Complex, but not complicated
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- เผยแพร่เมื่อ 15 ก.ย. 2024
- Here I solve a neat complex equation, both using algebra and geometry. Come and see why math is so beautiful!
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Lol I thought of the geometric way first and got the correct answer almost instantly. Goes to show how powerful our geometric reasoning is
I did the same. I love the complex plane. Whenever there's occasion for it I try to teach my students (American) a little bit about it but unfortunately it is almost completely absent from the high school curriculum.
@@toaster4693 u dont have basics of complex nos in high school? wat?
@@toaster4693 In my precal and algebra 2 class we covered complex numbers, but not deeply
@@divyanshaggarwal6243 In Algebra 2 the curriculum covers intro to imaginary numbers (basically arithmetic and powers of i) but usually almost nothing with the complex plane. The only time in high school they get a glimpse of it is being taught DeMoivres theorem but I've found high school teachers almost never explain why it works, they just teach it as a formula to memorize (I suspect they don't know much about the complex plane themselves).
@@toaster4693 Thats sad. I was always captivated by how many difficult problems could be solved inside the head by visualising the complex plane. Questions where algebraic methods give solution after many pages of calculations, geometric method can give answer in a few lines. Its a shame that high school students dont get to see this beauty.
Different way for the algebraic approach: divide both sides by |z-i|. This is okay because the only way for this to be zero is if z = i, but the RHS wouldn't be zero. Join the two absolute values, you get |(z+i)/(z-i)| = 1, so (z+i)/(z-i) = u, a complex number on the unit circle. Solving for z (easy because it's a Mobius transformation) gives z = i(u+1)/(u-1). Let r be a square root of u, and divide top and bottom by it, giving i(r+r^-1)/(r-r^-1). Divide again by 2, and take the i to the denominator of the denominator (no, I didn't mistakenly repeat myself). r can be written as exp(2pi*i*t) where t is real, plug that in, you see you have the complex formula for cos(t) on the top, and sin(t) at the bottom, and so this whole thing is z = cot(t), and the range of the cotangent function is the whole real line.
Fun fact about that Mobius transformation above: it gives a bijection between the Poincare half-plane and Poincare disk models of the hyperbolic plane.
3:54 "time for the geometric way" **sound effect of going through a portal to hell**
yeah sounds about right for my geometry skills
That was so cool!!! The geometric solution was so helpful.
What a fantastic geometric solution, it made me jump out of my seat. Thank you very much.
FWIW: Perpendicular bisector = Mittelsenkrechte in German ;-)
Thanks for such a great content with love from India
Simple problem, great explanation. Happy new year Dr. Peyam!
Hello friend! If you trace two perp. lines at i and -i, then you draw a circle with r generical at 0,0... it is seen the distance from the origin to the intersections means that the imaginary parts are equal. Happy new year!
The symmetry of the functional equation of the Riemann zeta function is due to the invariant form s=(1-s) so points satisfying the equation are equidistant to the critical line Re(s)=1/2. In particular, when they lie on this line they form a complex conjugate pair (having the same magnitude or modulus, of course).
Fun fact, Riemann made his famous conjecture about the location of the non trivial zeros of zeta using a slightly modified version of the invariant equation and mapping the critical line to coincide with the real axis...Happy holidays!
This remember me the type of sequences in topology.
this is a beautiful solution!!
4:26-4:38 Glitch in the matrix...they changed something...
Yes a Médiatrice, very sexy when it's you saying that though
Excelent , prof Peyam. Congratulations from Brazil.
My geometric interpretation: z represents some point in the complex plane. Adding i means moving up one unit, subtracting i means moving down one unit. Absolute value is the distance to origin. The question is therefore looking for all points that have the same distance to origin after being translated up or down one unit. So for the point z, you must be able to draw a circle around the origin, such that a parallel to the Im axis intersects the circle and z, and the two intersected points are exactly two units apart, with z in the middle of them. Or in other words, any point z is a solution, if it is the center of a two-unit line connecting two points of a circle around the origin, and the line is parallel to the Im axis. So we can hang our two-unit line on any point of the circle, but the only place where that line will connect to the another point of the circle is where the center of the line is on the real axis (as long as r >= 1, otherwise no point will be suitable). Therefore, only the real numbers are solutions to this question.
