If $$z$$ is a non-real complex number, then the minimum value of $$\frac{Im\ z^5}{(Im\ z)^5}$$ is (Where $$Im\ z$$ = Imaginary part of $$z$$)

Let $$ z $$ be a non-real complex number, so $$ z = x + iy $$ where $$ x $$ and $$ y $$ are real numbers and $$ y \neq 0 $$. The imaginary part of $$ z $$ is $$ \operatorname{Im} z = y $$. We need to find the minimum value of $$ \frac{\operatorname{Im} z^5}{(\operatorname{Im} z)^5} = \frac{\operatorname{Im} (x + iy)^5}{y^5} $$.

First, expand $$ (x + iy)^5 $$ using the binomial theorem:

$$ (x + iy)^5 = \sum_{k=0}^{5} \binom{5}{k} x^{5-k} (iy)^k $$

Compute each term:

$$ \binom{5}{0} x^5 (iy)^0 = x^5 $$

$$ \binom{5}{1} x^4 (iy)^1 = 5x^4 (iy) = 5i x^4 y $$

$$ \binom{5}{2} x^3 (iy)^2 = 10x^3 (i^2 y^2) = 10x^3 (-1) y^2 = -10x^3 y^2 $$

$$ \binom{5}{3} x^2 (iy)^3 = 10x^2 (i^3 y^3) = 10x^2 (-i) y^3 = -10i x^2 y^3 $$

$$ \binom{5}{4} x^1 (iy)^4 = 5x (i^4 y^4) = 5x (1) y^4 = 5x y^4 $$

$$ \binom{5}{5} x^0 (iy)^5 = (i^5 y^5) = i^5 y^5 = i^4 \cdot i \cdot y^5 = 1 \cdot i \cdot y^5 = i y^5 $$

Combine all terms:

$$ (x + iy)^5 = x^5 - 10x^3 y^2 + 5x y^4 + i(5x^4 y - 10x^2 y^3 + y^5) $$

The imaginary part of $$ z^5 $$ is the coefficient of $$ i $$:

$$ \operatorname{Im} z^5 = 5x^4 y - 10x^2 y^3 + y^5 $$

Now substitute into the expression:

$$ \frac{\operatorname{Im} z^5}{(\operatorname{Im} z)^5} = \frac{5x^4 y - 10x^2 y^3 + y^5}{y^5} $$

Since $$ y \neq 0 $$, divide each term in the numerator by $$ y^5 $$:

$$ = \frac{5x^4 y}{y^5} - \frac{10x^2 y^3}{y^5} + \frac{y^5}{y^5} = 5 \frac{x^4}{y^4} - 10 \frac{x^2}{y^2} + 1 $$

Set $$ t = \frac{x}{y} $$, which is a real number. The expression simplifies to:

$$ f(t) = 5t^4 - 10t^2 + 1 $$

To find the minimum value of $$ f(t) $$ for $$ t \in \mathbb{R} $$, compute the derivative:

$$ f'(t) = 20t^3 - 20t $$

Set the derivative equal to zero to find critical points:

$$ 20t^3 - 20t = 0 $$

$$ 20t(t^2 - 1) = 0 $$

$$ 20t(t - 1)(t + 1) = 0 $$

So the critical points are $$ t = 0 $$, $$ t = 1 $$, and $$ t = -1 $$.

Evaluate $$ f(t) $$ at these points:

At $$ t = 0 $$:

$$ f(0) = 5(0)^4 - 10(0)^2 + 1 = 1 $$

At $$ t = 1 $$:

$$ f(1) = 5(1)^4 - 10(1)^2 + 1 = 5 - 10 + 1 = -4 $$

At $$ t = -1 $$:

$$ f(-1) = 5(-1)^4 - 10(-1)^2 + 1 = 5(1) - 10(1) + 1 = 5 - 10 + 1 = -4 $$

As $$ t \to \infty $$ or $$ t \to -\infty $$, $$ f(t) = 5t^4 - 10t^2 + 1 \to \infty $$ because the $$ t^4 $$ term dominates. Therefore, the minimum value is $$-4$$, achieved at $$ t = 1 $$ and $$ t = -1 $$.

At $$ t = 1 $$, $$ x = y $$, so $$ z = x + ix = x(1 + i) $$, which is non-real since $$ y \neq 0 $$. Similarly, at $$ t = -1 $$, $$ z = x - ix = x(1 - i) $$, which is also non-real. Hence, the minimum value is $$-4$$.

Comparing with the options: A is $$-2$$, B is $$-4$$, C is $$-5$$, D is $$-1$$. The value $$-4$$ corresponds to option B.

Hence, the correct answer is Option B.