Solveeit Logo
Login

Question

Question: Let \(Z_{k}(k = 0,1,2,...............6)\) be the roots of the equation \((z + 1)^{7} + z^{7} = 0\), ...

Let Zk(k=0,1,2,...............6)Z_{k}(k = 0,1,2,...............6) be the roots of the equation (z+1)7+z7=0(z + 1)^{7} + z^{7} = 0, then ∑k=06Re(zk)\sum_{k = 0}^{6}{{Re}\left( z_{k} \right)} is equal to

A

3 - 2i

B

0

C

−72- \frac{7}{2}

D

3+2i3 + 2i

Answer

−72- \frac{7}{2}

Explanation

Solution

Sol. Let zk=xk+iyk,z_{k} = x_{k} + iy_{k}, we have (zk+1)7+z7k=0\left( z_{k} + 1 \right)^{7} + {z^{7}}_{k} = 0

⇒ (zk+1)7=−z7k\left( z_{k} + 1 \right)^{7} = - {z^{7}}_{k}

⇒ ∣zk+1∣7+∣zk∣7|z_{k} + 1|^{7} + |z_{k}|^{7}

⇒ ∣zk+1∣=∣zk∣|z_{k} + 1| = |z_{k}|

⇒ ∣xk+iyk+1∣2=∣xk+iyk∣2|x_{k} + iy_{k} + 1|^{2} = |x_{k} + iy_{k}|^{2}

⇒ (xk+1)2+yk2=x2k+y2k\left( x_{k} + 1 \right)^{2} + {y_{k}}^{2} = {x^{2}}_{k} + {y^{2}}_{k}

⇒ 2xk+1=02x_{k} + 1 = 0 or xk=−12x_{k} = - \frac{1}{2}

Thus, ∑k=06Re(zk)=∑k=06xk=−72\sum_{k = 0}^{6}{{Re}\left( z_{k} \right) =}\sum_{k = 0}^{6}{x_{k} = - \frac{7}{2}}