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Question Number 222462 by fantastic last updated on 27/Jun/25

$${i}^{{i}} =?? \\ $$

Answered by vnm last updated on 27/Jun/25

=(e^(i((π/2)+2nπ)) )^i =e^(π(−(1/2)+2n)) , n∈Z

$$=\left({e}^{{i}\left(\frac{\pi}{\mathrm{2}}+\mathrm{2}{n}\pi\right)} \right)^{{i}} ={e}^{\pi\left(−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2}{n}\right)} ,\:{n}\in\mathbb{Z} \\ $$

Commented by Frix last updated on 27/Jun/25

No. x^y is a simple calculation with a unique result. Recently people started to confuse these things. It′s like claiming (√4)=−2 which is also wrong. For x^y with x, y ∈C we need x=re^(iθ) ; r≥0∧ −π<θ≤π y=a+bi ⇒ unique result (re^(iθ) )^(a+ib) =r^(a+ib) e^(−bθ+iaθ) =e^(ln r (a+ib)) e^(−bθ+iaθ) = =e^(aln r −bθ) e^(i(aθ+bln r)) = =e^(aln r −bθ) (cos (aθ+bln r) +i sin (aθ+bln r)

$$\mathrm{No}. \\ $$$${x}^{{y}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{simple}\:\mathrm{calculation}\:\mathrm{with}\:\mathrm{a}\:\mathrm{unique} \\ $$$$\mathrm{result}. \\ $$$$\mathrm{Recently}\:\mathrm{people}\:\mathrm{started}\:\mathrm{to}\:\mathrm{confuse}\:\mathrm{these} \\ $$$$\mathrm{things}.\:\mathrm{It}'\mathrm{s}\:\mathrm{like}\:\mathrm{claiming}\:\sqrt{\mathrm{4}}=−\mathrm{2}\:\mathrm{which}\:\mathrm{is} \\ $$$$\mathrm{also}\:\mathrm{wrong}. \\ $$$$ \\ $$$$\mathrm{For} \\ $$$${x}^{{y}} \:\mathrm{with}\:{x},\:{y}\:\in\mathbb{C}\:\mathrm{we}\:\mathrm{need} \\ $$$${x}={r}\mathrm{e}^{\mathrm{i}\theta} ;\:{r}\geqslant\mathrm{0}\wedge\:−\pi<\theta\leqslant\pi \\ $$$${y}={a}+{b}\mathrm{i} \\ $$$$\Rightarrow\:\mathrm{unique}\:\mathrm{result} \\ $$$$\left({r}\mathrm{e}^{\mathrm{i}\theta} \right)^{{a}+\mathrm{i}{b}} ={r}^{{a}+\mathrm{i}{b}} \mathrm{e}^{−{b}\theta+\mathrm{i}{a}\theta} =\mathrm{e}^{\mathrm{ln}\:{r}\:\left({a}+\mathrm{i}{b}\right)} \mathrm{e}^{−{b}\theta+\mathrm{i}{a}\theta} = \\ $$$$=\mathrm{e}^{{a}\mathrm{ln}\:{r}\:−{b}\theta} \mathrm{e}^{\mathrm{i}\left({a}\theta+{b}\mathrm{ln}\:{r}\right)} = \\ $$$$=\mathrm{e}^{{a}\mathrm{ln}\:{r}\:−{b}\theta} \left(\mathrm{cos}\:\left({a}\theta+{b}\mathrm{ln}\:{r}\right)\:+\mathrm{i}\:\mathrm{sin}\:\left({a}\theta+{b}\mathrm{ln}\:{r}\right)\right. \\ $$

Commented by mr W last updated on 28/Jun/25

i agree with you. it′s like: cos^(−1) (1/2) is unique (=(π/3)). but if cos x=(1/2), x is not unique and x=2nπ±(π/3). we can not say cos^(−1) (1/2)=2nπ±(π/3).

$${i}\:{agree}\:{with}\:{you}.\:{it}'{s}\:{like}: \\ $$$$\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\:{is}\:{unique}\:\left(=\frac{\pi}{\mathrm{3}}\right).\: \\ $$$${but}\:{if}\:\mathrm{cos}\:{x}=\frac{\mathrm{1}}{\mathrm{2}},\:{x}\:{is}\:{not}\:{unique}\:{and} \\ $$$${x}=\mathrm{2}{n}\pi\pm\frac{\pi}{\mathrm{3}}. \\ $$$${we}\:{can}\:{not}\:{say}\: \\ $$$$\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}=\mathrm{2}{n}\pi\pm\frac{\pi}{\mathrm{3}}. \\ $$

Answered by Frix last updated on 27/Jun/25

$$\mathrm{i}^{\mathrm{i}} =\left(\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} \right)^{\mathrm{i}} =\mathrm{e}^{−\frac{\pi}{\mathrm{2}}} \\ $$

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