Complex Numbers (1 of 6: Solving Harder Complex Numbers Questions) [Student requested problem]

[@player-ne6hs]

I got upto solutions then completely blown when sir draw that graph🤯

[@williamnathanael412]

8:35 why are the nine solutions are either solution of the left factor or the right factor? Why can't any of them be solutions of BOTH factors?

[@XxStuart96xX]

We're solving for the nine 9th roots of unity; we know these to be 9 distinct values. z^3-1 gives the three distinct 3rd roots of unity, which are all also three of the 9th roots of unity. As z^9-1 = (z^3-1)(z^6+z^3+1), it follows the remaining six distinct 9th roots of unity are the roots of the other polynomial.

[@player-ne6hs]

@@XxStuart96xX z^3-1=0 will give us z=1

[@carultch]

@@player-ne6hs z^3 - 1 = 0 only has 1 real solution that is z=1. But it will also have two other solutions in general. z can also equal -1/2 + sqrt(3)/2 * i, and z can also equal -1/2 - sqrt(3)/2 * i.
z^n = 1 in general has n solutions, where n is a positive integer. One of the solutions will be on the positive real axis, while the remaining solutions will be uniformly distributed around the angles of the unit circle on the complex plane.
In polar form, the solution of z^n =1 in general, is:
z = e^(2*pi*k*i/n)
where:
e is Euler's number
i is the imaginary unit
k is an integer from 0 to n-1
n is any positive integer
pi is the circle constant
and
z is a complex number that solves the equation

[@satyavirsingh5855]

really intelligent way

[@winson7232]

which textbook are you getting these questions from?

[@joshuatamer1933]

Fitzpatrick

[@ronniesuazo7817]

how to solve complex equations like this one a+big=-12+7i

[@carultch]

@@ronniesuazo7817 Not sure what you mean by "big".
But if you want to solve:
a + b*i = -12 + 7*i
You simply equate the real sides and get a=-12, and likewise the imaginary sides, and get b*i = 7*i. Since i cannot be zero as it is defined to equal sqrt(-1), this means that b=7.

[@radiobum986]

I am not sure but this question is similar to a problem from AIME 1984. You can find such problems on AOPS

[@ems6964]

how come the firrst solution of unity is always unity itself?:) love ur vids!

[@bhargavgohil25]

Sir please solve this question
Z-i/Z+i where (z not equal to -i) is purely imaginary number then z×z` is equal to

[@carultch]

Question unclear.

[@bhargavgohil25]

Asked when I was preparing for My bachelor's University.
Will be completing my bachelor's this year. 😂

[@Sasukej2004]

@@bhargavgohil25 lmao what r you studying right now? How had the last 5 years been for you?

[@bhargavgohil25]

@@Sasukej2004 it's was quite a roller coaster tbh.
I am currently a software engineer.

[@Sasukej2004]

@@bhargavgohil25 bro im coming into engineering next yr any advice 😅

[@mustefamohammed5004]

Please do this question now
Find the complex number z such that z^5=5-5i

[@soumayaalami7616]

z^5 =5-5i => z^5=5*sqrt(2)*(e^i(-pi/4)) and u find ur z applying the rule of a complex number roots

[@cheofusi3562]

Thanks

[@joeblow4976]

Chr9ne enters chr9ne.....,.

[@farhadbaghirov3156]

OMAE OMOWUE NANI

[@amiyancandol4499]

These are hard?

[@kosingtayeng690]

Aryan Darad darad