MX2 Marathon (1 Viewer)

Luukas.2 · Dec 20, 2023

Yes, applying the corrections that have been noted above by @scaryshark09, the working should be:

$\begin{align*} \text{LHS} &= x\left(x^{k+2}\right) + (x + 1)^2(x + 1)^{2k + 1} \\ &= x\left(x^{k+2}\right) + \left(x^2 + 2x + 1\right)(x + 1)^{2k + 1} \\ &= x\left(x^{k+2}\right) + \left(x^2 + x + 1\right)(x + 1)^{2k + 1} + x(x + 1)^{2k + 1} \\ &= x\left[x^{k+2} + (x + 1)^{2k + 1}\right] + \left(x^2 + x + 1\right)(x + 1)^{2k + 1} \\ &= x\left[m\left(x^2 + x + 1\right)\right] + \left(x^2 + x + 1\right)(x + 1)^{2k + 1} \qquad \text{using the induction hypothesis $P(k)$} \\ &= \left(x^2 + x + 1\right)\left[mx + (x + 1)^{2k + 1}\right] \\ &= p\left(x^2 + x + 1\right) \qquad \text{where $p = mx + (x + 1)^{2k + 1} \in \mathbb{Z}$ as $m$, $x$, $k \in \mathbb{Z}$} \\ &= \text{RHS} \end{align*}$

HazzRat · Jan 16, 2024

Have I been smoking crack or what is this

carrotsss · Jan 16, 2024

HazzRat said:
Have I been smoking crack or what is this View attachment 42150

e^(-pi/2i) is just 1cis(-pi/2) which is -i, then multiplying that by -2i is pretty simple

Average Boreduser · Jan 16, 2024

HazzRat said:
Have I been smoking crack or what is this View attachment 42150

convert e^...sdjkjfhwkjehgkle to cis form and the i is basically xing cispi/2?

Luukas.2 · Jan 17, 2024

HazzRat said:
Have I been smoking crack or what is this View attachment 42150

Or...

$\begin{align*} -2i\left(e^{-\frac{\pi}{2}i}\right) &= -2 \times 1\left(e^{\frac{\pi}{2}i}\right) \times \left(e^{-\frac{\pi}{2}i}\right) \qquad \text{as $|i| = 1$ and $\arg{i} = \frac{\pi}{2}$} \\ &= -2\left(e^0\right) \\ &= -2 \times 1 \\ &= -2 \end{align*}$

Or...

$\begin{align*} -2i\left(e^{-\frac{\pi}{2}i}\right) &= 2 \times 1\left(e^{-\frac{\pi}{2}i}\right) \times \left(e^{-\frac{\pi}{2}i}\right) \qquad \text{as $|-i| = 1$ and $\arg{(-i)} = -\frac{\pi}{2}$} \\ &= 2\left(e^{-\frac{\pi}{2}i}\right)^2 \\ &= 2\left(e^{-i\pi}\right)\\ &= \frac{2}{e^{i\pi}} \\ &= \frac{2}{-1} \qquad \text{as $e^{i\pi} = -1$ is the Euler identity} \\ &= -2 \end{align*}$

HazzRat · Jan 17, 2024

Luukas.2 said:
Or...

$\begin{align*} -2i\left(e^{-\frac{\pi}{2}i}\right) &= -2 \times 1\left(e^{\frac{\pi}{2}i}\right) \times \left(e^{-\frac{\pi}{2}i}\right) \qquad \text{as $|i| = 1$ and $\arg{i} = \frac{\pi}{2}$} \\ &= -2\left(e^0\right) \\ &= -2 \times 1 \\ &= -2 \end{align*}$

Or...

$\begin{align*} -2i\left(e^{-\frac{\pi}{2}i}\right) &= 2 \times 1\left(e^{-\frac{\pi}{2}i}\right) \times \left(e^{-\frac{\pi}{2}i}\right) \qquad \text{as $|-i| = 1$ and $\arg{(-i)} = -\frac{\pi}{2}$} \\ &= 2\left(e^{-\frac{\pi}{2}i}\right)^2 \\ &= 2\left(e^{-i\pi}\right)\\ &= \frac{2}{e^{i\pi}} \\ &= \frac{2}{-1} \qquad \text{as $e^{i\pi} = -1$ is the Euler identity} \\ &= -2 \end{align*}$

In the end it was just cuz I didn't know eulers formula

Luukas.2 · Jan 17, 2024

HazzRat said:
In the end it was just cuz I didn't know eulers formula

Cool... I was just showing some other approaches

HazzRat · Jan 18, 2024

Does anyone have a cheat sheet for proving shapes in complex numbers? Whenever I'm given a question like "prove these complex points form a parallelogram" I never know how to prove it and the answer's always smthn random like "the diagonals bisect each other". So is there a method of knowing the proof for each shape?

