z-1/2z=7 then prove that z^2+1+4z^2=50
Please answer thid it's urgent!
Answers
Answer:
ye cvh
Step-by-step explanation:
ask again
image not clear
Answer:
Hope it will be helpful to you.
Step-by-step explanation:
Down vote
Accepted
It is well known that zn−1=(z−1)(∑n−1i=0zi). Substituting n=5 gives us z5−1=(z−1)(1+z+z2+z3+z4). Thus, we have the following:
1+z+z2+z3+z4=z5−1z−1 (z≠1)
The right side of the equation is 0 only when the numerator z5−1=0 and z≠1. Thus, we must solve the equation z5=1 when z≠1, which is just the complex fifth roots of unity.
By De Moivre's formula, we know these roots of unity are cos2nπ5+isin2nπ5 for n∈Z. However, if n is 0, then this number equals 1, which can not be a solution. Also, if n<0 or n≥5, then we can take the modulo of n by 5 to find an equivalent solution as adding or subtracting n by 5 simply adds or subtracts the angle by 2π, which does not change the answer. Thus, 1≤n≤4 and the solution set is {cos2nπ5+isin2nπ5:n∈{1,2,3,4}}