Question Number 189815 by sonukgindia last updated on 22/Mar/23
Answered by Frix last updated on 22/Mar/23
![z=e^(iθ) =cos θ +i sin θ =c+si [for typing less] z^z =(e^(iθ) )^(c+si) =e^(−θs+iθc) =(1/e^(θs) )e^(iθc) z^z =e^(−θsin θ) e^(iθcos θ) ∣z^z ∣=e^(−θsin θ) −π<θ≤π ⇒ mMaximum =1 at θ∈{−π, 0, π} The minimum is harder to find ((d∣z^z ∣)/dθ)=0 −(sin θ +θcos θ)e^(−θsin θ) =0 sin θ +θcos θ =0 Minimum ≈.162073435459 at θ≈±2.02875783811](https://www.tinkutara.com/question/Q189821.png)
$${z}=\mathrm{e}^{\mathrm{i}\theta} =\mathrm{cos}\:\theta\:+\mathrm{i}\:\mathrm{sin}\:\theta\:={c}+{s}\mathrm{i}\:\left[\mathrm{for}\:\mathrm{typing}\:\mathrm{less}\right] \\ $$$${z}^{{z}} =\left(\mathrm{e}^{\mathrm{i}\theta} \right)^{{c}+{s}\mathrm{i}} =\mathrm{e}^{−\theta{s}+\mathrm{i}\theta{c}} =\frac{\mathrm{1}}{\mathrm{e}^{\theta{s}} }\mathrm{e}^{\mathrm{i}\theta{c}} \\ $$$${z}^{{z}} =\mathrm{e}^{−\theta\mathrm{sin}\:\theta} \mathrm{e}^{\mathrm{i}\theta\mathrm{cos}\:\theta} \\ $$$$\mid{z}^{{z}} \mid=\mathrm{e}^{−\theta\mathrm{sin}\:\theta} \\ $$$$−\pi<\theta\leqslant\pi\:\Rightarrow\:\mathrm{mMaximum}\:=\mathrm{1}\:\mathrm{at}\:\theta\in\left\{−\pi,\:\mathrm{0},\:\pi\right\} \\ $$$$\mathrm{The}\:\mathrm{minimum}\:\mathrm{is}\:\mathrm{harder}\:\mathrm{to}\:\mathrm{find} \\ $$$$\frac{{d}\mid{z}^{{z}} \mid}{{d}\theta}=\mathrm{0} \\ $$$$−\left(\mathrm{sin}\:\theta\:+\theta\mathrm{cos}\:\theta\right)\mathrm{e}^{−\theta\mathrm{sin}\:\theta} =\mathrm{0} \\ $$$$\mathrm{sin}\:\theta\:+\theta\mathrm{cos}\:\theta\:=\mathrm{0} \\ $$$$\mathrm{Minimum}\:\approx.\mathrm{162073435459}\:\mathrm{at}\:\theta\approx\pm\mathrm{2}.\mathrm{02875783811} \\ $$