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Let's complex bash Part 2:P

 Okie so continuing from the previous post ( sorry for huge time gap, got stuck in Allen stuff)

A warning though if anyone from the "anti-bash" community is reading, sorry in advance and R.I.P. 


Notes:-

1. We have $z_1=r_1e^{i\theta_1}$ and $z_2=r_2e^{i\theta_2}$ then we have $z_1z_2=r_1r_2e^{i(\theta_1+\theta_2)}$ and we get $|z_1z_2|=|z_1||z_2|$ and $\arg z_1z_2=\arg z_1+\arg z_2 .$

SPIRAL SIMILARITIES and TRANSFORMATIONS:-

2. How to rotate a point about origin by 90.

  • If $90^{\circ}$ anti-clockwise then multiply the number by $i$ 
  • If $90^{\circ}$ clockwise then multiply by $-i.$
Proof:- Note that $i= 0+1\cdot i.$ So $|i|= \sqrt{0^2+1^2}=1.$ And when we locate $i$ in complex plane, clearly $\arg i=\pi/2.$

Similarly for the second, but note that $\arg -i=-\pi /2.$ (Since angles are measured anti-clockwise)

Example:- Here we have $z=1+2i$ then when we multiply $i$ we get $zi=i-2.$

Now, what about we want to dialate a point from one to another and scale it? (Note that that is spiral similarity only :P)



Like say we want to rotate point $z=4+5i$ about point $x=2+2i$ with dilation of $\pi/2.$

In this case what we do is we shift the whole diagram such that $x$ to $(0,0).$ Here we have $(2,2)\rightarrow (0,0),$ so $z$ goes to $z=(4,5)\rightarrow z'=(4-2,5-2)=(2,3)= 2+3i.$

Now we perform the same thing, so we will multiply $i$ and we get $z''=i-3=(-3,2).$
Then we transform it back, so we get $z''=(-3,2)\rightarrow z'''=(-3+2,2+2)=(-1,4).$



So in summary, what we are doing is shifting the $x$ to $O$ i.e in all the coordinates we subtracting $x$'s coordinates.. and then we get the new image of $z,$ we multiply $i,$ get new image, of $(z-x)$ then add the coordinates back .

So we get $\boxed{z\rightarrow i(z-x)+x.}$

Here $i$'s modulus is one, so there hasn't been a scaling thingy but we can do that too. 

$$z\rightarrow \alpha(z-w)+w$$ 

This is a spiral similarity centrerd at $w$ dilating $z$ which rotates by $\alpha $ and dilates by $|\alpha |.$

COMPLEX REFLECTION:- 

Let $W$ be the reflection of $Z$ over a given $AB.$ Then

$$w=\frac{  (a-b)\overline{z} + \overline{a}b - a\overline{b}}{(\overline{a} - \overline{b})}$$

Proof:- It's in egmo:P.

We first transform $z$ to $z-a$ ( We have the motive to turn $\overline{AB}$ in to the segment between $0$ and $1$ on the real axis, so that we can then apply complex conjugate thing that we have on reflection about real axis)

Then we down everything by $b-a.$

So we get $$\left(\frac{w-a}{b-a}\right) = \overline{\left(\frac{z-a}{b-a}\right)}$$

Using the basic conjugate properties, we get,

$$ \overline{\left(\frac{z-a}{b-a}\right)}=\left(\frac{\bar{z}-\bar {a}}{\bar{b}-\bar{a}}\right)$$

Then calcs done, we get $$w=\frac{  (a-b)\overline{z} + \overline{a}b - a\overline{b}}{(\overline{a} - \overline{b})}$$

Then we get this problem from egmo :P

Problem:- Show that the foot of the altitude from $Z$ to $\overline{AB}$ is given by

$$ \frac{(\overline{a} - \overline{b})z + (a-b)\overline{z} + \overline{a}b - a\overline{b}}{2(\overline{a} - \overline{b})} $$

Proof:-Consider the reflection of $z$ say $w.$ Then we know that the foot ( say $x$) will be the midpoint of $z$ and $w$ i.e $x=\frac{z+w}{2}.$

