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Question Number 5091 by FilupSmith last updated on 11/Apr/16

Part 1  S=ln(a+bi),      a, b∈R                                   i^2 =−1  is S∈C?       (b≠0)    Part 2  t=ln(ln(...(ln(k)))),    k∈R, k>1  ∴t∈C?  is t undefined?

$$\mathrm{Part}\:\mathrm{1} \\ $$ $${S}=\mathrm{ln}\left({a}+{bi}\right),\:\:\:\:\:\:{a},\:{b}\in\mathbb{R} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{i}^{\mathrm{2}} =−\mathrm{1} \\ $$ $$\mathrm{is}\:{S}\in\mathbb{C}?\:\:\:\:\:\:\:\left({b}\neq\mathrm{0}\right) \\ $$ $$ \\ $$ $$\mathrm{Part}\:\mathrm{2} \\ $$ $${t}=\mathrm{ln}\left(\mathrm{ln}\left(...\left(\mathrm{ln}\left({k}\right)\right)\right)\right),\:\:\:\:{k}\in\mathbb{R},\:{k}>\mathrm{1} \\ $$ $$\therefore{t}\in\mathbb{C}? \\ $$ $$\mathrm{is}\:{t}\:\mathrm{undefined}? \\ $$

Commented byprakash jain last updated on 11/Apr/16

S=x+iy  e^(x+iy) =a+bi  e^x (cos y+isin y)=a+bi

$${S}={x}+{iy} \\ $$ $${e}^{{x}+{iy}} ={a}+{bi} \\ $$ $${e}^{{x}} \left(\mathrm{cos}\:{y}+{i}\mathrm{sin}\:{y}\right)={a}+{bi} \\ $$

Commented by123456 last updated on 11/Apr/16

e^x cos y=a  e^x sin y=b  e^(2x) =a^2 +b^2   e^x =(√(a^2 +b^2 ))  x=ln (√(a^2 +b^2 ))  tan y=(b/a)  y=arg(a+ib)+2πk

$${e}^{{x}} \mathrm{cos}\:{y}={a} \\ $$ $${e}^{{x}} \mathrm{sin}\:{y}={b} \\ $$ $${e}^{\mathrm{2}{x}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$ $${e}^{{x}} =\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$ $${x}=\mathrm{ln}\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$ $$\mathrm{tan}\:{y}=\frac{{b}}{{a}} \\ $$ $${y}=\mathrm{arg}\left({a}+{ib}\right)+\mathrm{2}\pi{k} \\ $$

Commented byYozzii last updated on 12/Apr/16

After a certain number of  successive ln−applications to k>1, the result  is negative so that the ln−applications  thereafter give t having a complex  result. t would become undefined if  ever t_n =0 in t_(n+1) =lnt_n  , t_1 =lnk, n≥1.    Suppose after k+1 ln−applications  t_(k+1) <0.⇒lnt_k <0⇒0<t_k <1.  lnt_k =(−lnt_k )(−1+0)=(−lnt_k )(cosπ+isinπ)  lnt_k =(−lnt_k )e^(πi) =t_(k+1)   ⇒t_(k+2) =lnt_(k+1) =ln((−lnt_k )e^(πi) )  t_(k+2) =ln(−lnt_k )+πi  ⇒t_(k+2) ∈C.  If {t_n } converges we′d need to solve  u=e^u   where for sufficiently large n,  t_(n+1) =t_(n+2) =....=u.

$${After}\:{a}\:{certain}\:{number}\:{of} \\ $$ $${successive}\:{ln}−{applications}\:{to}\:{k}>\mathrm{1},\:{the}\:{result} \\ $$ $${is}\:{negative}\:{so}\:{that}\:{the}\:{ln}−{applications} \\ $$ $${thereafter}\:{give}\:{t}\:{having}\:{a}\:{complex} \\ $$ $${result}.\:{t}\:{would}\:{become}\:{undefined}\:{if} \\ $$ $${ever}\:{t}_{{n}} =\mathrm{0}\:{in}\:{t}_{{n}+\mathrm{1}} ={lnt}_{{n}} \:,\:{t}_{\mathrm{1}} ={lnk},\:{n}\geqslant\mathrm{1}. \\ $$ $$ \\ $$ $${Suppose}\:{after}\:{k}+\mathrm{1}\:{ln}−{applications} \\ $$ $${t}_{{k}+\mathrm{1}} <\mathrm{0}.\Rightarrow{lnt}_{{k}} <\mathrm{0}\Rightarrow\mathrm{0}<{t}_{{k}} <\mathrm{1}. \\ $$ $${lnt}_{{k}} =\left(−{lnt}_{{k}} \right)\left(−\mathrm{1}+\mathrm{0}\right)=\left(−{lnt}_{{k}} \right)\left({cos}\pi+{isin}\pi\right) \\ $$ $${lnt}_{{k}} =\left(−{lnt}_{{k}} \right){e}^{\pi{i}} ={t}_{{k}+\mathrm{1}} \\ $$ $$\Rightarrow{t}_{{k}+\mathrm{2}} ={lnt}_{{k}+\mathrm{1}} ={ln}\left(\left(−{lnt}_{{k}} \right){e}^{\pi{i}} \right) \\ $$ $${t}_{{k}+\mathrm{2}} ={ln}\left(−{lnt}_{{k}} \right)+\pi{i} \\ $$ $$\Rightarrow{t}_{{k}+\mathrm{2}} \in\mathbb{C}. \\ $$ $${If}\:\left\{{t}_{{n}} \right\}\:{converges}\:{we}'{d}\:{need}\:{to}\:{solve} \\ $$ $${u}={e}^{{u}} \:\:{where}\:{for}\:{sufficiently}\:{large}\:{n}, \\ $$ $${t}_{{n}+\mathrm{1}} ={t}_{{n}+\mathrm{2}} =....={u}. \\ $$

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