Complex Numbers III: Solving z^5-(z-i)^5=0

[Punch]

The first part of the question asked to find the roots of w^5=1 which I have found to be e^{2k\pi)i}

Hence show that the roots of the equation z^5-(z-i)^5=0, z not equal i, are \frac{1}{2}(cot{\frac({k\pi}{5})+i), where k=-2, -1, 0, 1, 2.

 

[CaptainBlack]

The first part of the question asked to find the roots of \(w^5=1\) which I have found to be \( e^{2k\pi\;i}\)

Those are not the 5-th roots of unity, whatever integer values k takes, they are all the same and =1

You find the 5-th roots of unity by putting:

\[w^5=1=e^{2\pi k\;i},\ k=0,\pm 1, ...\]

so:

\[ w=e^{ \frac{2 \pi k \; i}{5} },\ k= 0, \pm 1, ...\]

and any set of 5 consecutive values of k will give the 5 distinct roots of unity, so:\(w=e^{\frac{2\pi k\;i}{5}},\ k=-2,-1, 0,1,2 \)

Hence show that the roots of the equation \( z^5-(z-i)^5=0, z \) not equal \(i\), are \(\frac{1}{2} \left(cot\left(\frac{k\pi}{5}\right)+i\right)\), where \( k=-2, -1, 0, 1, 2.\)

Since either one of the purported roots is infinite oR with a different guess at where the brackets are supposed to be \( z^5-(z-i)^5=0, \) is a quartic and so has exactly 4 complex roots, but you list 5 distinct roots.

 

[Punch]

Those are not the 5-th roots of unity, whatever integer values k takes, they are all the same and =1

You find the 5-th roots of unity by putting:

\[w^5=1=e^{2\pi k\;i},\ k=0,\pm 1, ...\]

so:

\[ w=e^{ \frac{2 \pi k \; i}{5} },\ k= 0, \pm 1, ...\]

and any set of 5 consecutive values of k will give the 5 distinct roots of unity, so:\(w=e^{\frac{2\pi k\;i}{5}},\ k=-2,-1, 0,1,2 \)


Since either one of the purported roots is infinite oR with a different guess at where the brackets are supposed to be \( z^5-(z-i)^5=0, \) is a quartic and so has exactly 4 complex roots, but you list 5 distinct roots.

I understood your answer to the first quote. However, I didn't understand your answer to the second quote. I have checked the question and indeed, 5 distinct roots are listed.

 

[HallsofIvy]

$(z- i)^5= z^5- 5iz^4+ 10i^2z^3- 10i^3z^2+ 5i^4z- i^5= z^5- 5iz^4- 10z^3+ 10iz^2+ 5z- 1$
so that $z^5- (z- i)^5= z^5- (z^5- 5iz^4+ 10i^2z^3- 10i^3z^2+ 5i^4z- i^5= z^5- 5iz^4- 10z^3+ 10iz^2+ 5z- 1)= 5iz^4+ 10z^3- 10iz^20- 5z+ 1= 0$
That's a fourth degree equation and has 4 roots.

 

[CellCoree]

To solve this equation, we can use the fact that the roots of w^5=1 are given by w=e^{2k\pi i}, where k=0,1,2,3,4. We can then substitute this into the original equation to get:

z^5-(z-i)^5=0
(z-e^{2k\pi i})^5-(z-i-e^{2k\pi i})^5=0
(z^5-5ze^{2k\pi i}^4+10z^3e^{2k\pi i}^3-10z^2e^{2k\pi i}^2+5ze^{2k\pi i}-e^{2k\pi i})-(z^5-5z^4e^{2k\pi i}+10z^3e^{4k\pi i}-10z^2e^{6k\pi i}+5ze^{8k\pi i}-e^{10k\pi i})=0

Simplifying this, we get:

-5ze^{2k\pi i}^4+10z^3e^{2k\pi i}^3-10z^2e^{2k\pi i}^2+5ze^{2k\pi i}=0

We can then factor out a z and e^{2k\pi i} from this equation to get:

z(e^{2k\pi i}-1)(-5e^{2k\pi i}+10z^2+10z+5)=0

From this, we can see that the solutions for z are given by:

z=0 or e^{2k\pi i}-1=0 or -5e^{2k\pi i}+10z^2+10z+5=0

The first solution, z=0, is not valid since it is equal to i, which is not allowed in the original equation. To find the remaining solutions, we can solve the quadratic equation -5e^{2k\pi i}+10z^2+10z+5=0, which gives us:

z=\frac{-5\pm\sqrt{5^2-4(10)(5)}}{2(10)}=\frac{-5\pm i\sqrt{5}}{20}

S