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Question Number 179886 by yaojunon2t last updated on 03/Nov/22

Commented by yaojunon2t last updated on 03/Nov/22

∵e^(iθ) =cos θ+i sin θ,θ=(π/2)+2kπ ∴e^(i((π/2)+2kπ)) =cos (π/2)+i sin (π/2) ∴e^(i((π/2)+2kπ)) =i ∴i^i =(e^(i((π/2)+2kπ)) )^i =e^(i^2 ((π/2)+2kπ)) =e^(−((π/2)+2kπ)) (k∈Z) ⇒i^i =e^((3π)/2) ,e^(−(π/2)) ,e^(−((5π)/2)) ,... ∴i^i ≈111.32,0.2079,0.000388...

$$\because{e}^{{i}\theta} =\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta,\theta=\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi \\ $$$$\therefore{e}^{{i}\left(\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi\right)} =\mathrm{cos}\:\frac{\pi}{\mathrm{2}}+{i}\:\mathrm{sin}\:\frac{\pi}{\mathrm{2}} \\ $$$$\therefore{e}^{{i}\left(\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi\right)} ={i} \\ $$$$\therefore{i}^{{i}} =\left({e}^{{i}\left(\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi\right)} \right)^{{i}} ={e}^{{i}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi\right)} ={e}^{−\left(\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi\right)} \:\:\:\:\left({k}\in\mathbb{Z}\right) \\ $$$$\Rightarrow{i}^{{i}} ={e}^{\frac{\mathrm{3}\pi}{\mathrm{2}}} ,{e}^{−\frac{\pi}{\mathrm{2}}} ,{e}^{−\frac{\mathrm{5}\pi}{\mathrm{2}}} ,... \\ $$$$\therefore{i}^{{i}} \approx\mathrm{111}.\mathrm{32},\mathrm{0}.\mathrm{2079},\mathrm{0}.\mathrm{000388}... \\ $$

Commented by yaojunon2t last updated on 03/Nov/22

is the solution for i^i correct?

$${is}\:{the}\:{solution}\:{for}\:{i}^{{i}} \:{correct}? \\ $$

Commented by mr W last updated on 03/Nov/22

yes

$${yes} \\ $$

Commented by MJS_new last updated on 03/Nov/22

since when do we get several solutions for a simple calculation? it′s a similar misconception as (√4)=^? ±2 we′re not solving anything here, we′re just calculating a simple root i=e^(iπ/2) ⇒ i^i =(e^(iπ/2) )^i =e^(−π/2) it′s unique as 2^2 =4.

$$\mathrm{since}\:\mathrm{when}\:\mathrm{do}\:\mathrm{we}\:\mathrm{get}\:\mathrm{several}\:\mathrm{solutions}\:\mathrm{for}\:\mathrm{a} \\ $$$$\mathrm{simple}\:\mathrm{calculation}? \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{similar}\:\mathrm{misconception}\:\mathrm{as}\:\sqrt{\mathrm{4}}\overset{?} {=}\pm\mathrm{2} \\ $$$$\mathrm{we}'\mathrm{re}\:\mathrm{not}\:\mathrm{solving}\:\mathrm{anything}\:\mathrm{here},\:\mathrm{we}'\mathrm{re}\:\mathrm{just} \\ $$$$\mathrm{calculating}\:\mathrm{a}\:\mathrm{simple}\:\mathrm{root} \\ $$$$\mathrm{i}=\mathrm{e}^{\mathrm{i}\pi/\mathrm{2}} \:\Rightarrow\:\mathrm{i}^{\mathrm{i}} =\left(\mathrm{e}^{\mathrm{i}\pi/\mathrm{2}} \right)^{\mathrm{i}} =\mathrm{e}^{−\pi/\mathrm{2}} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{unique}\:\mathrm{as}\:\mathrm{2}^{\mathrm{2}} =\mathrm{4}. \\ $$

Commented by yaojunon2t last updated on 04/Nov/22

thanks sir

$${thanks}\:{sir} \\ $$

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