Find Minimum value of $\dfrac {\mathrm{Im}~ z^5} {(\mathrm{Im}~ z)^5}$ for $z = x+iy$.
Here I started with $z = r \exp(i \theta)$ solve the expression finally i got the answer in terms of $\tan$ which i find little difficult to get along with
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If we set $z=x+iy$
Using Binomial Theorem,
$z^5=(x+iy)^5=x^5+\binom51x^4(iy)+\binom52x^3(iy)^2+\binom53x^3(iy)^3+\binom53x(iy)^4+(iy)^5$
$=\cdots+i\{\binom51x^4y-\binom53x^2y^3+y^4\}$
So, img$(z^5)=\binom51x^4y-\binom53x^2y^3+y^4=5x^4y-10x^2y^3+y^4$
$$\implies \frac{\text{img}(z^5)}{(\text{img}(z))^5}=5\left(\frac xy\right)^4-10\left(\frac xy\right)^2+1=5\left(\left(\frac xy\right)^2-1\right)^2+1-5\ge-4$$
The equability occurs when $\left(\frac xy\right)^2-1=0\implies \frac xy=\pm1\iff x=\pm y$
If we start using polar coordinate,
as $\displaystyle \frac xy=\frac{r\cos\theta}{r\sin\theta}=\cot\theta$
we shall reach at $\displaystyle 5\cot^4\theta-10\cot^2\theta+1=5(\cot^2\theta-1)^2+1-5\ge-4$
lab bhattacharjee
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Yep this is correct . is it possible to solve this using derivative test for maxima minima. Like if i put x/y = t and then minimizing the function 5t^4-10t^2+1.. – Milan Amrut Joshi Sep 15 '13 at 09:07
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@MilanAmrutJoshi, have you tried http://en.wikipedia.org/wiki/Second_derivative_test ? – lab bhattacharjee Sep 15 '13 at 09:54
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Nice solution lab bhattacharjee...... – juantheron Dec 15 '13 at 11:13