Average Boreduser · Jan 18, 2024

HazzRat said:
Does anyone have a cheat sheet for proving shapes in complex numbers? Whenever I'm given a question like "prove these complex points form a parallelogram" I never know how to prove it and the answer's always smthn random like "the diagonals bisect each other". So is there a method of knowing the proof for each shape?

just look up properties lol. no way of getting around that

liamkk112 · Jan 18, 2024

HazzRat said:
Does anyone have a cheat sheet for proving shapes in complex numbers? Whenever I'm given a question like "prove these complex points form a parallelogram" I never know how to prove it and the answer's always smthn random like "the diagonals bisect each other". So is there a method of knowing the proof for each shape?

u just got to memorise the quadrilateral properties no way around it

usually though:
- parallelogram -> pairs of equal side lengths, parallel sides
- square -> equal side lengths, 90 degrees between sides, parallel sides
- rectangle -> 90 degrees between sides, parallel sides
- rhombus -> pairs of equal side lengths, parallel sides, diagonals meet at 90 degrees and bisect

there r also kites but i forget how those work and they're relatively uncommon

Luukas.2 · Jan 18, 2024

HazzRat said:
Does anyone have a cheat sheet for proving shapes in complex numbers? Whenever I'm given a question like "prove these complex points form a parallelogram" I never know how to prove it and the answer's always smthn random like "the diagonals bisect each other". So is there a method of knowing the proof for each shape?

There is always a purely algebraic method, which is usually awful. There are sometimes purely geometric methods (like for arg(z - i) = arg(z + 1) etc.). If there isn't an obvious purely geometric approach, the efficient answer is likely to involve:

treating the complex numbers as vectors
looking for geometric properties that proves the required result
demonstrating these properties through algebraic representation of vectors

For example... the complex number z represents a point A in the first quadrant. If O is the origin, B lies in the second quadrant, and OACB is a square, find the complex number representing point C. Under what conditions is C located in the second quadrant.

A diagram should make it obvious that side OB is adjacent to side OA in the square.

Properties of a square then dictate that OB = i.OA, and so the complex number iz represents B.

Then, using vector reasoning:

$\begin{align*} \overrightarrow{OC} &= \overrightarrow{OA} + \overrightarrow{AC} \\ &= \overrightarrow{OA} + \overrightarrow{AB} \qquad \text{as $\overrightarrow{AC} = \overrightarrow{OB}$ as opposite sides of a square} \\ &= z + iz \\ &= z(1 + i) \end{align*}$

Hence, the point C is represented by z(1 + i), and so is in the second quadrant if

$\frac{\pi}{4} < \arg{z} < \frac{\pi}{2}$

from the diagram (as A must be in quadrant 1 and, for C to be in quadrant 2 given angle COA is 45 degrees, OA must be inclined at at least 45 degrees above the real axis), or (algebraically), by solving:

$\begin{align*} \frac{\pi}{2} &< \arg{[z(1 + i)]} < \pi \\ \frac{\pi}{2} &< \arg{z} + \arg{(1 + i)} < \pi \\ \frac{\pi}{2} &< \arg{z} + \frac{\pi}{4} < \pi \\ \frac{\pi}{4} &< \arg{z} < \frac{3\pi}{4} \qquad \qquad \text{. . . (1)} \\ \\ \text{But, as $A$ lies in quadrant 1:} \qquad 0 &< \arg{z} < \frac{\pi}{2} \qquad \qquad \text{. . . (2)} \\ \\ \text{Thus, to satisfy both (1) and (2):} \qquad \frac{\pi}{4} &< \arg{z} < \frac{\pi}{2} \end{align*}$

