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

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)=2∏k=17(z−zk) 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

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 1≤k≤7, where wk=exp⁡((π+2kπ)i7) Write αk=π+2kπ14, so we have zk=1e2βki−1=e−βki2isin⁡βk and therefore zk=−i2sin⁡βk(cos⁡βk−isin⁡βk) and we obtain Rezk=12

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)=2∏k=17(z−zk) 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

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 1≤k≤7, where wk=exp⁡((π+2kπ)i7) Write αk=π+2kπ14, so we have zk=1e2βki−1=e−βki2isin⁡βk and therefore zk=−i2sin⁡βk(cos⁡βk−isin⁡βk) and we obtain Rezk=12

Answered: "If the equation (z+1)7+z7=0 has roots z1,z2,⋯z7 then..."

Answered question

2022-01-17

If the equation

{(z + 1)}^{7} + z^{7} = 0

has roots

z_{1}, z_{2}, \dots z_{7}

then what is the value of

\sum_{r = 1}^{7} R (Z_{r}) ?

Answer & Explanation

nick1337

Expert2022-01-17Added 777 answers

Step 1
I would choose a different approach, saving some labour:
The polynomial:
$p (z) = (z + 1)^{7} + z^{7} = 2 z^{7} + 7 z^{6} + \dots = 2 (z^{7} + \frac{7}{2 z^{6}} + \dots)$
And may also be written as:
$p (z) = 2 q (z) = 2 \prod_{k = 1}^{7} (z - z_{k})$
The sum of the roots of the monomial $q (z)$ is equal to minus the coefficient of $z^{6}$ , as can be seen by expanding $q (z)$
The sum of the roots of $q (z)$ equals the sum of roots of $p (z)$ :
$R e (\sum_{k = 1}^{7} z_{k}) = (\sum_{k = 1}^{7} z_{k}) = - \frac{7}{2}$

star233

Skilled2022-01-17Added 403 answers

Step 1
You can observe that z=0 is not a solution, so the equation can be rewritten as
$(1 + \frac{1}{z})^{7} = - 1$
Hence we get
$1 + \frac{1}{z_{k}} = w_{k}$
for $1 \leq k \leq 7$ , where $w_{k} = \exp ((π + 2 k π) \frac{i}{7})$
Write $α_{k} = \frac{π + 2 k π}{14}$ , so we have
$z_{k} = \frac{1}{e^{2 β_{k} i} - 1} = \frac{e^{- β_{k} i}}{2 i \sin β_{k}}$
and therefore
$z_{k} = - \frac{i}{2 \sin β_{k}} (\cos β_{k} - i \sin β_{k})$
and we obtain $R e z_{k} = \frac{1}{2}$

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