Menu Close

Solve-for-z-C-z-z-1-

Tinku Tara 0 Comments
Pin




Question Number 215994 by Frix last updated on 25/Jan/25
Solve for z∈C: ∣z^z ∣=1
$$\mathrm{Solve}\:\mathrm{for}\:{z}\in\mathbb{C}:\:\:\:\:\:\mid{z}^{{z}} \mid=\mathrm{1} \\ $$
Answered by mr W last updated on 25/Jan/25
say z=re^(iθ) =r(cos θ+i sin θ) with 0≤θ<2π, r>0 ∣z^z ∣=1 ⇒z^z =e^(iα) with 0≤α<2π z^z =(re^(iθ) )^(re^(iθ) ) =e^(iα) re^(iθ) (ln r+iθ)=iα r(cos θ+i sin θ)(ln r+iθ)=iα r(cos θ ln r−θ sin θ)+i(sin θ ln r+r θ cos θ)=iα ⇒r(cos θ ln r−θ sin θ)=0 ⇒sin θ ln r+r θ cos θ=α cos θ ln r−θ sin θ=0 ⇒ln r=θ tan θ ⇒r=e^(θ tan θ) ...(I) θ sin θ tan θ+θ cos θ e^(θ tan θ) =α θ cos θ (tan^2 θ+e^(θ tan θ) )=α ...(II)
$${say}\:{z}={re}^{{i}\theta} ={r}\left(\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta\right) \\ $$$${with}\:\mathrm{0}\leqslant\theta<\mathrm{2}\pi,\:{r}>\mathrm{0} \\ $$$$\mid{z}^{{z}} \mid=\mathrm{1}\:\Rightarrow{z}^{{z}} ={e}^{{i}\alpha} \:{with}\:\mathrm{0}\leqslant\alpha<\mathrm{2}\pi \\ $$$${z}^{{z}} =\left({re}^{{i}\theta} \right)^{{re}^{{i}\theta} } ={e}^{{i}\alpha} \\ $$$${re}^{{i}\theta} \left(\mathrm{ln}\:{r}+{i}\theta\right)={i}\alpha \\ $$$${r}\left(\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta\right)\left(\mathrm{ln}\:{r}+{i}\theta\right)={i}\alpha \\ $$$${r}\left(\mathrm{cos}\:\theta\:\mathrm{ln}\:{r}−\theta\:\mathrm{sin}\:\theta\right)+{i}\left(\mathrm{sin}\:\theta\:\mathrm{ln}\:{r}+{r}\:\theta\:\mathrm{cos}\:\theta\right)={i}\alpha \\ $$$$\Rightarrow{r}\left(\mathrm{cos}\:\theta\:\mathrm{ln}\:{r}−\theta\:\mathrm{sin}\:\theta\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:\theta\:\mathrm{ln}\:{r}+{r}\:\theta\:\mathrm{cos}\:\theta=\alpha \\ $$$$\mathrm{cos}\:\theta\:\mathrm{ln}\:{r}−\theta\:\mathrm{sin}\:\theta=\mathrm{0} \\ $$$$\Rightarrow\mathrm{ln}\:{r}=\theta\:\mathrm{tan}\:\theta\: \\ $$$$\Rightarrow{r}={e}^{\theta\:\mathrm{tan}\:\theta} \:\:\:\:\:…\left({I}\right) \\ $$$$\theta\:\mathrm{sin}\:\theta\:\mathrm{tan}\:\theta+\theta\:\mathrm{cos}\:\theta\:{e}^{\theta\:\mathrm{tan}\:\theta} =\alpha \\ $$$$\theta\:\mathrm{cos}\:\theta\:\left(\mathrm{tan}^{\mathrm{2}} \:\theta+{e}^{\theta\:\mathrm{tan}\:\theta} \right)=\alpha\:\:\:…\left({II}\right) \\ $$
Commented by Ghisom last updated on 25/Jan/25
yes but I think r=e^(θtan θ) ⇒ z=e^(θtan θ) e^(iθ) is enough ⇒ z^z =e^(i((θe^(θtan θ) )/(cos θ))) ⇒ ∣z^z ∣=1
$$\mathrm{yes} \\ $$$$\mathrm{but}\:\mathrm{I}\:\mathrm{think} \\ $$$${r}=\mathrm{e}^{\theta\mathrm{tan}\:\theta} \:\Rightarrow\:{z}=\mathrm{e}^{\theta\mathrm{tan}\:\theta} \mathrm{e}^{\mathrm{i}\theta} \\ $$$$\mathrm{is}\:\mathrm{enough} \\ $$$$\Rightarrow \\ $$$${z}^{{z}} =\mathrm{e}^{\mathrm{i}\frac{\theta\mathrm{e}^{\theta\mathrm{tan}\:\theta} }{\mathrm{cos}\:\theta}} \:\Rightarrow\:\mid{z}^{{z}} \mid=\mathrm{1} \\ $$
Commented by Frix last updated on 26/Jan/25
Thank you both!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{both}! \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *