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Question Number 4582 by FilupSmith last updated on 08/Feb/16
S=log_a i i^2 =−1 a∈R Solve S extra: what if a∈C?
$${S}=\mathrm{log}_{{a}} {i} \\ $$$${i}^{\mathrm{2}} =−\mathrm{1} \\ $$$${a}\in\mathbb{R} \\ $$$${Solve}\:{S} \\ $$$$ \\ $$$${extra}: \\ $$$$\mathrm{what}\:\mathrm{if}\:\:{a}\in\mathbb{C}? \\ $$
Answered by Yozzii last updated on 09/Feb/16
By change of base, S=log_a i=((lni)/(lna)). Now, i=0+i×1=cosα+isinα ∴ sinα=1⇒α=nπ+(−1)^n (π/2) (n∈Z) cosα=cos(nπ+(−1)^n 0.5π)=(−1)^n ×0+0×1=0 ∴ i=e^(i(nπ+(−1)^n (π/2))) ⇒S=((lnexp(i{nπ+(((−1)^n π)/2)}))/(lna)) S=(i/(lna))(nπ+((π(−1)^n )/2)) (1)If a>0, then lna∈R and indeed S=(i/(lna))(nπ+((π(−1)^n )/2)) (n∈Z). So, S can take infinitely many values for any given a>0 since ∣Z∣=ℵ_0 . (2)If a=0, S is undefined since the graph of y=lnx is discontinuous at x=0. However, lim_(x→0^+ ) lnx=−∞ ⇒ lim_(a→0^+ ) S=lim_(a→0^+ ) (i/(lna))(nπ+0.5π(−1)^n )=(i/(−∞))(nπ+0.5(−1)^n π)=0×i=0. (3)If a<0⇒lna=ln(−r)=ln(re^((2w+1)πi) ) lna=lnr+(2w+1)πi where r>0 and w∈Z. ∴S=((i(nπ+0.5π(−1)^n ))/((lnr+(2w+1)πi)))=((i(nπ+0.5π(−1)^n ))/((ln(−a)+(2w+1)πi))) S=((i(ln(−a)−(2w+1)πi)(nπ+0.5π(−1)^n ))/(({ln(−a)}^2 +(2w+1)^2 π^2 ))) S=((((2w+1)π+iln(−a))(nπ+0.5π(−1)^n ))/(2(ln^2 (−a)+(2w+1)^2 π^2 ))) (n,w∈Z) In this case, S is not unique for a given a<0, similar to when a>0. For a given a<0, S takes infinitely many values determined by the integer pairs (n,w) where n=w or n≠w. (4)For a∈C⇒let a=ce^(iθ) where c>0 and θ=arg(a)=argument of a. ∴ lna=lnce^(iθ) =lnc+iθ ∴ S=((i(nπ+0.5(−1)^n π))/(lnc+iθ)) S=((πi(lnc−iθ)(n+0.5(−1)^n ))/({lnc}^2 +θ^2 )) S=((π(n+0.5(−1)^n )(θ+ilnc))/({lnc}^2 +θ^2 )) (n∈Z)
$${By}\:{change}\:{of}\:{base},\: \\ $$$${S}={log}_{{a}} {i}=\frac{{lni}}{{lna}}. \\ $$$${Now},\:{i}=\mathrm{0}+{i}×\mathrm{1}={cos}\alpha+{isin}\alpha \\ $$$$\therefore\:{sin}\alpha=\mathrm{1}\Rightarrow\alpha={n}\pi+\left(−\mathrm{1}\right)^{{n}} \frac{\pi}{\mathrm{2}}\:\:\left({n}\in\mathbb{Z}\right) \\ $$$${cos}\alpha={cos}\left({n}\pi+\left(−\mathrm{1}\right)^{{n}} \mathrm{0}.\mathrm{5}\pi\right)=\left(−\mathrm{1}\right)^{{n}} ×\mathrm{0}+\mathrm{0}×\mathrm{1}=\mathrm{0} \\ $$$$\therefore\:{i}={e}^{{i}\left({n}\pi+\left(−\mathrm{1}\right)^{{n}} \frac{\pi}{\mathrm{2}}\right)} \\ $$$$\Rightarrow{S}=\frac{{lnexp}\left({i}\left\{{n}\pi+\frac{\left(−\mathrm{1}\right)^{{n}} \pi}{\mathrm{2}}\right\}\right)}{{lna}} \\ $$$${S}=\frac{{i}}{{lna}}\left({n}\pi+\frac{\pi\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}}\right) \\ $$$$\left(\mathrm{1}\right){If}\:{a}>\mathrm{0},\:{then}\:{lna}\in\mathbb{R}\:{and}\:{indeed}\:{S}=\frac{{i}}{{lna}}\left({n}\pi+\frac{\pi\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}}\right)\:\:\left({n}\in\mathbb{Z}\right). \\ $$$${So},\:{S}\:{can}\:{take}\:{infinitely}\:{many}\:{values} \\ $$$${for}\:{any}\:{given}\:{a}>\mathrm{0}\:{since}\:\mid\mathbb{Z}\mid=\aleph_{\mathrm{0}} . \\ $$$$ \\ $$$$\left(\mathrm{2}\right){If}\:{a}=\mathrm{0},\:{S}\:{is}\:{undefined}\:{since}\:{the}\: \\ $$$${graph}\:{of}\:{y}={lnx}\:{is}\:{discontinuous}\:{at}\: \\ $$$${x}=\mathrm{0}.