Menu Close

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-

Tinku Tara 0 Comments
Pin




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!
$$\mathrm{Suggested}\:\mathrm{solution}\:\mathrm{method}\:\mathrm{to} \\ $$$$\mathrm{question}\:\mathrm{203502} \\ $$$$ \\ $$$${z}\mathrm{e}^{{z}} =\mathrm{1} \\ $$$$\mathrm{Obviously}\:\mathrm{the}\:\mathrm{only}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{is} \\ $$$${z}={W}\left(\mathrm{1}\right)\approx.\mathrm{567143290} \\ $$$$ \\ $$$${z}={a}+{b}\mathrm{i}\wedge{b}\neq\mathrm{0} \\ $$$$\left({a}+{b}\mathrm{i}\right)\mathrm{e}^{{a}+{b}\mathrm{i}} =\mathrm{1} \\ $$$$\left({a}+{b}\mathrm{i}\right)\left(\mathrm{cos}\:{b}\:+\mathrm{isin}\:{b}\right)\mathrm{e}^{{a}} =\mathrm{1} \\ $$$$\mathrm{e}^{{a}} \left({a}\mathrm{cos}\:{b}\:−{b}\mathrm{sin}\:{b}\right)+\mathrm{e}^{{a}} \left({a}\mathrm{sin}\:{b}\:+{b}\mathrm{cos}\:{b}\right)\mathrm{i}=\mathrm{1} \\ $$$$\begin{cases}{\mathrm{e}^{{a}} \left({a}\mathrm{cos}\:{b}\:−{b}\mathrm{sin}\:{b}\right)=\mathrm{1}}\\{\mathrm{e}^{{a}} \left({a}\mathrm{sin}\:{b}\:+{b}\mathrm{cos}\:{b}\right)=\mathrm{0}\:\Rightarrow\:{a}=−{b}\mathrm{cot}\:{b}}\end{cases} \\ $$$$\mathrm{Inserting}\:\&\:\mathrm{transforming}\:\mathrm{leaves}\:\mathrm{us}\:\mathrm{with}: \\ $$$$\begin{cases}{{b}\mathrm{e}^{−{b}\mathrm{cot}\:{b}} +\mathrm{sin}\:{b}\:=\mathrm{0}}\\{{a}=−{b}\mathrm{cot}\:{b}}\end{cases} \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate}. \\ $$$$\mathrm{The}\:\mathrm{first}\:\mathrm{solutions}\:\mathrm{are}: \\ $$$${b}\approx\pm\mathrm{4}.\mathrm{37518515}\:\Rightarrow\:{a}\approx−\mathrm{1}.\mathrm{53391332} \\ $$$$\Rightarrow\:{z}\approx−\mathrm{1}.\mathrm{53391332}\pm\mathrm{4}.\mathrm{37518515i} \\ $$$${b}\approx\pm\mathrm{10}.\mathrm{7762995}\:\Rightarrow\:{a}\approx−\mathrm{2}.\mathrm{40158510} \\ $$$$\Rightarrow\:{z}\approx−\mathrm{2}.\mathrm{40158510}\pm\mathrm{10}.\mathrm{7762995i} \\ $$$$\mathrm{These}\:\mathrm{are}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{the}\:\mathrm{complex} \\ $$$$\mathrm{LambertW}−\mathrm{function}\:{W}_{{n}} \left(\mathrm{1}\right);\:{n}\in\mathbb{Z} \\ $$$$\mathrm{Following}\:\mathrm{this}\:\mathrm{path}\:\mathrm{we}\:\mathrm{can}\:\mathrm{solve} \\ $$$${z}\mathrm{e}^{{z}} ={w}\:\mathrm{with}\:{z},\:{w}\in\mathbb{Z} \\ $$$$\mathrm{I}\:\mathrm{hope}\:\mathrm{this}\:\mathrm{is}\:\mathrm{helpful}! \\ $$
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.
$${z}\mathrm{e}^{{z}} ={p}+{q}\mathrm{i} \\ $$$${z}={a}+{b}\mathrm{i} \\ $$$$\begin{cases}{\mathrm{e}^{{a}} \left({a}\mathrm{cos}\:{b}\:−{b}\mathrm{sin}\:{b}\right)={p}}\\{\mathrm{e}^{{a}} \left({a}\mathrm{sin}\:{b}\:+{b}\mathrm{cos}\:{b}\right)={q}}\end{cases} \\ $$$$\mathrm{Solve}\:\mathrm{for}\:\left(\mathrm{cos}\:{b};\:\mathrm{sin}\:{b}\right) \\ $$$$\mathrm{cos}\:{b}\:=\frac{{ap}+{bq}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\mathrm{e}^{−{a}} \\ $$$$\mathrm{sin}\:{b}\:=\frac{{aq}+{bp}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\mathrm{e}^{−{a}} \\ $$$$\Rightarrow\:\mathrm{tan}\:{b}\:=\frac{{ap}+{bq}}{{aq}−{bp}} \\ $$$$\Rightarrow\:{a}=−\frac{{q}+{p}\mathrm{tan}\:{b}}{{p}−{q}\mathrm{tan}\:{b}}{b} \\ $$$$\mathrm{Insert}\:\mathrm{in}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{and}\:\mathrm{approximate}. \\ $$
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 ...
$$\mathrm{Example} \\ $$$${z}\mathrm{e}^{{z}} =\mathrm{2}−\mathrm{3i} \\ $$$$\mathrm{leads}\:\mathrm{to} \\ $$$$\begin{cases}{{b}\mathrm{e}^{−{b}\frac{\mathrm{2}−\mathrm{3tan}\:{b}}{\mathrm{3}+\mathrm{2tan}\:{b}}} +\mathrm{3cos}\:{b}\:+\mathrm{2sin}\:{b}\:=\mathrm{0}}\\{{a}=−{b}\frac{\mathrm{2}−\mathrm{3tan}\:{b}}{\mathrm{3}+\mathrm{2tan}\:{b}}}\end{cases} \\ $$$${b}\approx−.\mathrm{530139721}\:\Rightarrow\:{a}\approx\mathrm{1}.\mathrm{09007653} \\ $$$${z}\approx\mathrm{1}.\mathrm{09007653}−.\mathrm{530139721i} \\ $$$${b}\approx\mathrm{3}.\mathrm{72110799}\:\Rightarrow\:{a}\approx−.\mathrm{0315828084} \\ $$$${z}\approx−.\mathrm{0315828084}+\mathrm{3}.\mathrm{72110799i} \\ $$$$… \\ $$

Pin

Leave a Reply

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