Question:
If \(z\) is a non-real complex number, then the minimum value of \(\dfrac{\operatorname{Im} z^5}{(\operatorname{Im} z)^5}\) is :
If \(z\) is a non-real complex number, then the minimum value of \(\dfrac{\operatorname{Im} z^5}{(\operatorname{Im} z)^5}\) is :
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Write \(z=r(\cos\theta+i\sin\theta)\) so the ratio becomes \(\sin5\theta/\sin^5\theta\) and reduce it to a quadratic in \(1/\sin^2\theta\).
Updated On: Nov 11, 2026
- \(-1\)
- \(-2\)
- \(-4\)
- \(-5\)
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The Correct Option is C
Solution and Explanation
Let \(z=r(\cos\theta+i\sin\theta)\), \(\sin\theta\ne 0\). Then \(\operatorname{Im} z^5=r^5\sin5\theta\) and \((\operatorname{Im} z)^5=r^5\sin^5\theta\).
Using \(\sin5\theta=5\sin\theta-20\sin^3\theta+16\sin^5\theta\), the ratio is \(\dfrac{5}{\sin^4\theta}-\dfrac{20}{\sin^2\theta}+16\).
Let \(t=\dfrac{1}{\sin^2\theta}\ge 1\). Ratio \(=5t^2-20t+16=5(t-2)^2-4\).
At \(t=2\) (allowed since \(t\ge1\)) the minimum is \(-4\).
Answer: (3) \(-4\)
Using \(\sin5\theta=5\sin\theta-20\sin^3\theta+16\sin^5\theta\), the ratio is \(\dfrac{5}{\sin^4\theta}-\dfrac{20}{\sin^2\theta}+16\).
Let \(t=\dfrac{1}{\sin^2\theta}\ge 1\). Ratio \(=5t^2-20t+16=5(t-2)^2-4\).
At \(t=2\) (allowed since \(t\ge1\)) the minimum is \(-4\).
Answer: (3) \(-4\)
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