Now, using the reflection lemma, we get $$w+z=\frac{ (a-b)\overline{z} + \overline{a}b - a\overline{b}}{(\overline{a} - \overline{b})}+ z=\frac{(\overline{a} - \overline{b})z + (a-b)\overline{z} + \overline{a}b - a\overline{b}}{(\overline{a} - \overline{b})} $$

So $$\frac{w+z}{2}=  \frac{(\overline{a} - \overline{b})z + (a-b)\overline{z} + \overline{a}b - a\overline{b}}{2(\overline{a} - \overline{b})}$$

Properties:-
  • $z=\bar{z}$ iff $z$ is real
  • $z+\bar{z}=0 $ iff $z$ is pure imaginary  
perpendicular criterion:- The compleplex numbers $a,b,c,d$ have the property $AB\perp CD$ if and only if
$$\frac{d-c}{b-a} + \overline{\left(\frac{d-c}{b-a}\right)}=0$$

Proof:- We transform $d\rightarrow d-c, c \rightarrow 0.$ Now, this transformation takes line $CD$ to a line parallel to it, passing through origin.

Similarly, we transform $d\rightarrow d-c, c \rightarrow 0.$ Now, this transformation takes line $CD$ to a line parallel to it, passing through origin.

Now, remember in the previous spiral similarity transformation, we noticed that multiplying the complex number with $xi$ transforms line to a perpendicular line. So if $d-c$ and $b-a$ are perpendicular then  $\frac{d-c}{b-a}$ must be pure imaginary.

Using the properties, we get that  $$\frac{d-c}{b-a} + \overline{\left(\frac{d-c}{b-a}\right)}=0.$$

Collinearity Lemma:- Prove that complex numbers $z$, $a$ and $b$ are collinear iff

$$ \frac{z-a}{z-b} = \overline{\left(\frac{z-a}{z-b}\right)} $$

Proof:- We consider the transformation $z-a$ and $z-b.$ These vectors have tails in $O$ and heads as $z-a, z-b$ respectively. Now note that if $z,a,b$ are collinear then these two vectors ( because the transformation is mapping into parallel line through $O$) 
So we have the angles between these two vectors as $0, \pi.$

Hence the two vectors' quotient must be a real number. Using Proposition 6.4, we get that 

$$\frac{z-a}{z-b} + \overline{\left(\frac{z-a}{z-b}\right)}=0.$$

Complex shoelace Formula:- If $a,b,c$ are complex numbers, then the signed area of triangle $ABC$ is given by 





THE UNIT CIRCLE:-

The unit circle is the set of complex numbers $z$ with $|z|=1,$ which is centred at $0$ with radius $1.$

We have for any $z$ on the unit circle, $\boxed{\bar {z}=\frac{1}{z}}.$
This is because $z\cdot \bar{z}=|z|^2.$

----

Yeah now lemme do polynomials :P  

See ya next time! Also do check my other blog (if you get free time tmrw, new post coming there )

Sunaina 💜

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Comments

  1. Bandwala pagol chhagolJanuary 9, 2022 at 1:17 AM

    Where actually is the bash you just posted the thm's right?

    ReplyDelete
    Replies
    1. Sunaina Pati January 9, 2022 at 8:56 AM

      I realised Bashing is bad :)

      Delete
    2. Bandwala pagol chhagolJanuary 16, 2022 at 10:52 PM

      No bashing is the best,also even bashes are beautiful sometimes for eg check Mr Oreo Juice/nathantareep's (shameless advertising)/jj_ca888 solution to Romania TST 2007 Day 6 P2(by Cosmin Pohoata)

      Delete
    3. Sunaina Pati January 22, 2022 at 5:34 PM

      lol!! Now I know who you are :sunglasses:.. I tbh cant bash in an exam condition, so i felt it's not that useful to learn plus I dont even have specific interest! Also Oreo doing bash :O dang he must be high!

      Delete
    4. Bandwala pagol chhagolJanuary 23, 2022 at 4:20 PM

      @bove lol tru :rotfl:!I ve never literally talked to him(maybe bcoz I don't use discord) but I have seen him write that in his posts :P.Yah I am not that good too in bashes.Just learning.

      Also complex bash is tough :(

      Delete

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