Question

If $Z$ is non-real complex number, then find the minimum value of $\frac{Im{z}^5}{{Im}^{5}z}$.

Answer 1

Let z = x + iy

$\frac{Im{z}^5}{Im{z}^5}=\frac{Im(x+i{y}^5}{\left(Im\right(x+iy){)}^5}$

We have $\frac{Im(x+iy{)}^5}{\left[y^5\right]}$

$=\frac{1}{y^5}{[}^5{C}_0{x}^5{+}^5{C}_1{x}^4\left(iy\right){+}^5{C}_2{x}^3\left(iy{)}^2{+}^5{C}_3{x}^2\right(iy{)}^3{+}^5{C}_{4}x\left(iy{)}^4{+}^5{C}_5\right(iy{)}^5]$

$=\frac{1}{y^5}\left[Im\left\{{}^5{C}_0{x}^5{+}^5{C}_1{x}^4\right(iy\left){+}^5{C}_2{x}^3\right(iy{)}^2{+}^5{C}_3{x}^2\left(iy{)}^3{+}^5{C}_{4}x\right(iy{)}^4{+}^5{C}_5\left(iy\right)\}\right]$

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

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

$=5\left[\right(\frac{x}{y}{)}^4-2(\frac{x}{y}{)}^2+1]-5+1$

$=5\left[\right(\frac{x}{y}{)}^2-1{]}^2-4$

$\therefore (\frac{x}{y}{)}^2-1{)}^2\ge 0\Rightarrow ((\frac{x}{y}{)}^2-1{)}^2-4\ge 0\ge -4$

$\therefore \frac{Im{2}^5}{(I{m}^2{)}^5}=-4$ is the minimum value