The Answer Is n^2 Part II

Let roots are $a,b,c$ (i am too lazy to type \delta sorry) Expression can be written as:-

$3+15(\frac{1}{\frac{1}{2}-a}+\frac{1}{\frac{1}{2}-b}+\frac{1}{\frac{1}{2}-b})......(1)$

According to FTA:-

$x^4-1=(x-1)(x-a)(x-b)(x-c)$

Taking $log$ both sides of above equation:-

$log(x^4-1)=log(x-1)+log(x-a)+log(x-b)+log(x-c)$

Differentiating both sides:-

$\frac{4x^3}{x^4-1}=\frac{1}{x-1}+\frac{1}{x-a}+\frac{1}{x-b}+\frac{1}{x-c}$

Put $x=\frac{1}{2}$ in above equation:-

$\frac{1}{\frac{1}{2}-a}+\frac{1}{\frac{1}{2}-b}+\frac{1}{\frac{1}{2}-c}=\frac{22}{15}$

Putting this value in equation (1):-

$given expression = 3 + 15×\frac{22}{15} = \boxed{25}$

Let $z = a + bi$

The fourth roots of unity of $z$ are $1, \delta_1, \delta_2, \delta_3$

We only know one root to be $1$ so

$\sqrt [4] {z} = 1$

This means that

$z = 1^4 = 1$

$\sqrt [4] {(1)} = 1, i, -1, -i$

Now we substitute these values in

$\frac {31 - 2(i)}{1 - 2(i)} + \frac {31 - 2(-1)}{1 - 2(-1)} + \frac {31 - 2(-i)}{1 - 2(-i)}$

Simplifying this gives us

$\frac {(35 + 60i) + (35 - 60i)}{5} + 11$

This then goes to

$\frac {70}{5} + 11 = 25$

So the answer is $25$

The set {d1, d2, d3} is the same as the set {-1, i, -i} Substituting the values in the expression and solving algebraically gives the sum to be 25.

Let $x = a + bi$ The fourth roots of unity of $z$ are $1, \delta_1, \delta_2, \delta_3$

But we know one root as $1$

$~~~~~~~~~~~~$ $\sqrt [4] {z} = 1$

This means that

$~~~~~~~~~~~~$ $z = 1^4 = 1$

$~~~~~~~~~~~~$ $\sqrt [4] {(1)} = 1, i, -1, -i$

Now we will substitute

$~~~~~~~~~~~~$ $\dfrac {31 - 2(i)}{1 - 2(i)} + \frac {31 - 2(-1)}{1 - 2(-1)} + \frac {31 - 2(-i)}{1 - 2(-i)}$

Which gives us

$~~~~~~~~~~~~$ $\frac {70}{5} + 11 = 25$

$\Rightarrow\huge\boxed{25}$

$i^4=1$ which means iota is a fourth root of unity. Similarly $-i,1,-1$ will be the fourth roots of unity. Plugging these value in the expression we get $\boxed{25}$ .

Alternatively, sum of $n^{th}$ roots of unity is 0. One of the given roots is -1 ,let's say $\delta_{1}$ . Which implies the other two will be the complex roots and will be conjugate of each other.

Or $\delta_{2}=-\delta_{3}$ . Also, $\delta_{2}^{4} =1$ which means $\delta_{2}^{2}$ is either 1 or -1.But $\delta_{2}^{2}$ cannot be 1 as it is complex. So $\delta_{2}^{2}=-1$ . Simplifying the expression and putting the values we get $\boxed {25}$