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.
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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 💜
Where actually is the bash you just posted the thm's right?
ReplyDeleteI realised Bashing is bad :)
DeleteNo 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)
Deletelol!! 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@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.
DeleteAlso complex bash is tough :(