qquad \text{and} \qquad P(-2i) = R(-2i) = -2Ai + B$

as you know $P(2i)$ and $P(-2i)$ from the earlier parts.

Luukas.2 · Jan 15, 2024

For the last one, the max and min values of arg z occur when the ray for each is a tangent to the circle, and so perpendicular to the radius at the point of contact. Use right-angled triangle trigonometry on the resulting triangles.

Luukas.2 · Jan 15, 2024

In any triangle, the longest side is opposite the largest angle.

Or... draw the circle with the red line as radius, centred where all the coloured lines meet. This circle and the one you have drawn touch with a common tangent at the end of your red line, and nowhere else... yet, your coloured lines must cross the new circle if they are longer than the red line. Thus, as they don't, they must be shorter.

ISAM77 · Jan 15, 2024

Luukas.2 said:
For the first one...

From any point outside a circle, the closest and furthest points must lie on the line through the external point and the centre of the circle.

Got it. Is this year 10 circle geometry?

Luukas.2 said:
For 12G, you could say that

Got it! thankyou.

Luukas.2 said:
For the last one, the max and min values of arg z occur when the ray for each is a tangent to the circle, and so perpendicular to the radius at the point of contact. Use right-angled triangle trigonometry on the resulting triangles.

too easy. I'm embarrassed I didn't see this one, thanks for your time.

Luukas.2 said:
In any triangle, the longest side is opposite the largest angle.

Or...

I constructed triangles by connecting the different colours to red, however, it's difficult to tell which angle is the largest.

The circles with a touching tangent visualisation worked better for me.

---

Appreciate all the help! Thankyou.

ISAM77 · Jan 17, 2024

Here's a question and solution.

Originally, I only considered when arg(z) is +ve (or when Im(z) > 0). After checking the answers and some thought, here is my solution.

My way of thinking is sufficient to answer the question, however, it feels strange. I was wondering if anyone has any thoughts on this question, or has anything to add so that it doesn't feel so fuzzy to me. Maybe something to do with circles? As arg((z-z1)/(z-z2)) = theta becomes the arc of a circle.

If not, I suppose it's just a new type of question I'm being exposed to, and to look out for in future.

Luukas.2 · Jan 17, 2024

Interpreting $\arg{(z - i)} = \arg{(z + 1)}$ in vector terms, the equation means:

the direction of the vector from $i$ to $z$ is the same as the direction from $-1$ to $z$

This includes every point on the line through $i$ and $-1$ EXCEPT FOR:
(a) those points themselves, as they lead to a term $\arg{0}$ , which is undefined; and,
(b) the points on the line that are between those points, as then the directions of the two vectors are opposite and so the arguments differ by $\pi$ .

Hence, the locus are the two rays from the line through $z = i$ to $z = -1$ , but excluding those points and continuing to infinity in each direction.

If you solve the problem algebraically by setting $z = x + iy$ , you should get $y = x + 1$ on the domain $x \in (-\infty,\ -1) \cup (0,\ \infty)$ .

If you want to check that the result does make sense, try some values:

$\begin{align*} \text{Test $z = 1 + 2i$:} \qquad \text{LHS} &= \arg{(z - i)} \\ &= \arg{(1 + 2i - i)} \\ &= \arg{(1 + i) }\\ &= \frac{\pi}{4} \\ \\ \text{RHS} &= \arg{(z + 1)} \\ &= \arg{(1 + 2i + 1)} \\ &= \arg{(2 + 2i)} \\ &= \frac{\pi}{4} \\ &= \text{LHS} \qquad \implies \qquad z = 1 + 2i\ \text{does lie in the locus} \end{align*}$

$\begin{align*} \text{Test $z = -\frac{1}{2} + \frac{i}{2}$:} \qquad \text{LHS} &= \arg{(z - i)} \\ &= \arg{\left(-\frac{1}{2} + \frac{i}{2} - i\right)} \\ &= \arg{\left(-\frac{1}{2} - \frac{i}{2}\right)}\\ &= \frac{-3\pi}{4} \\ \\ \text{RHS} &= \arg{(z + 1)} \\ &= \arg{\left(-\frac{1}{2} + \frac{i}{2} + 1\right)} \\ &= \arg{\left(\frac{1}{2} + \frac{i}{2}\right)} \\ &= \frac{\pi}{4} \\ &\neq \text{LHS} \qquad \implies \qquad z = -\frac{1}{2} + \frac{i}{2}\ \text{does NOT lie in the locus} \end{align*}$

$\begin{align*} \text{Test $z = -2 - i$:} \qquad \text{LHS} &= \arg{(z - i)} \\ &= \arg{(-2 - i - i)} \\ &= \arg{(-2 - 2i) }\\ &= \frac{-3\pi}{4} \\ \\ \text{RHS} &= \arg{(z + 1)} \\ &= \arg{(-2 - i + 1)} \\ &= \arg{(-1 - i)} \\ &= \frac{-3\pi}{4} \\ &= \text{LHS} \qquad \implies \qquad z = -2 - i\ \text{does lie in the locus} \end{align*}$

ISAM77 · Jan 17, 2024

Trying to solve algebraically feels garbage. (or I'm garbage. one or the other)

The visual explanation makes it very easy though.