1:45 Why not multiply out (y+1)² = (y-1)² to obtain y²+2y+1 = y²-2y+1, cancel like terms, and use 2y = -2y to get y=0?
He use a shortcut
You could also have , from (y-1)^2 = (y+1)^2 --> (y-1)^2 - (y+1)^2 = 0 so 2y = 0 too.
Could this geometric solution be generalized? Suppose Q and W are each complex numbers. Also suppose they each have a real part that is a (possibly unique) function of X. For example:
Q = 5X + 2 + 6i,
W = 8X - 3 - 9i
Would the solution set of |Q| = |W| still be the perpendicular bisector of segment QW?
I was going to criticize your use of the term "absolute value" rather than "modulus", but since you later spoke correctly, I suppose I'll let it go... this time.😁
Neat problem indeed great video
Pagal
tameez seekho aur zubaan ko lagaam do
@@umerfarooq4831 👌👌
By your geometric argument actually you could find for ANY two complex numbers such a line where |z-a|=|z-b| z= xm+n for m,n complex number solutions and x is a real free variable
n is the average of a and b, and m is any complex number with an angle perpendicular to the line connecting a,b... trivially: i(a-b). So we get z= i(a-b)x + (a+b)/2 should work. To test it: | iax-ibx+a/2+b/2 -a | - | iax-ibx+a/2+b/2 -b | = | iax-ibx+(b-a)/2 | - | iax-ibx+(a-b)/2 |... we can factor an |a-b| from both to get | 1/2+ix | - | 1/2-ix | which is always zero as you showed.
Yes
Actually a brilliant video, especially the geometric way!
Beautiful as always
The most beautiful matrix equation
e^(π𝕚)+1⋅eye(2)=0
where,
eye(2)=[1,0;0,1]% the identity matrix in ℝ² ̽²
𝕚=[0,-1;1,0];
or, which points have the property such that, if you move up or move down by the same amount, you end up the same distance away from the origin
"By construction"
I tried a geometric method but instead taking the distances of z+i and z-i from the origin and I got a longer (though quite similar) solution 😅
"i" caramba.
I liked the geometric way
Nice video
05:07 Try to pronounce this in Polish:
"symetralna" ;)
To i tak nie takie straszne słowo w porównaniu do całego polskiego :))
Very nice! 😃
Are you french ? Because your pronouncation of "médiatrice" is perfect. Great video !
Merci! :) I went to a French school
Please solve more problems on complex no's☺️
Elegant.
Bold words by Dr Peyam
nice prononciation of mediatrice hehe
Before watching the video: My calculations show that z must be real.
After: Yay, I did it!
Square both sides first: |z - i|^2 = |z + i|^2
Remember, for all complex z: |z|^2 = z z*
(z - i)(z* + i) = (z + i)(z* - i)
z z* - i z* + i z + 1 = z z* + i z* - i z + 1
- i z* + i z = i z* - i z
2i z = 2i z*
z = z*
Thus, z is real.
"Mediatriz" in spanish
Probably taken from french
Update: from Late Latin term «mediātrix» or «mediātrīcis» which means "mediator"
Update: from Late Latin term «mediātrix» or «mediātrīcis» actually, which means "mediator"
👍👍
Please prove that,
|z1 + z2 |< or = |z1| + |z2| , where z1 and z2 are any two complex numbers .
Can be proved by cauchy schwartz inequality
3:29 I don't get that | x + i | = sqr( x^2 + (1)^2 ). I thought that | x + i | = sqr( x^2 + (i)^2 ) = sqr( x^2 *-* 1 ). You say it's sqr ( x^2 *+* 1 ). Why do yo say that (i)^2 is equal to (1)^2? The number one is not equal to the number i.
| a + b | = a^2 + b^2 (Pythagoras)
(a + b)^2 = a^2 + 2ab + b^2
(a - b)(a + b) = a^2 - b^2
No |a+bi| = sqrt(a^2 + b^2), you’re confusing it with something else
@@drpeyam Aha. That way. Thanks.
Isnt z just 0? | i | = | -i |
More than that
Math makes people going cringe