Complex numbers (1 Viewer)

sunmoonlight · Oct 21, 2023

If z = x+iy, how to prove ii?

Life'sHard · Oct 21, 2023

Been a while since I've done this
$\begin{align*}\frac{|z|+z}{|z|+\bar{z}}\cdot\frac{|z|+z}{|z|+z} &=\frac{|z|^2+z^2+2z|z|}{|z|^2+z\bar{z}+z|z|-\bar{z}|z|} \\ &=\frac{z(z+z|z|)+|z|^2}{|z|(2z|z|+(z+\bar{z}))} \\ &=\frac{z(z+2z|z|+\frac{|z|^2}{z})}{|z|(2z|z|+(z+\bar{z}))} \\ &=\frac{z((z+\bar{z})+2z|z|)}{|z|(2z|z|+(z+\bar{z}))} \\ &= \frac{z}{|z|}\end{align*}$

scaryshark09 · Oct 25, 2023

There's a much easier and more elegant solution if you just multiply by z/z instead.

Screen Shot 2023-10-25 at 8.30.03 pm.png

scaryshark09 · Oct 25, 2023

also i think you made a mistake with your second line of working
2z|z| doesn't equal z^2|z|

sunmoonlight · Oct 26, 2023

How about this question?

Screenshot 2023-10-26 at 12.17.09 am.png

scaryshark09 · Oct 26, 2023

here are both the solutions i wrote up

Screen Shot 2023-10-26 at 12.46.30 am.png

Screen Shot 2023-10-26 at 12.53.33 am.png

note that:
2cosθ = exp(-iθ) + exp(iθ)
2isinθ = exp(iθ) - exp(-iθ). <=> -2isinθ = exp(-iθ) - exp(iθ)
these can be proven by expanding the RHS

synthesisFR · Oct 26, 2023

scaryshark09 said:
here are both the solutions i wrote up
View attachment 41111 View attachment 41114
note that:
2cosθ = exp(-iθ) + exp(iθ)
2isinθ = exp(iθ) - exp(-iθ). <=> -2isinθ = exp(-iθ) - exp(iθ)
these can be proven by expanding the RHS

Why did u become smart after the hsc ended
Is this a cannon event
Was jimmy also like this????!!

scaryshark09 · Oct 26, 2023

synthesisFR said:
Why did u become smart after the hsc ended
Is this a cannon event
Was jimmy also like this????!!

ik right
its just cause there is so much pressure that surrounds the hsc, and now that im done, i can actually do math without any stress and its way more fun than it already was.
like i was generally pretty good throughout the year, i just stuffed up both the exams that mattered most (3u and 4u HSCs)

HazzRat · Nov 3, 2023

Can someone plz explain the answer of this for me:

I've got the worked solution here for the answer. I just don't get how they're equal. So using an example that lies on the curve, say for arg(z + i) the number for it is 3 + 2i. Then if arg(z - 1) would be 2 + i. But arg(3 + 2i) doesn't = arg(2 + i). Please explain what I'm doing wrong.

kendricklamarlover101 · Nov 3, 2023

HazzRat said:
Can someone plz explain the answer of this for me:
View attachment 41373

I've got the worked solution here for the answer. I just don't get how they're equal. So using an example that lies on the curve, say for arg(z + i) the number for it is 3 + 2i. Then if arg(z - 1) would be 2 + i. But arg(3 + 2i) doesn't = arg(2 + i). Please explain what I'm doing wrong.
View attachment 41374

the difference in arguments would be zero (just do lhs-rhs) so they are collinear. if z is in between 1 and -1, the argument is pi not 0 (you can see this by either testing values or drawing it out).

kendricklamarlover101 · Nov 3, 2023

HazzRat said:
Can someone plz explain the answer of this for me:
View attachment 41373

I've got the worked solution here for the answer. I just don't get how they're equal. So using an example that lies on the curve, say for arg(z + i) the number for it is 3 + 2i. Then if arg(z - 1) would be 2 + i. But arg(3 + 2i) doesn't = arg(2 + i). Please explain what I'm doing wrong.
View attachment 41374

also about your z=3+i example, 3+i does not lie on the curve. the cartesian equation would be y=x-1 and (3,1) does not lie on this line.

