Question

If Z is a non-real complex number, then the minimum value of $\frac{Im{Z}^5}{I{m}^{5}Z}$ is

• A. -1
• B. -2
• C. -4
• D. -5

Answer 1

The correct option is C -4

Let Z =a+ib, $b\ne 0$ where Im Z =b
$Z^5=(a+ib{)}^5=a^5{+}^5{C}_1{a}^{4}bi{+}^5{C}_2{a}^3{b}^2{i}^2{+}^5{C}_3{a}^2{b}^3{i}^3{+}^5{C}_4{i}^4+i^5{b}^5$
$Im{Z}^5=5a^{4}b-10a^2{b}^3+b^{5}y=\frac{Im{Z}^5}{I{m}^{5}z}=5{\left(\frac{a}{b}\right)}^4-10{\left(\frac{a}{b}\right)}^2+1$
Let ${\left(\frac{a}{b}\right)}^2=x\left(say\right),x\in R^{+}$
$y=5x^2-10x+1=5[x^2-2x]+1=5\left[\right(x-1{)}^2]-4$
Hence $y_{min}=-4.$