# Complex numbers HELPP pLEASE (1 Viewer)

#### pencilspanker

##### Member
Hey guys im struggling with a few concepts on complex numbers and just wondering if anyone could clear them up because im legit abt to neck. I know a lot of basic facts but I don't fully understand how they work which is really bugging me... Ok so, for the question:
i)
I get that you go arg(z1-z2) - arg (z1+z2) =90o
But what im struggling to understand is why the angle in between those two diagonals in that parallelogram is going to be 90. Like im trying to manipulate the angles that are forming with the positive x axis but i just dont get it...

ii) Why can we just randomly say that the vector in the form z - (+z0) will always be the vector running from z0 to z?

iii) If i have the vector --> OC = z, is the vector ---> CO = -z???? And if this is true, how does this work because isnt it that when you put a negative out the front you are multiplying by i2 which means you rotate it by 180o. And if that is true do we call that 180 rotation the position vector whilst the vector ---> CO is just the free vector with the exact same magnitude and direction? Here is a retarded diagram to explain exactly what I mean.

iv) help with this q plsss

THANK U VERY MUCH TO ANYONE THAT HELPS I REALLLLY FREAKING APPRECIATE IT!!!

#### fan96

##### 617 pages
i)

$\bg_white \frac{z_1 - z_2}{z_1 + z_2} = ki$

$\bg_white z_1 - z_2 = ki(z_1 + z_2)\quad ( z_1 + z_2 \neq 0)$

What does this mean?

It means that the vector $\bg_white z_1 - z_2$ is perpendicular to $\bg_white z_1 + z_2$. That is basically the condition given to you.

$\bg_white z_1 - z_2$ and $\bg_white z_1 + z_2$ are the diagonals of the parallelogram with vertices $\bg_white 0, \, z_1, \, z_2, \, (z_1+z_2)$.

But, a parallelogram with perpendicular diagonals is a rhombus.

ii)

iv)

b)

What is $\bg_white \tan \alpha/2$ ?

(It doesn't matter which point is $\bg_white z_1$, although this will be considered later.)

c)

$\bg_white z_2 = z_1 ( \cos \alpha + i \sin \alpha) \quad {\rm OR}\quad z_1 = z_2 ( \cos \alpha + i \sin \alpha)$.

But, $\bg_white z_1+ z_2 = 2i (z_1 -z_2)$ implies that $\bg_white z_1$ is always in the clockwise direction of $\bg_white z_2$.

$\bg_white \therefore z_2 = z_1 ( \cos \alpha + i \sin \alpha)$.

$\bg_white \tan \frac \alpha 2 = \frac 1 2 \implies \tan \alpha = \frac 43$

What are $\bg_white \cos \alpha$ and $\bg_white \sin \alpha$?

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#### fan96

##### 617 pages
Also the "identity"

$\bg_white \arg\left(\frac z w\right) = \arg z - \arg w$

isn't actually true because $\bg_white \arg z - \arg w$ isn't guaranteed to be in the range $\bg_white [\pi, \, -\pi)$.

If $\bg_white \arg z - \arg w$ was instead expressed as the principal value for that angle then I believe it would be true.

So just be careful when using it - if possible find another method.

#### peter ringout

##### New Member
Hey guys im struggling with a few concepts on complex numbers and just wondering if anyone could clear them up because im legit abt to neck. I know a lot of basic facts but I don't fully understand how they work which is really bugging me... Ok so, for the question:
i) View attachment 34699
I get that you go arg(z1-z2) - arg (z1+z2) =90o
But what im struggling to understand is why the angle in between those two diagonals in that parallelogram is going to be 90. Like im trying to manipulate the angles that are forming with the positive x axis but i just dont get it...

ii) Why can we just randomly say that the vector in the form z - (+z0) will always be the vector running from z0 to z?

iii) If i have the vector --> OC = z, is the vector ---> CO = -z???? And if this is true, how does this work because isnt it that when you put a negative out the front you are multiplying by i2 which means you rotate it by 180o. And if that is true do we call that 180 rotation the position vector whilst the vector ---> CO is just the free vector with the exact same magnitude and direction? Here is a retarded diagram to explain exactly what I mean.View attachment 34700

iv) help with this q plsss View attachment 34701

THANK U VERY MUCH TO ANYONE THAT HELPS I REALLLLY FREAKING APPRECIATE IT!!!