HazzRat · Nov 3, 2023

kendricklamarlover101 said:
also about your z=3+i example, 3+i does not lie on the curve. the cartesian equation would be y=x-1 and (3,1) does not lie on this line.

ok that makes sense. It's just a collinear line with the exception of 0 < x < 1.

Luukas.2 · Nov 4, 2023

HazzRat said:
Can someone plz explain the answer of this for me:
View attachment 41373

I've got the worked solution here for the answer. I just don't get how they're equal. So using an example that lies on the curve, say for arg(z + i) the number for it is 3 + 2i. Then if arg(z - 1) would be 2 + i. But arg(3 + 2i) doesn't = arg(2 + i). Please explain what I'm doing wrong.
View attachment 41374

The vector approach is definitely the way to go here, but we can derive it with algebra:

$\begin{align*} \arg{(z + i)} &= \arg{(z - 1)} \\ \arg{(z + i)} - \arg{(z - 1)} &= 0 \\ \arg{\left(\frac{z + i}{z - 1}\right)} &= 0 \qquad \qquad \text{(1)} \\ \\ \text{Taking $z = x + iy$, consider:} \qquad Z &= \frac{z + i}{z - 1} \\ &= \frac{x + iy + i}{x + iy - 1} \\ &= \frac{x + i(y + 1)}{(x - 1) + iy} \times \frac{(x - 1) - iy}{(x - 1) - iy} \\ &= \frac{x(x - 1) - i^2y(y + 1) + i[(y + 1)(x - 1) - xy]}{(x - 1)^2 - i^2y^2} \\ &= \frac{\left(x^2 - x + y^2 + y\right) + i\left(xy + x - y - 1 - xy\right)}{(x - 1)^2 + y^2} \\ &= \frac{\left(x^2 - x + y^2 + y\right) + i\left(x - y - 1\right)}{(x - 1)^2 + y^2} \end{align*}$

Now, a complex number $Z$ has a argument of zero if, and only if, $Z \in \mathbb{R}^+$ .

$\begin{align*} \text{Consider, in (1), that $\arg{(Z)} = 0$ requires (firstly) that} \qquad \Im{(Z)} &= 0 \\ \frac{x - y - 1}{(x - 1)^2 + y^2} &= 0 \\ y &= x - 1 \qquad \text{provided $(x - 1)^2 + y^2 \neq 0$, that is, $z \neq 1$} \end{align*}$

$\begin{align*} \text{Consider, further, that $\Re{(Z)} > 0$ and so, from (1):} \qquad \frac{x^2 - x + y^2 + y}{(x - 1)^2 + y^2} &> 0 \\ \left(x - \cfrac{1}{2}\right)^2 + \left(y + \cfrac{1}{2}\right)^2 - \cfrac{1}{2} &> 0 \qquad \text{provided $(x - 1)^2 + y^2$, that is, $z \neq 1$} \\ \left(x - \cfrac{1}{2}\right)^2 + \left(y + \cfrac{1}{2}\right)^2 &> \cfrac{1}{2} \end{align*}$

In other words, the required locus lies along the line $y = x - 1$ , but is required to lie outside the circle centred at $z = \frac{1}{2}(1 - i)$ and with radius of $\frac{1}{\sqrt{2}}$ . This second requirement excludes the values where $0 \leq \Re{(z)} \leq 1$ and $0 \leq \Im{(z)} \leq 1$ as the points inside this circle correspond to when $\Re{(Z)} < 0$ and where $\arg{(z + i)} - \arg{(z - 1)} = \pm\pi$ .