\:{However},\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{lnx}=−\infty \\ $$$$\Rightarrow\:\underset{{a}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{S}=\underset{{a}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{{i}}{{lna}}\left({n}\pi+\mathrm{0}.\mathrm{5}\pi\left(−\mathrm{1}\right)^{{n}} \right)=\frac{{i}}{−\infty}\left({n}\pi+\mathrm{0}.\mathrm{5}\left(−\mathrm{1}\right)^{{n}} \pi\right)=\mathrm{0}×{i}=\mathrm{0}. \\ $$$$ \\ $$$$\left(\mathrm{3}\right){If}\:\:{a}<\mathrm{0}\Rightarrow{lna}={ln}\left(−{r}\right)={ln}\left({re}^{\left(\mathrm{2}{w}+\mathrm{1}\right)\pi{i}} \right) \\ $$$${lna}={lnr}+\left(\mathrm{2}{w}+\mathrm{1}\right)\pi{i}\:{where}\:{r}>\mathrm{0}\:{and}\:{w}\in\mathbb{Z}. \\ $$$$\therefore{S}=\frac{{i}\left({n}\pi+\mathrm{0}.\mathrm{5}\pi\left(−\mathrm{1}\right)^{{n}} \right)}{\left({lnr}+\left(\mathrm{2}{w}+\mathrm{1}\right)\pi{i}\right)}=\frac{{i}\left({n}\pi+\mathrm{0}.\mathrm{5}\pi\left(−\mathrm{1}\right)^{{n}} \right)}{\left({ln}\left(−{a}\right)+\left(\mathrm{2}{w}+\mathrm{1}\right)\pi{i}\right)}\: \\ $$$${S}=\frac{{i}\left({ln}\left(−{a}\right)−\left(\mathrm{2}{w}+\mathrm{1}\right)\pi{i}\right)\left({n}\pi+\mathrm{0}.\mathrm{5}\pi\left(−\mathrm{1}\right)^{{n}} \right)}{\left(\left\{{ln}\left(−{a}\right)\right\}^{\mathrm{2}} +\left(\mathrm{2}{w}+\mathrm{1}\right)^{\mathrm{2}} \pi^{\mathrm{2}} \right)} \\ $$$${S}=\frac{\left(\left(\mathrm{2}{w}+\mathrm{1}\right)\pi+{iln}\left(−{a}\right)\right)\left({n}\pi+\mathrm{0}.\mathrm{5}\pi\left(−\mathrm{1}\right)^{{n}} \right)}{\mathrm{2}\left({ln}^{\mathrm{2}} \left(−{a}\right)+\left(\mathrm{2}{w}+\mathrm{1}\right)^{\mathrm{2}} \pi^{\mathrm{2}} \right)}\:\:\:\left({n},{w}\in\mathbb{Z}\right) \\ $$$${In}\:{this}\:{case},\:{S}\:{is}\:{not}\:{unique}\:{for}\:{a}\:{given}\:{a}<\mathrm{0}, \\ $$$${similar}\:{to}\:{when}\:{a}>\mathrm{0}. \\ $$$${For}\:{a}\:{given}\:{a}<\mathrm{0},\:{S}\:{takes}\:{infinitely}\: \\ $$$${many}\:{values}\:{determined}\:{by}\:{the}\:{integer} \\ $$$${pairs}\:\left({n},{w}\right)\:{where}\:{n}={w}\:{or}\:{n}\neq{w}. \\ $$$$ \\ $$$$\left(\mathrm{4}\right){For}\:{a}\in\mathbb{C}\Rightarrow{let}\:{a}={ce}^{{i}\theta} \:{where}\:{c}>\mathrm{0} \\ $$$${and}\:\theta={arg}\left({a}\right)={argument}\:{of}\:{a}. \\ $$$$\therefore\:{lna}={lnce}^{{i}\theta} ={lnc}+{i}\theta \\ $$$$\therefore\:{S}=\frac{{i}\left({n}\pi+\mathrm{0}.\mathrm{5}\left(−\mathrm{1}\right)^{{n}} \pi\right)}{{lnc}+{i}\theta}\: \\ $$$${S}=\frac{\pi{i}\left({lnc}−{i}\theta\right)\left({n}+\mathrm{0}.\mathrm{5}\left(−\mathrm{1}\right)^{{n}} \right)}{\left\{{lnc}\right\}^{\mathrm{2}} +\theta^{\mathrm{2}} }\:\: \\ $$$${S}=\frac{\pi\left({n}+\mathrm{0}.\mathrm{5}\left(−\mathrm{1}\right)^{{n}} \right)\left(\theta+{ilnc}\right)}{\left\{{lnc}\right\}^{\mathrm{2}} +\theta^{\mathrm{2}} }\:\:\left({n}\in\mathbb{Z}\right) \\ $$$$ \\ $$

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