If the equation (z+1)7+z7=0 has roots z1,z2,⋯z7 then what is the value of ∑r=17R(Zr)?

Answered

2022-01-17

If the equation (z+1)7+z7=0 has roots z1,z2,z7
then what is the value of r=17R(Zr)?

Answer & Explanation

nick1337

nick1337

Expert

2022-01-17Added 573 answers

Step 1
I would choose a different approach, saving some labour:
The polynomial:
p(z)=(z+1)7+z7=2z7+7z6+=2(z7+72z6+)
And may also be written as:
p(z)=2q(z)=2k=17(zzk)
The sum of the roots of the monomial q(z) is equal to minus the coefficient of z6, as can be seen by expanding q(z)
The sum of the roots of q(z) equals the sum of roots of p(z):
Re(k=17zk)=(k=17zk)=72

star233

star233

Expert

2022-01-17Added 238 answers

Step 1
You can observe that z=0 is not a solution, so the equation can be rewritten as
(1+1z)7=1
Hence we get
1+1zk=wk
for 1k7, where wk=exp((π+2kπ)i7)
Write αk=π+2kπ14, so we have
zk=1e2βki1=eβki2isinβk
and therefore
zk=i2sinβk(cosβkisinβk)
and we obtain Rezk=12

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