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 Yearling
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Sep
23
comment Show with DCT: $\lim_{R \to \infty} \int_0^{2\pi} \frac{id\theta}{\sqrt{1+R^{-2}e^{-2i\theta}}}=\int_0^{2\pi} id\theta$
@did This is the radius of the complex number $1+iR^{-1}e^{-i\theta}$.
Sep
23
asked Show with DCT: $\lim_{R \to \infty} \int_0^{2\pi} \frac{id\theta}{\sqrt{1+R^{-2}e^{-2i\theta}}}=\int_0^{2\pi} id\theta$
Sep
16
accepted Does $\int_a^b \text{Im}(f(t))e^{int}dt=0$ implies $f=0$ on $[a,b]$
Sep
16
revised Does $\int_a^b \text{Im}(f(t))e^{int}dt=0$ implies $f=0$ on $[a,b]$
corrected question per Alex Becker comment
Sep
16
comment Does $\int_a^b \text{Im}(f(t))e^{int}dt=0$ implies $f=0$ on $[a,b]$
@AlexBecker Good eyes, I meant $\text{Im}(f(t))$ is identically zero. Thank you.
Sep
16
asked Does $\int_a^b \text{Im}(f(t))e^{int}dt=0$ implies $f=0$ on $[a,b]$
Sep
16
accepted Does the complex conjugate of an integral equal the integral of the conjugate?
Sep
16
asked Does the complex conjugate of an integral equal the integral of the conjugate?
Sep
10
awarded  Civic Duty
Aug
29
accepted Find $\displaystyle{\int_C \left(1+ \frac{2}{z}\right) dz}$ , where $C(\theta)= e^{i\theta}, 0 \le \theta \le \pi$
Aug
28
asked Find $\displaystyle{\int_C \left(1+ \frac{2}{z}\right) dz}$ , where $C(\theta)= e^{i\theta}, 0 \le \theta \le \pi$
Aug
1
awarded  Yearling
Jun
28
comment Show $g(a)= \operatorname*{arg}_{b \in B} \, f(a,b)=0$ is continuous when $g(a)$ is single-valued and $f$ continuous
Our answers are complementary. But You were of precious help and I think you deserve the points. I learn a lot. Thank you.
Jun
28
accepted Show $g(a)= \operatorname*{arg}_{b \in B} \, f(a,b)=0$ is continuous when $g(a)$ is single-valued and $f$ continuous
Jun
28
comment Show $g(a)= \operatorname*{arg}_{b \in B} \, f(a,b)=0$ is continuous when $g(a)$ is single-valued and $f$ continuous
@CameronBuie I improved the first paragraph of the proof. You are of great help. Thank you for sharing your insight.
Jun
28
revised Show $g(a)= \operatorname*{arg}_{b \in B} \, f(a,b)=0$ is continuous when $g(a)$ is single-valued and $f$ continuous
First paragraph of the proof improved per Cameron comment
Jun
28
comment Show $g(a)= \operatorname*{arg}_{b \in B} \, f(a,b)=0$ is continuous when $g(a)$ is single-valued and $f$ continuous
@CameronBuie I think I can correct the flaw as follow. Assuming $g$ is discontinuous at $a$, there is a sequence $a_n$ in $A$ converging to $a$ and an $\epsilon >0$ such that for each $n$, $b_n=g(a_n)$ and $|b_n-g(a)|>\epsilon$.
Jun
28
answered Show $g(a)= \operatorname*{arg}_{b \in B} \, f(a,b)=0$ is continuous when $g(a)$ is single-valued and $f$ continuous
Jun
27
comment Show $g(a)= \operatorname*{arg}_{b \in B} \, f(a,b)=0$ is continuous when $g(a)$ is single-valued and $f$ continuous
@CameronBuie Thanks for your edit. You read my mind. I don't understand why, though, the $\arg_{b \in B}$ notation is not equivalent to the $B_a$ notation.
Jun
27
asked Show $g(a)= \operatorname*{arg}_{b \in B} \, f(a,b)=0$ is continuous when $g(a)$ is single-valued and $f$ continuous