Luukas.2 · Jan 17, 2024

ISAM77 said:
Trying to solve algebraically feels garbage. (or I'm garbage. one or the other)

The visual explanation makes it very easy though.

That algebra is not correct.

However, I agree with you that this sort of problem is much easier to do geometrically.

The problem with algebraic solutions to problems like these is spotting the limitations / exclusions.

Consider, for example, the following working for this problem, which does produce the equation for the locus... can you see what is wrong, or missing, or problematic, or needs changing, or ...?

$\begin{align*} \arg{(z - i)} &= \arg{(z + 1)} \\ \arg{(x + iy - i)} &= \arg{(x + iy + 1)} \qquad \text{on putting $z = x + iy$ where $x$, $y \in \mathbb{R}$} \\ \arg{[x + i(y - 1)]} &= \arg{[(x + 1) + iy]} \\ \tan^{-1}{\left(\frac{y - 1}{x}\right)} &= \tan^{-1}{\left(\frac{y}{x + 1}\right)} \\ \frac{y - 1}{x} &= \frac{y}{x + 1} \\ (x + 1)(y - 1) &= xy \\ xy + y - x - 1 &= xy \\ y &= x + 1 \end{align*}$

ISAM77 · Jan 17, 2024

The locus equation should be y=x-1 (not y=x+1), right? Your algebra checks out, but it doesn't give the correct equation of the locus.

My consideration for what's problematic would be the value of inverse tan as it isn't clear what quadrant each complex number is in.

Luukas.2 · Jan 17, 2024

ISAM77 said:
The locus equation should be y=x-1 (not y=x+1), right? Your algebra checks out, but it doesn't give the correct equation of the locus.

My consideration for what's problematic would be the value of inverse tan as it isn't clear what quadrant each complex number is in.

Oops... your question was $\arg{(z + i)} = \arg{(z - 1)}$ . I have done a different question.

You are right that one of the issues is in quadrants, because inverse tan can only return a value between -pi/2 and pi/2.

So, with that restriction recognised, in my working, the LHS is restricted to $x \ge 0$ and the RHS is restricted to $x \ge -1$ . So, making that correction, we now have the locus as $y = x + 1$ for $x \ge 0$ . We need a separate set of working for solutions where $x \le 0$ . We also need to figure out why we need to exclude $x = 0$ .

Luukas.2 · Jan 18, 2024

A related question...

In addition to figuring out the answer, you might give equations for each of the graphs shown.

ISAM77 · Jan 18, 2024

Sweet as. Please confirm my answers

The equation for each graph:
A. Arg(z-i) = Arg(z)+pi
B. Arg(z-i) = Arg(z)
C. Arg(z) = Arg(i)
D. |z| = |z-i|

Luukas.2 · Jan 18, 2024

ISAM77 said:
Sweet as. Please confirm my answers

The equation for each graph:
A. Arg(z-i) = Arg(z)+pi
B. Arg(z-i) = Arg(z)
C. Arg(z) = Arg(i)
D. |z| = |z-i|

3/4... (A) is not correct.

ISAM77 · Jan 18, 2024

Luukas.2 said:
3/4... (A) is not correct.

What's the equation for (A)?

Luukas.2 · Jan 18, 2024

ISAM77 said:
What's the equation for (A)?

How about this... if I help you see why your answer is wrong, maybe you can see what an answer should be... (I say "an" answer as there are more than one possible equations...)

You have said that the equation for the locus of $\left\{z = iy \in \mathbb{C}:\ 0 < y < 1,\ y \in \mathbb{R}\right\}$ is $\text{Arg$(z - i) =$ Arg$(z) + \pi$$ :

$\begin{align*} \text{Test $z = \frac{i}{2}$:} \qquad \text{LHS} &= \text{Arg}(z - i) \\ &= \text{Arg}\left(\frac{i}{2} - i\right) \\ &= \text{Arg}\left(-\frac{i}{2}\right) \\ &= -\frac{\pi}{2} \\ \\ \text{RHS} &= \text{Arg}(z) + \pi \\ &= \text{Arg}\left(\frac{i}{2}\right) + \pi \\ &= \frac{\pi}{2} + \pi \\ &= \frac{3\pi}{2} \\ &\neq \text{LHS} \qquad \implies \qquad z = \frac{i}{2}\ \text{is}\ \textbf{NOT}\ \text{in the locus} \end{align*}$

ISAM77 · Jan 19, 2024

Luukas.2 said:
How about this... if I help you see why your answer is wrong, maybe you can see what an answer should be...