It's always nice to use geometry but it is unfortunate that the question is so ham-fisted. Parts a) and b) are heavy machinery to get to a trivial result. A simple direct algebraic approach for c) is:

c) Divide top and bottom by z1 and let w=z2/z1. Then (1+w)/(1-w)=2i. Just solve for w.

Still you must answer the question on the page!

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#### fan96

##### 617 pages
but it is unfortunate that double angle formulae became necessary.
Why is that unfortunate?

#### peter ringout

##### New Member
Why is that unfortunate?
A poorly written question is always unfortunate. When it comes to writing complex analysis problems I always feel that geometry should be forced to provide elegant proofs for deep results, not difficult proofs for trivial results. Presumably part c) is the point of the question and unfortunately the examiner chose a very tortuous path for the candidates. Once you have c), a) and b) are trivial if you really want them.

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#### pencilspanker

##### Member
Forum) do you then translate that green vector to start at w and run to z? So that means it will have the same magnitude and direction but just won’t represent the exact point z-w. It is a free vector?

#### fan96

##### 617 pages
Forum) do you then translate that green vector to start at w and run to z? So that means it will have the same magnitude and direction but just won’t represent the exact point z-w. It is a free vector?
The green vector and the vector from $\bg_white w$ to $\bg_white z$ are the same vector. The position doesn't matter.

#### pencilspanker

##### Member
The green vector and the vector from $\bg_white w$ to $\bg_white z$ are the same vector. The position doesn't matter.
Right I see. I semi get it now, this might sound retarded, but how can we just magically translate the vectors to wherever we want and say that they represent the same thing?

#### fan96

##### 617 pages
Right I see. I semi get it now, this might sound retarded, but how can we just magically translate the vectors to wherever we want and say that they represent the same thing?
Vectors are defined by two things:

- a magnitude e.g. $\bg_white |z|$

- a direction e.g. $\bg_white \arg z$

So if two vectors have the same magnitude and direction then they are equal.

#### HeroWise

##### Active Member
Well it depends on what you are doing. In maths and Physics we see it as same and equal even if they move around but they must have same magnitude and direction. But smthn like in CS, Vectors are rather for storing information, so a translated vector would be different to the original vector

#### aa180

##### Member
An alternative approach for part (i) can be done as follows:

\bg_white \begin{align*} \frac{z_{1}-z_{2}}{z_{1}+z_{2}} &= ki \text{ } (1) \\[0.4cm] \Longrightarrow \overline{(\frac{z_{1}-z_{2}}{z_{1}+z_{2}})} &= -ki \\[0.4cm] \Longrightarrow \frac{\overline{z_{1}}-\overline{z_{2}}}{\overline{z_{1}}+\overline{z_{2}}} &= -ki \text{ } (2) \\[0.4cm] (1) + (2): \\[0.4cm] \frac{z_{1}-z_{2}}{z_{1}+z_{2}} + \frac{\overline{z_{1}}-\overline{z_{2}}}{\overline{z_{1}}+\overline{z_{2}}} &= 0 \\[0.4cm] \Longrightarrow (z_{1}-z_{2})(\overline{z_{1}}+\overline{z_{2}}) + (\overline{z_{1}} - \overline{z_{2}})(z_{1}+z_{2}) &= 0 \\[0.2cm] \Longrightarrow z_{1}\overline{z_{1}} + z_{1}\overline{z_{2}} - \overline{z_{1}}z_{2} - z_{2}\overline{z_{2}} + z_{1}\overline{z_{1}} + \overline{z_{1}}z_{2} - z_{1}\overline{z_{2}} - z_{2}\overline{z_{2}} &= 0 \\[0.2cm] \Longrightarrow 2|z_{1}|^{2} - 2|z_{2}|^{2} &= 0 \\[0.2cm] \therefore |z_{1}| &= |z_{2}| \end{align*}

#### Skuxxgolfer

##### New Member
\left|z+4\right|+2\left|z-3\right|=15

can someone solve this for me please