<s>[INST] Minimum Value of a complex expression 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 
 [/INST] 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 $\displaystyle 5\cot^4\theta-10\cot^2\theta+1=5(\cot^2\theta-1)^2+1-5\ge-4$
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