Suggested solution method to question 203502 ze^z =1 Obviously the only real solution is z=W(1)≈.567143290 z=a+bi∧b≠0 (a+bi)e^(a+bi) =1 (a+bi)(cos b +isin b)e^a =1 e^a (acos b −bsin b)+e^a (asin b +bcos b)i=1 { ((e^a (acos b −bsin b)=1)),((e^a (asin b +bcos b)=0 ⇒ a=−bcot b)) :} Inserting & transforming leaves us with: { ((be^(−bcot b) +sin b =0)),((a=−bcot b)) :} We can only approximate. The first solutions are: b≈±4.37518515 ⇒ a≈−1.53391332 ⇒ z≈−1.53391332±4.37518515i b≈±10.7762995 ⇒ a≈−2.40158510 ⇒ z≈−2.40158510±10.7762995i These are the values of the complex LambertW−function W_n (1); n∈Z Following this path we can solve ze^z =w with z, w∈Z I hope this is helpful!


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Question Number 203570 by Frix last updated on 22/Jan/24
$Suggested solution method to question 203502 ze^z =1 Obviously the only real solution is z=W(1)≈.567143290 z=a+bi∧b≠0 (a+bi)e^(a+bi) =1 (a+bi)(cos b +isin b)e^a =1 e^a (acos b −bsin b)+e^a (asin b +bcos b)i=1 { ((e^a (acos b −bsin b)=1)),((e^a (asin b +bcos b)=0 ⇒ a=−bcot b)) :} Inserting & transforming leaves us with: { ((be^(−bcot b) +sin b =0)),((a=−bcot b)) :} We can only approximate. The first solutions are: b≈±4.37518515 ⇒ a≈−1.53391332 ⇒ z≈−1.53391332±4.37518515i b≈±10.7762995 ⇒ a≈−2.40158510 ⇒ z≈−2.40158510±10.7762995i These are the values of the complex LambertW−function W_n (1); n∈Z Following this path we can solve ze^z =w with z, w∈Z I hope this is helpful!$
$Suggested solution method to$ $question 203502$ $z e^{z} = 1$ $Obviously the only real solution is$ $z = W (1) \approx . 567143290$ $z = a + b i \land b \neq 0$ $(a + b i) e^{a + b i} = 1$ $(a + b i) (\cos b + isin b) e^{a} = 1$ $e^{a} (a \cos b - b \sin b) + e^{a} (a \sin b + b \cos b) i = 1$ ${\begin{cases} e^{a} (a \cos b - b \sin b) = 1 \\ e^{a} (a \sin b + b \cos b) = 0 \Rightarrow a = - b \cot b \end{cases}$ $Inserting & transforming leaves us with :$ ${\begin{cases} b e^{- b \cot b} + \sin b = 0 \\ a = - b \cot b \end{cases}$ $We can only approximate .$ $The first solutions are :$ $b \approx \pm 4.37518515 \Rightarrow a \approx - 1.53391332$ $\Rightarrow z \approx - 1.53391332 \pm 4 . 37518515 i$ $b \approx \pm 10.7762995 \Rightarrow a \approx - 2.40158510$ $\Rightarrow z \approx - 2.40158510 \pm 10 . 7762995 i$ $These are the values of the complex$ $LambertW - function W_{n} (1); n \in Z$ $Following this path we can solve$ $z e^{z} = w with z, w \in Z$ $I hope this is helpful!$
Commented by Frix last updated on 22/Jan/24
$ze^z =p+qi z=a+bi { ((e^a (acos b −bsin b)=p)),((e^a (asin b +bcos b)=q)) :} Solve for (cos b; sin b) cos b =((ap+bq)/(a^2 +b^2 ))e^(−a) sin b =((aq+bp)/(a^2 +b^2 ))e^(−a) ⇒ tan b =((ap+bq)/(aq−bp)) ⇒ a=−((q+ptan b)/(p−qtan b))b Insert in one of the equations and approximate.$
$z e^{z} = p + q i$ $z = a + b i$ ${\begin{cases} e^{a} (a \cos b - b \sin b) = p \\ e^{a} (a \sin b + b \cos b) = q \end{cases}$ $Solve for (\cos b; \sin b)$ $\cos b = \frac{a p + b q}{a^{2} + b^{2}} e^{- a}$ $\sin b = \frac{a q + b p}{a^{2} + b^{2}} e^{- a}$ $\Rightarrow \tan b = \frac{a p + b q}{a q - b p}$ $\Rightarrow a = - \frac{q + p \tan b}{p - q \tan b} b$ $Insert in one of the equations and approximate .$
Commented by Frix last updated on 22/Jan/24
$Example ze^z =2−3i leads to { ((be^(−b((2−3tan b)/(3+2tan b))) +3cos b +2sin b =0)),((a=−b((2−3tan b)/(3+2tan b)))) :} b≈−.530139721 ⇒ a≈1.09007653 z≈1.09007653−.530139721i b≈3.72110799 ⇒ a≈−.0315828084 z≈−.0315828084+3.72110799i ...$
$Example$ $z e^{z} = 2 - 3 i$ $leads to$ ${\begin{cases} b e^{- b \frac{2 - 3 \tan b}{3 + 2 \tan b}} + 3 \cos b + 2 \sin b = 0 \\ a = - b \frac{2 - 3 \tan b}{3 + 2 \tan b} \end{cases}$ $b \approx - . 530139721 \Rightarrow a \approx 1.09007653$ $z \approx 1.09007653 - . 530139721 i$ $b \approx 3.72110799 \Rightarrow a \approx - . 0315828084$ $z \approx - . 0315828084 + 3 . 72110799 i$ $. . .$
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