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Question Number 2874 by RasheedAhmad last updated on 29/Nov/15
What is the meaning of (i) z→0 (ii) z→z_0 z,z_0 ∈C
$${What}\:{is}\:{the}\:{meaning}\:{of} \\ $$$$\left({i}\right)\:{z}\rightarrow\mathrm{0}\:\:\:\left({ii}\right)\:{z}\rightarrow{z}_{\mathrm{0}} \:\:\:{z},{z}_{\mathrm{0}} \in\mathbb{C} \\ $$
Answered by Filup last updated on 29/Nov/15
(1) z approaches 0 z→0^+ approaches from right side z→0^− approaches from left side e.g. f(x)= { ((x≥0, 1)),((x<0, 2)) :} lim_(x→0^+ ) f(x)=1 lim_(x→0^− ) f(x)=2 (2) z, z_0 ∈C means that both z and z_0 are within the complex set. Simply, they are complex functions. Such as: z=1+2i if z→z_0 , then for complex z, it approaches complex z_0
$$\left(\mathrm{1}\right) \\ $$$${z}\:\mathrm{approaches}\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:{z}\rightarrow\mathrm{0}^{+} \:\mathrm{approaches}\:\mathrm{from}\:\mathrm{right}\:\mathrm{side} \\ $$$$\:\:\:\:\:\:\:\:\:{z}\rightarrow\mathrm{0}^{−} \:\mathrm{approaches}\:\mathrm{from}\:\mathrm{left}\:\mathrm{side} \\ $$$$\mathrm{e}.\mathrm{g}. \\ $$$${f}\left({x}\right)=\begin{cases}{{x}\geqslant\mathrm{0},\:\:\mathrm{1}}\\{{x}<\mathrm{0},\:\:\mathrm{2}}\end{cases} \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{f}\left({x}\right)=\mathrm{1} \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}{f}\left({x}\right)=\mathrm{2} \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$${z},\:{z}_{\mathrm{0}} \in\mathbb{C}\:\:\mathrm{means}\:\mathrm{that}\:\mathrm{both}\:{z}\:\mathrm{and}\:{z}_{\mathrm{0}} \\ $$$$\mathrm{are}\:\mathrm{within}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{set}.\:\mathrm{Simply}, \\ $$$$\mathrm{they}\:\mathrm{are}\:\mathrm{complex}\:\mathrm{functions}. \\ $$$$\mathrm{Such}\:\mathrm{as}:\:\:\:{z}=\mathrm{1}+\mathrm{2}{i} \\ $$$$ \\ $$$$\mathrm{if}\:\:{z}\rightarrow{z}_{\mathrm{0}} ,\:\mathrm{then}\:\mathrm{for}\:\mathrm{complex}\:{z},\:\mathrm{it}\: \\ $$$$\mathrm{approaches}\:\mathrm{complex}\:{z}_{\mathrm{0}} \\ $$
Answered by Yozzi last updated on 29/Nov/15
For z∈C defined simply as z=x+iy where x,y∈R , (i)z→0⇒x+iy→0⇒(x→0)∧(y→0) (ii)z→z_0 ⇒x+iy→x_0 +iy_0 ⇒(x→x_0 )∧(y→y_0 ). E.g Define f(z)=z^2 , z∈C. ⇒lim_(z→0) f(z)=lim_(z→0) z^2 =0^2 =0 But, if z=x+iy, x,y∈R, f(x)=(x+iy)^2 =x^2 −y^2 +2iyx ∴lim_(z→0) f(z)=lim_((x,y)→(0,0)) (x^2 −y^2 +2ixy) =0^2 −0^2 +2i×0×0 =0 Similarly, suppose z_0 =3+i. lim_(z→z_0 ) f(z)=lim_(z→z_0 ) z^2 =(3+i)^2 =8+6i Alternatively, lim_(z→z_0 ) f(z)=lim_((x,y)→(3,1)) x^2 −y^2 +2ixy =9−1+2i×1×3 =8+6i
$${For}\:{z}\in\mathbb{C}\:\:{defined}\:{simply}\:{as}\:{z}={x}+{iy} \\ $$$${where}\:{x},{y}\in\mathbb{R}\:, \\ $$$$\left({i}\right){z}\rightarrow\mathrm{0}\Rightarrow{x}+{iy}\rightarrow\mathrm{0}\Rightarrow\left({x}\rightarrow\mathrm{0}\right)\wedge\left({y}\rightarrow\mathrm{0}\right) \\ $$$$\left({ii}\right){z}\rightarrow{z}_{\mathrm{0}} \Rightarrow{x}+{iy}\rightarrow{x}_{\mathrm{0}} +{iy}_{\mathrm{0}} \\ $$$$\Rightarrow\left({x}\rightarrow{x}_{\mathrm{0}} \right)\wedge\left({y}\rightarrow{y}_{\mathrm{0}} \right). \\ $$$$ \\ $$$${E}.