Question Number 189786 by sonukgindia last updated on 21/Mar/23
Answered by a.lgnaoui last updated on 22/Mar/23
![A•z^z =(e^(iθ) )^e^(iθ) =(cos θ+isin θ)^z =e^(zlnz) =e^(e^(iθ) ln(e^(iθ) )) =e^(ln[(e^(iθ) )^((cos θ+isin θ)) ]) =e^([ln(e^(iθcos θ) ×e^(−θsin θ) ]) =e^([ln(e^(iθcos θ) ×ln(e^(−θsin θ)) ]) =((ln(e^(iθcos θ) ))/(ln(e^(θsin θ) )))=((cos (θcos θ)+isin (θcos θ))/(θsin θ)) =((cos (θcos θ))/(θsin θ))+i((sin( θcos θ))/(θsin θ)) ⇒ z^z =e^([((cos (θcos θ)/(θsin θ))) ×e^(i((sin (θcos θ)/(θsin θ))) =e^((cos (θcos θ))/(θsin θ)) [cos (((sin (θcos θ))/(θsin θ)))+sin (((sin( θcos θ)/(θsin θ)))] ⇒ R(z^z )=e^((cos (θcos θ))/(θsin θ)) ×cos^((((sin (θcos θ))/(θsin θ))))](https://www.tinkutara.com/question/Q189839.png)
$${A}\bullet{z}^{{z}} =\left({e}^{{i}\theta} \right)^{{e}^{{i}\theta} } =\left(\mathrm{cos}\:\theta+{i}\mathrm{sin}\:\theta\right)^{{z}} \\ $$$$={e}^{{z}\mathrm{lnz}} =\mathrm{e}^{\mathrm{e}^{\mathrm{i}\theta} \mathrm{ln}\left(\mathrm{e}^{\mathrm{i}\theta} \:\right)} ={e}^{\mathrm{ln}\left[\left(\mathrm{e}^{\mathrm{i}\theta} \right)^{\left(\mathrm{cos}\:\theta+{i}\mathrm{sin}\:\theta\right)} \right]} \: \\ $$$$={e}^{\left[\mathrm{ln}\left({e}^{{i}\theta\mathrm{cos}\:\theta} ×{e}^{−\theta\mathrm{sin}\:\theta} \right]\right.} ={e}^{\left[\mathrm{ln}\left(\mathrm{e}^{\mathrm{i}\theta\mathrm{cos}\:\theta} ×\mathrm{ln}\left(\mathrm{e}^{\left.−\theta\mathrm{sin}\:\theta\right)} \right]\right.\right.} \\ $$$$=\frac{\mathrm{ln}\left(\mathrm{e}^{\mathrm{i}\theta\mathrm{cos}\:\theta} \right)}{\mathrm{ln}\left(\mathrm{e}^{\theta\mathrm{sin}\:\theta} \right)}=\frac{\mathrm{cos}\:\left(\theta\mathrm{cos}\:\theta\right)+{i}\mathrm{sin}\:\left(\theta\mathrm{cos}\:\theta\right)}{\theta\mathrm{sin}\:\theta} \\ $$$$=\frac{\mathrm{cos}\:\left(\theta\mathrm{cos}\:\theta\right)}{\theta\mathrm{sin}\:\theta}+{i}\frac{\mathrm{sin}\left(\:\theta\mathrm{cos}\:\theta\right)}{\theta\mathrm{sin}\:\theta} \\ $$$$\Rightarrow \\ $$$${z}^{{z}} ={e}^{\left[\frac{\mathrm{cos}\:\left(\theta\mathrm{cos}\:\theta\right.}{\theta\mathrm{sin}\:\theta}\right.} ×{e}^{{i}\frac{\mathrm{sin}\:\left(\theta\mathrm{cos}\:\theta\right.}{\theta\mathrm{sin}\:\theta}} \\ $$$$={e}^{\frac{\mathrm{cos}\:\left(\theta\mathrm{cos}\:\theta\right)}{\theta\mathrm{sin}\:\theta}} \left[\mathrm{cos}\:\left(\frac{\mathrm{sin}\:\left(\theta\mathrm{cos}\:\theta\right)}{\theta\mathrm{sin}\:\theta}\right)+\mathrm{sin}\:\left(\frac{\mathrm{sin}\left(\:\theta\mathrm{cos}\:\theta\right.}{\theta\mathrm{sin}\:\theta}\right)\right] \\ $$$$\Rightarrow \\ $$$$\:\:\boldsymbol{{R}}\left(\boldsymbol{{z}}^{\boldsymbol{{z}}} \right)={e}^{\frac{\mathrm{cos}\:\left(\theta\mathrm{cos}\:\theta\right)}{\theta\mathrm{sin}\:\theta}} ×\mathrm{cos}\:^{\left(\frac{\mathrm{sin}\:\left(\theta\mathrm{cos}\:\theta\right)}{\theta\mathrm{sin}\:\theta}\right)} \\ $$$$ \\ $$