Question Number 222031 by fantastic last updated on 15/Jun/25
$${x}^{{x}} =−\mathrm{1} \\ $$$${Number}\:{of}\:{solutions}?? \\ $$
Answered by mr W last updated on 15/Jun/25
![solve z^z =−1for z∈C. say z=e^(x+yi) =e^x (cos y+i sin y) z^z =−1=e^((2k+1)πi) z ln z=(2k+1)πi e^x (cos y+i sin y)(x+yi)=(2k+1)πi e^x [(x cos y−y sin y)+i (x sin y+y cos y)]=(2k+1)πi ⇒x cos y−y sin y=0 ⇒x=y tan y (I) ⇒e^x (x sin y+y cos y)=(2k+1)π ⇒((ye^(y tan y) )/(cos y))=(2k+1)π (II) there are infinite solutions. example: x≈0.924539370742 y≈0.835778625924 z≈1.690386757164+1.869907964027i](https://www.tinkutara.com/question/Q222032.png)
$${solve}\:\boldsymbol{{z}}^{\boldsymbol{{z}}} =−\mathrm{1}{for}\:\boldsymbol{{z}}\in\boldsymbol{{C}}. \\ $$$${say}\:\boldsymbol{{z}}=\boldsymbol{{e}}^{\boldsymbol{{x}}+\boldsymbol{{yi}}} ={e}^{{x}} \left(\mathrm{cos}\:{y}+{i}\:\mathrm{sin}\:{y}\right) \\ $$$${z}^{{z}} =−\mathrm{1}={e}^{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi{i}} \\ $$$${z}\:\mathrm{ln}\:{z}=\left(\mathrm{2}{k}+\mathrm{1}\right)\pi{i} \\ $$$${e}^{{x}} \left(\mathrm{cos}\:{y}+{i}\:\mathrm{sin}\:{y}\right)\left({x}+{yi}\right)=\left(\mathrm{2}{k}+\mathrm{1}\right)\pi{i} \\ $$$${e}^{{x}} \left[\left({x}\:\mathrm{cos}\:{y}−{y}\:\mathrm{sin}\:{y}\right)+{i}\:\left({x}\:\mathrm{sin}\:{y}+{y}\:\mathrm{cos}\:{y}\right)\right]=\left(\mathrm{2}{k}+\mathrm{1}\right)\pi{i} \\ $$$$\Rightarrow{x}\:\mathrm{cos}\:{y}−{y}\:\mathrm{sin}\:{y}=\mathrm{0}\:\Rightarrow{x}={y}\:\mathrm{tan}\:{y}\:\:\left({I}\right) \\ $$$$\Rightarrow{e}^{{x}} \left({x}\:\mathrm{sin}\:{y}+{y}\:\mathrm{cos}\:{y}\right)=\left(\mathrm{2}{k}+\mathrm{1}\right)\pi \\ $$$$\Rightarrow\frac{{ye}^{{y}\:\mathrm{tan}\:{y}} }{\mathrm{cos}\:{y}}=\left(\mathrm{2}{k}+\mathrm{1}\right)\pi\:\:\:\left({II}\right) \\ $$$${there}\:{are}\:\underline{\boldsymbol{{infinite}}}\:{solutions}. \\ $$$$ \\ $$$${example}: \\ $$$${x}\approx\mathrm{0}.\mathrm{924539370742} \\ $$$${y}\approx\mathrm{0}.\mathrm{835778625924} \\ $$$${z}\approx\mathrm{1}.\mathrm{690386757164}+\mathrm{1}.\mathrm{869907964027}{i} \\ $$
Commented by fantastic last updated on 15/Jun/25
$${Sir}\:{can}\:{you}\:{please}\:{tell}\:{me}\:{what}\:{application} \\ $$$${do}\:{you}\:{use}\:{to}\:{draw}\:{geomwtrical}\:{problems} \\ $$
Commented by AntonCWX8 last updated on 15/Jun/25
$${The}\:{last}\:{time}\:{I}\:{asked},\:{he}\:{suggested}\:{Lekh}. \\ $$
Commented by mr W last updated on 15/Jun/25
$${Lekh}\:{is}\:{good}\:{for}\:{creating}\:{accurate} \\ $$$${diagrams},\:{like}\:{those}\:{from}\:{sir}\:{ajfour}. \\ $$$${but}\:{usually}\:{i}\:{only}\:{make}\:{freehand} \\ $$$${drawings},\:{using}\:{the}\:{app}\:{INKredible} \\ $$$${or}\:{the}\:{app}\:{Photo}\:{editor}. \\ $$
Commented by fantastic last updated on 15/Jun/25
$${I}\:{installed}\:{it}\:{now} \\ $$$$ \\ $$
Answered by AntonCWX8 last updated on 15/Jun/25
$${x}^{{x}} =−\mathrm{1} \\ $$$${xln}\left({x}\right)={ln}\left(−\mathrm{1}\right) \\ $$$${ln}\left({x}\right){e}^{{ln}\left({x}\right)} ={ln}\left(−\mathrm{1}\right) \\ $$$${ln}\left({x}\right)={W}_{{n}} \left({i}\pi\right) \\ $$$${x}={e}^{{W}_{{n}} \left({i}\pi\right)} \\ $$$${where}\:{n}={any}\:{integer} \\ $$$$\Rightarrow{Infinite}\:{solution} \\ $$
Answered by wewji12 last updated on 15/Jun/25
$${f}\left({x}\right)={x}^{{x}} \:\mathrm{and}\:\: \\ $$$$\mathrm{inverse}\:\mathrm{function}\:\mathrm{of}\:{x}^{{x}} \:\mathrm{is}\:{f}^{−\mathrm{1}} \left({x}\right)=\frac{\mathrm{ln}\left({x}\right)}{{W}\left(\mathrm{ln}\left({x}\right)\right)} \\ $$$$\frac{\mathrm{ln}\left(−\mathrm{1}\right)}{{W}\left(\mathrm{ln}\left(−\mathrm{1}\right)\right)}=\frac{\boldsymbol{{i}}\pi}{{W}\left(\boldsymbol{{i}}\pi\right)}\approx\mathrm{1}.\mathrm{690386757}…+\mathrm{1}.\mathrm{86990796}\boldsymbol{{i}} \\ $$