# MX2 Marathon (1 Viewer)

#### Luukas.2

##### Well-Known Member
Yes, applying the corrections that have been noted above by @scaryshark09, the working should be:

\bg_white \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

##### H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪
Have I been smoking crack or what is this

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#### Luukas.2

##### Well-Known Member
Have I been smoking crack or what is this View attachment 42150
Or...
\bg_white \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...
\bg_white \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

##### H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪
Or...
\bg_white \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...
\bg_white \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

##### Well-Known Member
In the end it was just cuz I didn't know eulers formula
Cool... I was just showing some other approaches

#### HazzRat

##### H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪
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

##### Rising Renewal
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

##### Well-Known Member
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

##### Well-Known Member
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:
\bg_white \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

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

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:
\bg_white \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

##### H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪
I this question so much

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

#### HazzRat

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

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#### liamkk112

##### Well-Known Member
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

##### Well-Known Member
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

##### Well-Known Member
also there should be an open circle at (-1, 0) as this would make the denominator 0
And $\bg_white z = i$ is also impossible, as it makes
$\bg_white \frac{z - i}{z+1} = \frac{i - i}{i+1} = \frac{0}{i+1} = 0$

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

#### Luukas.2

##### Well-Known Member
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 $\bg_white z = -1$ and $\bg_white z = i$ excluded, and is quicker:

\bg_white \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 $\bg_white -1$ to $\bg_white z$ and the vector from $\bg_white i$ to $\bg_white z$ is a right angle.

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

The two end points of the diameter, $\bg_white z = -1$ and $\bg_white z = i$, 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 $\bg_white \arg{0}$ and thus undefined.

---

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

#### HazzRat

##### H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪
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

##### o_o
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 $\bg_white z_2=\alpha{z_1}$ so then $\bg_white z_1^2+z_2^2=z_1^2+(\alpha{z_1})^2=z_1^2(1+\alpha^2)$