HazzRat · Jan 24, 2024

I

this question so much

(doesn't need solving. I was a big boy and did it myself)

HazzRat · Jan 24, 2024

Also for complex number revision I rec these booklets: https://drive.google.com/drive/folders/1LBtygQwi9cZtiosWOCodci1_6NjGPdmQ?usp=sharing

(ik many u have already been tested on this but like I've got a test coming up and I found it useful)

HazzRat · Jan 24, 2024

what bs is this. Can someone plz explain to me how these r equal

liamkk112 · Jan 24, 2024

HazzRat said:
what bs is this. Can someone plz explain to be how these r equal
View attachment 42231

if z-1/z+1 is purely imaginary, then the real part is zero, aka that x(x+1) + y(y-1) = 0
=> x^2 + x + y^2 - y = 0
=> x^2 + x + 1/4 + y^2 - y + 1/4 = 1/4 + 1/4 (completing the square)
=> (x+1/2)^2 + (y-1/2)^2 = 1/2
so basically it should be = 1/2, u can see they made a mistake because they noted that radius is 1/sqrt(2) which makes sense if
(x+1/2)^2 + (y-1/2)^2 = 1/2

liamkk112 · Jan 24, 2024

liamkk112 said:
if z-1/z+1 is purely imaginary, then the real part is zero, aka that x(x+1) + y(y-1) = 0
=> x^2 + x + y^2 - y = 0
=> x^2 + x + 1/4 + y^2 - y + 1/4 = 1/4 + 1/4 (completing the square)
=> (x+1/2)^2 + (y-1/2)^2 = 1/2
so basically it should be = 1/2, u can see they made a mistake because they noted that radius is 1/sqrt(2) which makes sense if
(x+1/2)^2 + (y-1/2)^2 = 1/2

also there should be an open circle at (-1, 0) as this would make the denominator 0

Luukas.2 · Jan 25, 2024

liamkk112 said:
also there should be an open circle at (-1, 0) as this would make the denominator 0

And $z = i$ is also impossible, as it makes

$\frac{z - i}{z+1} = \frac{i - i}{i+1} = \frac{0}{i+1} = 0$

which is not purely imaginary - so two open circles, at opposite ends of a diameter.

Luukas.2 · Jan 28, 2024

HazzRat said:
what bs is this. Can someone plz explain to me how these r equal
View attachment 42231

Note that a purely geometric approach would provide the graph, with the points $z = -1$ and $z = i$ excluded, and is quicker:

$\begin{align*} \text{If $\frac{z -i}{z + 1}$ is purely imaginary, then} \quad \frac{z -i}{z + 1} &= ki \qquad \text{for some $k \in \mathbb{R}$, where $k \neq 0$} \\ \arg{\left(\frac{z -i}{z + 1}\right)} &= \pm \frac{\pi}{2} \\ \arg{(z -i)} - \arg{(z + 1)} &= \pm \frac{\pi}{2} \end{align*}$

Geometrically, this statement tells us that the angle between the vector from $-1$ to $z$ and the vector from $i$ to $z$ is a right angle.

Applying the converse of the angle in a semicircle theorem, it follows that $z$ lies on one of the two semi-circles whose diameter is the interval joining $z = -1$ and $z = i$ .

The two end points of the diameter, $z = -1$ and $z = i$ , are excluded from the locus as each results in one of the two vectors being the zero vector, and hence one of the two arguments being $\arg{0}$ and thus undefined.

---

Taking the algebraic approach, the two constraints that lead to points being excluded are that $(x + 1)^2 + y^2 \neq 0$ and $(y - 1)(x + 1) - xy \neq 0$ . The point (-1, 0) violates both constraints, whilst the point (0, 1) violates only the second constraint.

HazzRat · Feb 1, 2024

Can someone plz explain to me the second last line of reasoning? That is: z1^2 + z2^2 = z1^2*(1+a^2). How did they get that?

Lith_30 · Feb 1, 2024

HazzRat said:
Can someone plz explain to me the second last line of reasoning? That is: z1^2 + z2^2 = z1^2*(1+a^2). How did they get that?
View attachment 42294

they said that $z_2=\alpha{z_1}$ so then $z_1^2+z_2^2=z_1^2+(\alpha{z_1})^2=z_1^2(1+\alpha^2)$

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