{g}\:{Define}\:{f}\left({z}\right)={z}^{\mathrm{2}} ,\:{z}\in\mathbb{C}. \\ $$$$\Rightarrow\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\left({z}\right)=\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}{z}^{\mathrm{2}} =\mathrm{0}^{\mathrm{2}} =\mathrm{0} \\ $$$${But},\:{if}\:{z}={x}+{iy},\:{x},{y}\in\mathbb{R}, \\ $$$${f}\left({x}\right)=\left({x}+{iy}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} −{y}^{\mathrm{2}} +\mathrm{2}{iyx} \\ $$$$\therefore\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\left({z}\right)=\underset{\left({x},{y}\right)\rightarrow\left(\mathrm{0},\mathrm{0}\right)} {\mathrm{lim}}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} +\mathrm{2}{ixy}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0}^{\mathrm{2}} −\mathrm{0}^{\mathrm{2}} +\mathrm{2}{i}×\mathrm{0}×\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0} \\ $$$$ \\ $$$${Similarly},\:{suppose}\:{z}_{\mathrm{0}} =\mathrm{3}+{i}. \\ $$$$\underset{{z}\rightarrow{z}_{\mathrm{0}} } {\mathrm{lim}}{f}\left({z}\right)=\underset{{z}\rightarrow{z}_{\mathrm{0}} } {\mathrm{lim}}{z}^{\mathrm{2}} =\left(\mathrm{3}+{i}\right)^{\mathrm{2}} =\mathrm{8}+\mathrm{6}{i} \\ $$$${Alternatively}, \\ $$$$\underset{{z}\rightarrow{z}_{\mathrm{0}} } {\mathrm{lim}}{f}\left({z}\right)=\underset{\left({x},{y}\right)\rightarrow\left(\mathrm{3},\mathrm{1}\right)} {\mathrm{lim}}{x}^{\mathrm{2}} −{y}^{\mathrm{2}} +\mathrm{2}{ixy} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{9}−\mathrm{1}+\mathrm{2}{i}×\mathrm{1}×\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{8}+\mathrm{6}{i} \\ $$$$ \\ $$
Commented by Yozzi last updated on 29/Nov/15
I see. Thanks!
$${I}\:{see}.\:{Thanks}! \\ $$
Commented by 123456 last updated on 29/Nov/15
however there infinite ways to aproach z_0 on C, so if the limit depends on path that you taked, them it doenst exist [like in the R, lim_(x→0^− ) f(x)≠lim_(x→0^+ ) f(x),lim_(x→0) f(x)=∄]
$$\mathrm{however}\:\mathrm{there}\:\mathrm{infinite}\:\mathrm{ways}\:\mathrm{to}\:\mathrm{aproach} \\ $$$${z}_{\mathrm{0}} \:\mathrm{on}\:\mathbb{C},\:\mathrm{so}\:\mathrm{if}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{depends}\:\mathrm{on}\:\mathrm{path} \\ $$$$\mathrm{that}\:\mathrm{you}\:\mathrm{taked},\:\mathrm{them}\:\mathrm{it}\:\mathrm{doenst}\:\mathrm{exist} \\ $$$$\left[\mathrm{like}\:\mathrm{in}\:\mathrm{the}\:\mathbb{R},\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}{f}\left({x}\right)\neq\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{f}\left({x}\right),\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\left({x}\right)=\nexists\right] \\ $$
Commented by prakash jain last updated on 29/Nov/15
The same concept (path dependency applies) when computing limits for functions of more than one variable. for z→0 or z→z_0 limits it is sometimes easier to convert to polar coordinates then consider limit on r with no constraints on θ.
