solve-for-z-C-zln-z-z-2-

Question Number 205147 by Ghisom last updated on 10/Mar/24

solve for z \in C

z \ln z = z - 2

Answered by pi314 last updated on 10/Mar/24

z = e^{y}

\Leftrightarrow {y e}^{y} = e^{y} - 2

(y - 1) e^{y - 1} = - 2 e^{- 1}

y - 1 = W (- 2 e^{- 1})

y = 1 + W (- 2 e^{- 1})

z = e^{1 + W (- 2 e^{- 1})}

Commented by Frix last updated on 10/Mar/24

{What}^{'} s the approximate value of z ?

Answered by Frix last updated on 10/Mar/24

Different method :

z \ln z - z + 2 = 0

z (\ln z - 1) + 2 = 0

z = r e^{i θ}

r e^{i θ} (\ln r - 1 + i θ) + 2 = 0

{\begin{cases} r (\cos θ \ln r - \cos θ - θ \sin θ) + 2 = 0 \\ i r (\sin θ \ln r + θ \cos θ - \sin θ) = 0 \end{cases}

(2) \Rightarrow r = e^{1 - \frac{θ}{\tan θ}}

(1) 2 - e^{1 - \frac{θ}{\tan θ}} \frac{θ}{\sin θ} = 0

θ \approx \pm 1.13267249760

r \approx 1.59896034818

z \approx . 678344933205 \pm 1 . 44793727304 i

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