/tech/ - Soyence and Technology

ask claude or something

Actual snca holy shit nobody cares about maths

>Actual snca holy shit nobody cares about maths
fuck you

>>29358 (OP)
maybr back when i took complex analysis. forgot everything now cause snca

>>29358 (OP)
i dont know how to count

geg im studying this atm

nigga use de moivre's formula and the formula for the roots of unity
do you even read your notes or not nigger

>>29358 (OP)
Mathchan captcha aah question

>>29537
It died in 2024 or early 2025 btw ;_;

hi just visiting not an aryan intellectual

a is 0, 2pi/5, 4pi/5, 6pi/5, 8pi/5
cos theta + i sin theta is r=1, angle=theta in polar coordinates: the point on the unit circle with angle theta (bc cos^2 t + sin^2 t = 1)
Complex multiplication adds the angles and multiplies the radii, so the fifth roots of 1 are 1/5 around the unit circle, 2/5 around the unit circle, etc. When you take the fifth power they end up 0 around the unit circle

Now I'm not sure how you get b algebraically (update post if you find out), but I did it kind of geometrically?

We know (z+1)/z needs to have radius 1 and angle 2kpi/5 because those are the solutions to the fifth root of 1. Because it has radius 1, z+1 needs to have the same radius as z. Since their imaginary parts (y coordinates) are the same, they need to have opposite real parts: Re(z + 1) = - Re(z). We're flipping z around the y axis. Thus Re(z)=-1/2 and Re(z+1)=1/2.

Because the angle of (z+1)/z must be 2kpi/5, the difference in their angles must be 2kpi/5. However, we already discussed how adding 1 to z flips it around the y axis. This means <(z+1)=pi/2-(<(z)-pi/2)=pi-<(z). Thus <(z+1)-<(z) = pi-2<(z)= 2kpi/5, so <(z)=(5-2k)pi/10. Since Re(z) is negative, <(z) is strictly between pi/2 and 3pi/2, so it must be 7pi/10, 9pi/10, 11pi/10, 13pi/10.

This means that the valid solutions to z are the points of intersection of x=-1/2 and the lines through the origin at those angles, y/x=-2y=tan(<(z)). If we look at the triangles of those lines and the y axis instead, we can also describe them as y/x=cot(<(z)-pi), or y=Im(z)=-1/2 * cot(n*pi/10) where n is 2, 4, 6, or 8. Simplifying, this is -1/2 * cot(k*pi/5) where k is 1, 2, 3, or 4. Thus we have found the imaginary part and the answer is -1/2(1+ i cot(k*pi/5))

wait actually i think i messed something up
-x/y = tan(<(z)-pi/2) so cot(<(z)-pi/2)= y/-x
but that would make it -1/2(1-icot(k*pi/5))

>>29571
i will shove triangles up your ass :)