HazzRat · Nov 4, 2023

Luukas.2 said:
The vector approach is definitely the way to go here, but we can derive it with algebra:

$\begin{align*} \arg{(z + i)} &= \arg{(z - 1)} \\ \arg{(z + i)} - \arg{(z - 1)} &= 0 \\ \arg{\left(\frac{z + i}{z - 1}\right)} &= 0 \qquad \qquad \text{(1)} \\ \\ \text{Taking $z = x + iy$, consider:} \qquad Z &= \frac{z + i}{z - 1} \\ &= \frac{x + iy + i}{x + iy - 1} \\ &= \frac{x + i(y + 1)}{(x - 1) + iy} \times \frac{(x - 1) - iy}{(x - 1) - iy} \\ &= \frac{x(x - 1) - i^2y(y + 1) + i[(y + 1)(x - 1) - xy]}{(x - 1)^2 - i^2y^2} \\ &= \frac{\left(x^2 - x + y^2 + y\right) + i\left(xy + x - y - 1 - xy\right)}{(x - 1)^2 + y^2} \\ &= \frac{\left(x^2 - x + y^2 + y\right) + i\left(x - y - 1\right)}{(x - 1)^2 + y^2} \end{align*}$

Now, a complex number $Z$ has a argument of zero if, and only if, $Z \in \mathbb{R}^+$ .

$\begin{align*} \text{Consider, in (1), that $\arg{(Z)} = 0$ requires (firstly) that} \qquad \Im{(Z)} &= 0 \\ \frac{x - y - 1}{(x - 1)^2 + y^2} &= 0 \\ y &= x - 1 \qquad \text{provided $(x - 1)^2 + y^2 \neq 0$, that is, $z \neq 1$} \end{align*}$

$\begin{align*} \text{Consider, further, that $\Re{(Z)} > 0$ and so, from (1):} \qquad \frac{x^2 - x + y^2 + y}{(x - 1)^2 + y^2} &> 0 \\ \left(x - \cfrac{1}{2}\right)^2 + \left(y + \cfrac{1}{2}\right)^2 - \cfrac{1}{2} &> 0 \qquad \text{provided $(x - 1)^2 + y^2$, that is, $z \neq 1$} \\ \left(x - \cfrac{1}{2}\right)^2 + \left(y + \cfrac{1}{2}\right)^2 &> \cfrac{1}{2} \end{align*}$

In other words, the required locus lies along the line $y = x - 1$ , but is required to lie outside the circle centred at $z = \frac{1}{2}(1 - i)$ and with radius of $\frac{1}{\sqrt{2}}$ . This second requirement excludes the values where $0 \leq \Re{(z)} \leq 1$ and $0 \leq \Im{(z)} \leq 1$ as the points inside this circle correspond to when $\Re{(Z)} < 0$ and where $\arg{(z + i)} - \arg{(z - 1)} = \pm\pi$ .

tysm. But yeah, I think if this was a test I'd solve this geometrically.

Luukas.2 · Nov 4, 2023

HazzRat said:
tysm. But yeah, I think if this was a test I'd solve this geometrically.

Absolutely, the intention is to approach the problem geometrically. However, getting the answer algebraically is better than not getting the answer at all. Further, understanding why the restriction exists and how it emerges from the algebra is a useful skill in constructing solid proofs.

Complex numbers (1 Viewer)

sunmoonlight

New Member

Life'sHard

Well-Known Member

scaryshark09

∞∆ who let 'em cook dis long ∆∞

scaryshark09

∞∆ who let 'em cook dis long ∆∞

sunmoonlight

New Member

scaryshark09

∞∆ who let 'em cook dis long ∆∞

synthesisFR

afterhscivemostlybeentrollingdonttakeitsrsly

scaryshark09

∞∆ who let 'em cook dis long ∆∞

HazzRat

H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪

kendricklamarlover101

Member

kendricklamarlover101

Member

HazzRat

H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪

Luukas.2

Well-Known Member

HazzRat

H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪

Luukas.2

Well-Known Member

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