Sydney boys 2023 questions (1 Viewer)

scaryshark09 · Oct 13, 2023

Screen Shot 2023-10-13 at 11.17.31 am.png

how do they go from the first line to the second?

scaryshark09 · Oct 13, 2023

Screen Shot 2023-10-13 at 11.18.56 am.png

how do you do part (iii)?
also im not super confident with the rest, so i would appreciate if someone could go through them all please

liamkk112 · Oct 13, 2023

scaryshark09 said:
View attachment 40452
how do they go from the first line to the second?

as they say its the fundamental theorem of calculus:

its just switching x and t and using t^2 instead of t

liamkk112 · Oct 13, 2023

scaryshark09 said:
View attachment 40453
how do you do part (iii)?
also im not super confident with the rest, so i would appreciate if someone could go through them all please

for i, b>1, meaning that arg(z1) is between pi/4 and pi/2 since when b = 1 the argument is pi/4 (tan inverse 1) and otherwise it increases.
for z2, 0<c<1: so arg(z2) is between 0 and -pi/4 since when c = 0, argument is 0, and when c=1 the argument is -pi/4
u can just pick any points where those conditions are met basically

for ii), u need to pick z3 such that there is aline from z2 to z3 passing through the origin with same modulus of z1 that u drew, there should be two (notice that z2 modulus is less than z1, so u ahve a point in quadrant 2 and quadrant 4, just draw it and it should make more sense)

for iii), essentially it boils down to noticing that the points are collinear. In the case where z3 is in quadrant 4, k is simply -(alpha +beta), because the complex number needs to rotate z1 clockwise to have the same argument as z2, so thats just the argument of z1 + argument of z2, but negated bc clockwise rotation.
in the case where z3 is in quadrant 2; using some trig, in this case z3 is a 180 degree rotation of z2 (collinear, if that makes sense? idk how to say this), and in this case 180 = k + alpha + beta, meaning that k = 180 - alpha - beta.

i think this is correct, if someone could double check

Luukas.2 · Oct 13, 2023

The Fundamental Theorem of Calculus states that

$\frac{d}{dx} \int_a^x f(t)\ \!dt = f(x)$

Applied to this problem:

$\begin{align*} \frac{d}{dt} \int_a^{t^2} xf(x)\ \!dx &= \frac{d}{dt^2} \int_a^{t^2} xf(x)\ \!dx \times \frac{d(t^2)}{dt} \qquad \text{using the Chain Rule} \\ &= t^2f\left(t^2\right) \times 2t \\ &= 2t^3f\left(t^2\right) \end{align*}$

If the above is unclear, you can think of what I am doing here as making a substitution of $u = t^2$ so that the upper boundary matches the form required by the FTC. Then, the Chain Rule can be applied to modifying the derivative to match the new variable: $\frac{d}{dt}$ as $\frac{du}{dt} \times \frac{d}{du} = 2t\frac{d}{du}$ . Hence:

$\begin{align*} \frac{d}{dt} \int_a^{t^2} xf(x)\ \!dx &= \frac{d}{dt} \int_a^u xf(x)\ \!dx \qquad \text{on substituting $u = t^2$} \\ &= \frac{du}{dt} \times \frac{d}{du} \int_a^u xf(x)\ \!dx \qquad \text{using the Chain Rule} \\ &= 2t \times uf(u) \qquad \text{applying the Fundamental Theorem of Calculus} \\ &= 2t \times t^2f\left(t^2\right) \qquad \text{on back-substituting $t^2$ for $u$} \\ &= 2t^3f\left(t^2\right) \end{align*}$

Note that the solution provided has a mistake i the second last line, where it shows $f(t)^2$ but should show $f\left(t^2\right)$ , as it does in the final line of the solution.

Luukas.2 · Oct 13, 2023

liamkk112 said:
for i, b>1, meaning that arg(z1) is between pi/4 and pi/2 since when b = 1 the argument is pi/4 (tan inverse 1) and otherwise it increases.
for z2, 0<c<1: so arg(z2) is between 0 and -pi/4 since when c = 0, argument is 0, and when c=1 the argument is -pi/4
u can just pick any points where those conditions are met basically

for ii), u need to pick z3 such that there is aline from z2 to z3 passing through the origin with same modulus of z1 that u drew, there should be two (notice that z2 modulus is less than z1, so u ahve a point in quadrant 2 and quadrant 4, just draw it and it should make more sense)

for iii), essentially it boils down to noticing that the points are collinear. In the case where z3 is in quadrant 4, k is simply -(alpha +beta), because the complex number needs to rotate z1 clockwise to have the same argument as z2, so thats just the argument of z1 + argument of z2, but negated bc clockwise rotation.
in the case where z3 is in quadrant 2; using some trig, in this case z3 is a 180 degree rotation of z2 (collinear, if that makes sense? idk how to say this), and in this case 180 = k + alpha + beta, meaning that k = 180 - alpha - beta.

i think this is correct, if someone could double check

To help with the drawing, Z₃ lies on a circle with OZ₁ as radius and O as the centre (because of the equal moduli). Z₂ lies in the fourth quadrant and is inside this circle, so after extending OZ₂ in both directions, the two possible locations of Z₃ are the intersections of this extended line with the circle.

