Question

Solve the following equation
8 z /3z =2, z≠0​

Answer 1

Answer:

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}}

Answer 2

Answer:

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Step-by-step explanation:

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