Interesting... Is the equation Arg(z-i) = Arg(z) - pi ?

I had thought +pi or -pi didn't matter because they both ended up at the same place. Since the Complex plane is restricted between -pi and pi, I thought a result of 3pi/2 simply means the same as -pi/2. I simply saw +pi as a rotation. However, I do remember the Cambridge textbook specifying: -pi<= arg (z) <= pi , so it makes sense.

Maths is extremely accurate, and there is no guessing. Sometimes, my logic isn't specific or accurate enough and it leads to mistakes and misunderstandings. I'm glad I joined this forum to help me out. Appreciate your effort to help

Luukas.2 · Jan 19, 2024

ISAM77 said:
Interesting... Is the equation Arg(z-i) = Arg(z) - pi ?

I had thought +pi or -pi didn't matter because they both ended up at the same place. Since the Complex plane is restricted between -pi and pi, I thought a result of 3pi/2 simply means the same as -pi/2. I simply saw +pi as a rotation. However, I do remember the Cambridge textbook specifying: -pi<= arg (z) <= pi , so it makes sense.

Maths is extremely accurate, and there is no guessing. Sometimes, my logic isn't specific or accurate enough and it leads to mistakes and misunderstandings. I'm glad I joined this forum to help me out. Appreciate your effort to help

$\text{Arg}(z - i) = \text{Arg}(z) - \pi$

is, indeed, one possible correct answer. Two other possible answers are

$\arg{(z - i)} = \arg{(z)} - \pi$

and

$\arg{(z - i)} = \arg{(z)} + \pi$

The difference is the difference in the meaning of "arg" and "Arg".

Both $\arg{z}$ and $\arg{(z)}$ refer to the direction of the vector from the origin to $z$ . As you note, because angles can be equivalent under rotation, the statements $\arg{(-1)} = \pi$ and $\arg{(-1)} = 3\pi$ are equivalent... and there are infinitely many other values of $\arg{(-1)}.$ .

Hence, for the question that I have posed, it doesn't actually matter which pair of vectors differ by $\pi$ , as:

$\arg{(z - i)] = \arg{(z)} + \pi \qquad \implies \qquad \arg{\left(\frac{z - i}{z}\right) = \pi$

and

$\arg{(z - i)] = \arg{(z)} - \pi \qquad \implies \qquad \arg{\left(\frac{z}{z - i}\right) = \pi$

However, with a capital "A", $\text{Arg}(z)$ refers to the principal argument of the vector from the origin to $z$ , which is explicitly required to fall within the range $\text{Arg}(z) \in (-\pi,\ \pi]$ .

So, had you written your answer with lower case "a"'s, I would have said it was correct... but with a capital "A" and being restricted to principal arguments, the equation that you wrote actually has no solutions.

There are situations where restricting the domain of an angle to a single rotation is important. One classic case (of, actually, half a rotation) is seen in switching from trigonometry to inverse trigonometry. The functions $f(x) = \sin{\left(\sin^{-1}{x}\right)}$ and $g(x) = \sin^{-1}{\left(\sin{x}\right)}$ are identical in the domain $-1 \le x \le 1$ but different everywhere else, and that is because there is a strict restriction on the angles that $\sin^{-1}{x}$ that can output, but none on what $\sin{x}$ can accept as input.

We see above, also, that

$\arg{z} = \arg{(x + iy)} = \tan^{-1}{\left(\frac{y}{x}\right)}$

is only true if $z = x + iy$ is in the first or fourth quadrants.

I am sure that some teachers / markers would consider the distinction that I am highlighting here is trivial. There are also some that would consider accepting the answer that you originally gave as absolutely incorrect. Because of the former group, and human fallibility, you will see examples of mistakes in this area in questions and solutions, from time to time. I hope this all makes sense.

ISAM77 · Jan 20, 2024

Luukas.2 said:
$\text{Arg}(z - i) = \text{Arg}(z) - \pi$

is, indeed, one possible correct answer. Two other possible answers are

$\arg{(z - i)} = \arg{(z)} - \pi$

and

$\arg{(z - i)} = \arg{(z)} + \pi$

The difference is the difference in the meaning of "arg" and "Arg".

It does make sense, and I'm thankful for the clarity. I struggle with these subtleties, but I think they are very important and try and have them sink in whenever I come across them.

Cambridge EXT 2 - Chapter 1 | Q+As (1 Viewer)

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