scaryshark09 · Oct 13, 2023

liamkk112 said:
for iii), essentially it boils down to noticing that the points are collinear. In the case where z3 is in quadrant 4, k is simply -(alpha +beta), because the complex number needs to rotate z1 clockwise to have the same argument as z2, so thats just the argument of z1 + argument of z2, but negated bc clockwise rotation.
in the case where z3 is in quadrant 2; using some trig, in this case z3 is a 180 degree rotation of z2 (collinear, if that makes sense? idk how to say this), and in this case 180 = k + alpha + beta, meaning that k = 180 - alpha - beta.

i think this is correct, if someone could double check

Screen Shot 2023-10-13 at 11.51.31 am.png

thanks, for part iii i got the same answer as you, but the solutions have this. is it possible the solutions are wrong, or, maybe is it becuase beta is actually negative?

Luukas.2 · Oct 13, 2023

From the definition in the question, $\beta < 0$

liamkk112 · Oct 13, 2023

scaryshark09 said:
View attachment 40455
thanks, for part iii i got the same answer as you, but the solutions have this. is it possible the solutions are wrong, or, maybe is it becuase beta is actually negative?

oh thats maybe why, bc beta represents the argument instead of the actual angle

scaryshark09 · Oct 13, 2023

scaryshark09 said:
View attachment 40453

for part i, would the angle between the positive x-axis and z2 be 'beta' OR '-beta'?

like what is the difference between an argument and principal argumnet?

liamkk112 · Oct 13, 2023

Luukas.2 said:
From the definition in the question, $\beta < 0$

so in this case, would the answer given be correct or wrong?

Luukas.2 · Oct 13, 2023

scaryshark09 said:
for part i, would the angle between the positive x-axis and z2 be 'beta' OR '-beta'?

like what is the difference between an argument and principal argumnet?

Principal Argument is the argument $\in(-\pi,\ \pi]$

And, the acute angle you mention is $-\beta$

scaryshark09 · Oct 13, 2023

liamkk112 said:
so in this case, would the answer given be correct or wrong?

i think it would be correct cause beta just becomes negative

but what is the differnce between an argument and principal argumnet?

scaryshark09 · Oct 13, 2023

Luukas.2 said:
Principal Argument is the argument $\in(-\pi,\ \pi]$

so like how does this chance things?

so like a normal argument of 7pi/4 would be a principal argument of -pi/4 ??

carrotsss · Oct 13, 2023

scaryshark09 said:
so like how does this chance things?

so like a normal argument of 7pi/4 would be a principal argument of -pi/4 ??

yes

liamkk112 · Oct 13, 2023

scaryshark09 said:
so like how does this chance things?

so like a normal argument of 7pi/4 would be a principal argument of -pi/4 ??

yes, basically. all it means is that in this case beta can ONLY be between -pi/4 and 0, instead of between 7pi/4 and 2pi, 15pi/4 and 4pi, etc..

scaryshark09 · Oct 13, 2023

is the prinicple argumant ever the same as the normal argument?

if so, when is it the same?

carrotsss · Oct 13, 2023

scaryshark09 said:
is the prinicple argumant ever the same as the normal argument?

if so, when is it the same?

the normal argument encompasses all values, so between -pi and pi it is the same as the principal argument but otherwise it is not

Luukas.2 · Oct 13, 2023

liamkk112 said:
so in this case, would the answer given be correct or wrong?

The answers given are correct.

The angle between OZ₁ and OZ₂ is $\alpha + -\beta = \alpha - \beta$ . So, to get to the direction of OZ₂ from OZ₁ requires a clockwise rotation by $\alpha - \beta$ and hence is achieved by multiplying by $e^{-i(\alpha - \beta)} = e^{i(\beta - \alpha)}$ .

The angle between OZ₁ and OZ₃ (in quadrant 2, which is the direction of -OZ₂) is $\pi - (\alpha + -\beta) = \pi - \alpha + \beta$ . So, to get to the direction of the second quadrant OZ₃ from OZ₁ requires an anti-clockwise rotation by $\pi + \beta - \alpha$ and hence is achieved by multiplying by $e^{i(\pi + \beta - \alpha)}$ .

Luukas.2 · Oct 13, 2023

scaryshark09 said:
is the prinicple argumant ever the same as the normal argument?

if so, when is it the same?

I can write $1 + i = \sqrt{2}\left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) = \sqrt{2}e^{\frac{i\pi}{4}}$ using the Principal Argument of $1 + i$ , which is $\frac{\pi}{4}$ .

I can express the same number as

$1 + i = \sqrt(2)e^{i\left(\frac{\pi}{4} + 2\pi\right)} = \sqrt(2)e^{i\left(\frac{\pi}{4} - 2\pi\right)} = \sqrt(2)e^{i\left(\frac{\pi}{4} + 2024\pi\right)} = ...$

These are all the same because the argument represents the same angle.

In other words:

$\arg{(1 + i)} = \frac{\pi}{4} + 2k\pi \qquad \text{for any $k \in \mathbb{Z}$}$

and the complex number has infinitely many equally-valid arguments... but only one of them satisfies the domain requirement to be the Principal Argument, which is

$\text{Arg}(1 + i) = \frac{\pi}{4}$

Most of the time, if we report an argument, we will choose the principal argument, but not necessarily always.

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