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Question Number 120195 by SOMEDAVONG last updated on 30/Oct/20

C .lim_(z→0) ((Re(z).Im(z))/(Re(z)+Im(z))) =?

$$\mathbb{C}\:.\underset{\mathrm{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{Re}\left(\mathrm{z}\right).\mathrm{Im}\left(\mathrm{z}\right)}{\mathrm{Re}\left(\mathrm{z}\right)+\mathrm{Im}\left(\mathrm{z}\right)}\:=? \\ $$

Answered by snipers237 last updated on 30/Oct/20

z_n =(1/n_ )−i(1/n) →_∞ 0 and s_n =(1/n)→_∞ 0 ((Re(z_n )Im(z_n ))/(Re(z_n )+Im(z_n )))→_0 ∞ and ((Re(s_n )Im(s_n ))/(Re(s_n )+Im(s_n )))→_0 0 That allows to conclude that the limit doesn′t exist.

$${z}_{{n}} =\frac{\mathrm{1}}{{n}_{} }−{i}\frac{\mathrm{1}}{{n}}\:\underset{\infty} {\rightarrow}\mathrm{0}\:\:\:\:\:\:{and}\:\:{s}_{{n}} =\frac{\mathrm{1}}{{n}}\underset{\infty} {\rightarrow}\mathrm{0} \\ $$$$\frac{{Re}\left({z}_{{n}} \right){Im}\left({z}_{{n}} \right)}{{Re}\left({z}_{{n}} \right)+{Im}\left({z}_{{n}} \right)}\underset{\mathrm{0}} {\rightarrow}\infty\:\:\:\:\:{and}\:\:\frac{{Re}\left({s}_{{n}} \right){Im}\left({s}_{{n}} \right)}{{Re}\left({s}_{{n}} \right)+{Im}\left({s}_{{n}} \right)}\underset{\mathrm{0}} {\rightarrow}\mathrm{0} \\ $$$${That}\:{allows}\:{to}\:{conclude}\:{that}\:{the}\:{limit}\:{doesn}'{t}\:{exist}. \\ $$

Commented by SOMEDAVONG last updated on 30/Oct/20

Thanks sir

$$\mathrm{Thanks}\:\mathrm{sir}\: \\ $$

Answered by MJS_new last updated on 30/Oct/20

z=re^(iθ) =rcos θ +irsin θ z→0 ⇒ r→0 but θ has no effect lim_(z→0) ((Re (z) Im (z))/(Re (z) +Im (z))) = =lim_(r→0) ((r^2 cos θ sin θ)/(r(cos θ +sin θ))) =((cos θ sin θ)/(cos θ +sin θ))lim_(r→0) (r^2 /r) =0

$${z}={r}\mathrm{e}^{\mathrm{i}\theta} ={r}\mathrm{cos}\:\theta\:+\mathrm{i}{r}\mathrm{sin}\:\theta \\ $$$${z}\rightarrow\mathrm{0}\:\Rightarrow\:{r}\rightarrow\mathrm{0}\:\mathrm{but}\:\theta\:\mathrm{has}\:\mathrm{no}\:\mathrm{effect} \\ $$$$\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{Re}\:\left({z}\right)\:\mathrm{Im}\:\left({z}\right)}{\mathrm{Re}\:\left({z}\right)\:+\mathrm{Im}\:\left({z}\right)}\:= \\ $$$$=\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{r}^{\mathrm{2}} \mathrm{cos}\:\theta\:\mathrm{sin}\:\theta}{{r}\left(\mathrm{cos}\:\theta\:+\mathrm{sin}\:\theta\right)}\:=\frac{\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta\:+\mathrm{sin}\:\theta}\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{r}^{\mathrm{2}} }{{r}}\:=\mathrm{0}\: \\ $$

Answered by MJS_new last updated on 30/Oct/20

let z=(1/a)+(1/b)i z→0 means a→∞∧b→∞ lim_( { ((a→∞)),((b→∞)) :}) (((1/a)×(1/b))/((1/a)+(1/b))) =lim_( { ((a→∞)),((b→∞)) :}) (1/(a+b)) =0

$$\mathrm{let}\:{z}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\mathrm{i} \\ $$$${z}\rightarrow\mathrm{0}\:\mathrm{means}\:{a}\rightarrow\infty\wedge{b}\rightarrow\infty \\ $$$$\underset{\begin{cases}{{a}\rightarrow\infty}\\{{b}\rightarrow\infty}\end{cases}} {\mathrm{lim}}\:\frac{\frac{\mathrm{1}}{{a}}×\frac{\mathrm{1}}{{b}}}{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}}\:=\underset{\begin{cases}{{a}\rightarrow\infty}\\{{b}\rightarrow\infty}\end{cases}} {\mathrm{lim}}\:\frac{\mathrm{1}}{{a}+{b}}\:=\mathrm{0} \\ $$

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