Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{both}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com