# complex numbers III

#### Punch

##### New member
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

##### Well-known member
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.

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#### Punch

##### New member
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

##### Well-known member
MHB Math Helper
$(z- i)^5= z^5- 5iz^4+ 10i^2z^3- 10i^3z^2+ 5i^4z- i^5= z^5- 5iz^4- 10z^3+ 10iz^2+ 5z- 1$
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.