$$\mathrm{The}\:\mathrm{same}\:\mathrm{concept}\:\left(\mathrm{path}\:\mathrm{dependency}\:\mathrm{applies}\right) \\ $$$$\mathrm{when}\:\mathrm{computing}\:\mathrm{limits}\:\mathrm{for}\:\mathrm{functions}\:\mathrm{of} \\ $$$$\mathrm{more}\:\mathrm{than}\:\mathrm{one}\:\mathrm{variable}. \\ $$$$\mathrm{for}\:{z}\rightarrow\mathrm{0}\:\mathrm{or}\:{z}\rightarrow{z}_{\mathrm{0}} \:\mathrm{limits}\:\mathrm{it}\:\mathrm{is}\:\mathrm{sometimes}\:\mathrm{easier}\:\mathrm{to} \\ $$$$\mathrm{convert}\:\mathrm{to}\:\mathrm{polar}\:\mathrm{coordinates}\:\mathrm{then}\:\mathrm{consider} \\ $$$$\mathrm{limit}\:\mathrm{on}\:{r}\:\mathrm{with}\:\mathrm{no}\:\mathrm{constraints}\:\mathrm{on}\:\theta. \\ $$
Commented by Rasheed Soomro last updated on 29/Nov/15
What are other ways, for example, of approaching z_0 and what is ′path dependancy′?
$${What}\:{are}\:{other}\:{ways},\:{for}\:{example},\:{of}\:{approaching}\:{z}_{\mathrm{0}} \\ $$$${and}\:{what}\:{is}\:'{path}\:{dependancy}'? \\ $$
Commented by 123456 last updated on 29/Nov/15
lets z_0 =0 we can aproach z by z=te^(ıt) ,t→0^+ (path 1) z=re^(ıθ) ,r→0 (path 2) the first path is a spiral and the second is a stray line with angle θ with real line path depedency=limits depends on the path you take to aproach z_0 in real case with one varriable we have only the paths x→0^+ x→0^− however in C we have infinites path as you see in above a example of two path, there are more paths that you can take.
$$\mathrm{lets}\:{z}_{\mathrm{0}} =\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{aproach}\:{z}\:{by} \\ $$$${z}={te}^{\imath{t}} ,{t}\rightarrow\mathrm{0}^{+} \:\left(\mathrm{path}\:\mathrm{1}\right)\:\: \\ $$$${z}={re}^{\imath\theta} ,{r}\rightarrow\mathrm{0}\:\left(\mathrm{path}\:\mathrm{2}\right) \\ $$$$\mathrm{the}\:\mathrm{first}\:\mathrm{path}\:\mathrm{is}\:\mathrm{a}\:\mathrm{spiral}\:\mathrm{and}\:\mathrm{the}\:\mathrm{second} \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{stray}\:\mathrm{line}\:\mathrm{with}\:\mathrm{angle}\:\theta\:\mathrm{with}\:\mathrm{real}\:\mathrm{line} \\ $$$$\mathrm{path}\:\mathrm{depedency}=\mathrm{limits}\:\mathrm{depends}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{path}\:\mathrm{you}\:\mathrm{take}\:\mathrm{to}\:\mathrm{aproach}\:{z}_{\mathrm{0}} \\ $$$$\mathrm{in}\:\mathrm{real}\:\mathrm{case}\:\mathrm{with}\:\mathrm{one}\:\mathrm{varriable}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{only}\:\mathrm{the}\:\mathrm{paths} \\ $$$${x}\rightarrow\mathrm{0}^{+} \\ $$$${x}\rightarrow\mathrm{0}^{−} \\ $$$$\mathrm{however}\:\mathrm{in}\:\mathbb{C}\:\mathrm{we}\:\mathrm{have}\:\mathrm{infinites}\:\mathrm{path} \\ $$$$\mathrm{as}\:\mathrm{you}\:\mathrm{see}\:\mathrm{in}\:\mathrm{above}\:\mathrm{a}\:\mathrm{example}\:\mathrm{of}\:\mathrm{two} \\ $$$$\mathrm{path},\:\mathrm{there}\:\mathrm{are}\:\mathrm{more}\:\mathrm{paths}\:\mathrm{that}\:\mathrm{you} \\ $$$$\mathrm{can}\:\mathrm{take}. \\ $$
Commented by Rasheed Soomro last updated on 29/Nov/15
T HANK^(Sss) for all of YOU ! I always have learnt from you.
$$\mathcal{T}\:\mathcal{HANK}^{\mathcal{S}\boldsymbol{{s}}{s}} \:\boldsymbol{{for}}\:\boldsymbol{{all}}\:\boldsymbol{{of}}\:\boldsymbol{\mathcal{Y}}\mathcal{OU}\:! \\ $$$$\mathcal{I}\:{always}\:{have}\:{learnt}\:{from}\:{you}. \\ $$

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