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# Complex #'s

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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+0 Answers

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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