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Let  \(z\)  be a nonreal complex number. Find the smallest possible value of   

                                                                                                                    \(\frac{\text{Im}(z^5)}{[\text{Im}(z)]^5}. \)


Note: For a complex number \(z,\)\(\text{Im}(z)\) denotes the imaginary part of \(z.\)

 Nov 18, 2019
 #1
avatar+118 
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Let \(z=a+bi\). Then

 

\(z^5=(a+bi)^5=a^5+{5\choose 1}a^4bi+{5\choose 2}a^3(bi)^2+{5\choose3}a^2(bi)^3+{5\choose4}a(bi)^4+(bi)5\)

 

                           =\(a^5+5a^4bi-10a^3b^2-10a^2b^3i+5ab^4+b^5i\)\(=(a^5-10a^3b^2+5ab^4)+(5a^4b-10a^2b^3+b^5)i\)

 

S0\(Im(z^5)=5a^4b-10a^2b^3+b^5\) and

 

\(\frac{Im(z^5)} {[Im(z)]^5}=\frac{5a^4b-10a^2b^3+b^5}{b^5}\)\(=\frac{5a^4b}{b^5}-\frac{10a^2b^3}{b^5}+\frac{b^5}{b^5}\)=\(5 ( \frac{a}{b})^4-10(\frac{a}{b})^2 + 1\).

 

This is a quadratic-type expression that can be converted to a quadratic expression by replacing \((\frac{a}{b})^2\)with x. So we need to find the minimum value of \(5x^2-10x+1\), which can be found at the vertex of \(y=5x^2-10x+1\). Since \( \frac{-b}{2a}=1\), the minimum value is \(5(1)^2-10(1)+1 =-4\)

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 Nov 18, 2019

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