0-i-

Question Number 222441 by fantastic last updated on 27/Jun/25

$$\mathrm{0}^{{i}} \\ $$

Answered by wewji12 last updated on 27/Jun/25

0 because f(z)=0^z for all z∈R\{0} f(z)=0 const. f(z)= { (( 1 , z=0)),(( 0 , z∈R\{0})) :}

$$\mathrm{0}\:\mathrm{because}\:{f}\left({z}\right)=\mathrm{0}^{{z}} \:\mathrm{for}\:\mathrm{all}\:{z}\in\mathbb{R}\backslash\left\{\mathrm{0}\right\} \\ $$$${f}\left({z}\right)=\mathrm{0}\:\mathrm{const}. \\ $$$${f}\left({z}\right)=\begin{cases}{\:\mathrm{1}\:,\:{z}=\mathrm{0}}\\{\:\mathrm{0}\:,\:{z}\in\mathbb{R}\backslash\left\{\mathrm{0}\right\}}\end{cases} \\ $$

Commented by Ghisom last updated on 27/Jun/25

f_1 (z)=z^0 lim_(z→0) f_1 (z) =1 ⇒^? 0^0 =1 f_2 (z)=0^z lim_(z→0) f_2 (z) =0 ⇒^? 0^0 =0 an existing limit ≠ a general definition

$${f}_{\mathrm{1}} \left({z}\right)={z}^{\mathrm{0}} \:\:\:\:\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{f}_{\mathrm{1}} \left({z}\right)\:=\mathrm{1}\:\overset{?} {\Rightarrow}\:\mathrm{0}^{\mathrm{0}} =\mathrm{1} \\ $$$${f}_{\mathrm{2}} \left({z}\right)=\mathrm{0}^{{z}} \:\:\:\:\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{f}_{\mathrm{2}} \left({z}\right)\:=\mathrm{0}\:\overset{?} {\Rightarrow}\:\mathrm{0}^{\mathrm{0}} =\mathrm{0} \\ $$$$\mathrm{an}\:\mathrm{existing}\:\mathrm{limit}\:\neq\:\mathrm{a}\:\mathrm{general}\:\mathrm{definition} \\ $$

Answered by Ghisom last updated on 27/Jun/25

let a, b, r∈R, z∈C r^z =r^(a+bi) =r^a r^(bi) =r^a e^(iln b) in the given case r=0, a=0, b=1 0^i =0^(0+1i) =0^0 e^(iln 1) =0^0 e^0 =0^0 ×1 0^0 is not defined ⇒ 0^i is not defined

$$\mathrm{let}\:{a},\:{b},\:{r}\in\mathbb{R},\:{z}\in\mathbb{C} \\ $$$${r}^{{z}} ={r}^{{a}+{b}\mathrm{i}} ={r}^{{a}} {r}^{{b}\mathrm{i}} ={r}^{{a}} \mathrm{e}^{\mathrm{iln}\:{b}} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{given}\:\mathrm{case}\:{r}=\mathrm{0},\:{a}=\mathrm{0},\:{b}=\mathrm{1} \\ $$$$\mathrm{0}^{\mathrm{i}} =\mathrm{0}^{\mathrm{0}+\mathrm{1i}} =\mathrm{0}^{\mathrm{0}} \mathrm{e}^{\mathrm{iln}\:\mathrm{1}} =\mathrm{0}^{\mathrm{0}} \mathrm{e}^{\mathrm{0}} =\mathrm{0}^{\mathrm{0}} ×\mathrm{1} \\ $$$$\mathrm{0}^{\mathrm{0}} \:\mathrm{is}\:\mathrm{not}\:\mathrm{defined}\:\Rightarrow\:\mathrm{0}^{\mathrm{i}} \:\mathrm{is}\:\mathrm{not}\:\mathrm{defined} \\ $$

Commented by wewji12 last updated on 27/Jun/25

$$\mathrm{0}^{\mathrm{0}} =\mathrm{1}….. \\ $$$$\mathrm{yeah}….\:\mathrm{if}\:{f}\left({z}\right)={z}^{{z}} \:…. \\ $$

Commented by Ghisom last updated on 27/Jun/25

your method in a different case: f(x)=(x/x) ⇒ f(0)=(0/0)=1 great! g(x)=(x^3 /((1−cos x)sin x)) ⇒ g(0)=(0/0)=1 but lim_(x→0) g(x) =2 ...

$$\mathrm{your}\:\mathrm{method}\:\mathrm{in}\:\mathrm{a}\:\mathrm{different}\:\mathrm{case}: \\ $$$${f}\left({x}\right)=\frac{{x}}{{x}}\:\Rightarrow\:{f}\left(\mathrm{0}\right)=\frac{\mathrm{0}}{\mathrm{0}}=\mathrm{1} \\ $$$$\mathrm{great}! \\ $$$${g}\left({x}\right)=\frac{{x}^{\mathrm{3}} }{\left(\mathrm{1}−\mathrm{cos}\:{x}\right)\mathrm{sin}\:{x}}\:\Rightarrow\:{g}\left(\mathrm{0}\right)=\frac{\mathrm{0}}{\mathrm{0}}=\mathrm{1} \\ $$$$\mathrm{but}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{g}\left({x}\right)\:=\mathrm{2} \\ $$$$… \\ $$

Commented by wewji12 last updated on 27/Jun/25

yeah... you are right... so i told you ′′if f(z)=z^z ′′

$$\mathrm{yeah}…\:\mathrm{you}\:\mathrm{are}\:\mathrm{right}…\:\mathrm{so}\:\mathrm{i}\:\mathrm{told}\:\mathrm{you} \\ $$$$''\mathrm{if}\:{f}\left({z}\right)={z}^{{z}} '' \\ $$

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