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Question Number 221680 by ajfour last updated on 09/Jun/25
I suspect π=i∫_i ^(−1) (((z−1)dz)/( z(√(z^2 +1)))) someone please help confirm or reject!
$${I}\:{suspect} \\ $$$$\pi={i}\underset{\boldsymbol{{i}}} {\overset{−\mathrm{1}} {\int}}\frac{\left({z}−\mathrm{1}\right){dz}}{\:{z}\sqrt{{z}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${someone}\:{please}\:{help}\:{confirm}\:{or}\:{reject}! \\ $$
Commented by fantastic last updated on 09/Jun/25
what application are you using to draw??
$${what}\:{application}\:{are}\:{you}\:{using}\:{to}\:{draw}?? \\ $$
Commented by ajfour last updated on 09/Jun/25
https://youtu.be/T6ekucR7wrQ?si=FQQ2CbIeJztx2TKa uploaded a short question i thought on youtube minutes bsck.
Commented by ajfour last updated on 09/Jun/25
Flexible Whiteboard
$${Flexible}\:{Whiteboard} \\ $$
Commented by mr W last updated on 09/Jun/25
((ra)/2)=((πr^2 )/4)⇒(r/a)=(2/π) ((r−r sin θ)/(r cos θ))=(r/a)=(2/π) ((1−sin θ)/(cos θ))=(2/π) 2 cos θ+π sin θ=π sin (θ+tan^(−1) (2/π))=(π/( (√(π^2 +4)))) θ=tan^(−1) (π/( 2))−tan^(−1) (2/π) =(π/2)−2 tan^(−1) (2/π)≈25°
$$\frac{{ra}}{\mathrm{2}}=\frac{\pi{r}^{\mathrm{2}} }{\mathrm{4}}\Rightarrow\frac{{r}}{{a}}=\frac{\mathrm{2}}{\pi} \\ $$$$\frac{{r}−{r}\:\mathrm{sin}\:\theta}{{r}\:\mathrm{cos}\:\theta}=\frac{{r}}{{a}}=\frac{\mathrm{2}}{\pi} \\ $$$$\frac{\mathrm{1}−\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}=\frac{\mathrm{2}}{\pi} \\ $$$$\mathrm{2}\:\mathrm{cos}\:\theta+\pi\:\mathrm{sin}\:\theta=\pi \\ $$$$\mathrm{sin}\:\left(\theta+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}}{\pi}\right)=\frac{\pi}{\:\sqrt{\pi^{\mathrm{2}} +\mathrm{4}}} \\ $$$$\theta=\mathrm{tan}^{−\mathrm{1}} \frac{\pi}{\:\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}}{\pi} \\ $$$$\:\:\:=\frac{\pi}{\mathrm{2}}−\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}}{\pi}\approx\mathrm{25}° \\ $$
Answered by wewji12 last updated on 09/Jun/25
∫ ((z−1)/(z(√(z^2 +1)))) dz= Let′s Substitute z=tan(t) (dz/dt)=sec^2 (t) → dz=sec^2 (t)dt (√(z^2 +1))=^(z=tan(t)) (√(tan^2 (t)+1))=sec(t) ∫ ((tan(t)−1)/(tan(t)))∙cos(t)∙sec^2 (t)dt=∫ ((((sin(t))/(cos(t)))−1)/((sin(t))/(cos(t))))∙cos(t)sec^2 (t)dt ∫ ((sin(t)−cos(t))/(sin(t)))∙(1/(cos(t))) dt=∫ (1−((cos(t))/(sin(t))))(1/(cos(t))) dt ∫ (sec(t)−csc(t))dt Ehhh...... (dρ/dt)=sec^2 (t)+sec(t)tan(t) → dρ=sec^2 (t)+sec(t)tan(t)dt (∵ (((sec(t)+tan(t))sec(t))/(sec(t)+tan(t)))=sec(t)) and ρ=^(substitute) sec(t)+tan(t) (dρ/dt)=sec^2 (t)+sec(t)tan(t) → dρ=sec^2 (t)+sec(t)tan(t)dt ∫ (dρ/ρ)=ln(ρ)+const∙∙∙∙∙∙∙(A) and −∫ csc(t)dt=ln(cot(t)+csc(t)) ∴ ln(tan(t)+sec(t))+ln(cot(t)+csc(t))+const t=tan^(−1) (z) ln(tan(tan^(−1) (z))+sec(tan^(−1) (z)))+ln(cot(tan^(−1) (z))+csc(tan^(−1) (z))) cot(tan^(−1) (z))=(1/z) , csc(tan^(−1) (z))=((√(z^2 +1))/z) sec(tan^(−1) (z))=(√(z^2 +1)) , tan(tan^(−1) (z))=z ∴ln(((1+(√(z^2 +1)))/z))+ln(z+(√(z^2 +1)))=ln(((((√(z^2 +1))+1)((√(z^2 +1))+z))/z)) ∴ln(1+(√(z^2 +1)))−ln(z)+sinh^(−1) (z)+Const... Final!!!! ∴∫_(−1) ^( i) =−iπ i∙(−iπ)=π Siuuuuuuu
$$\int\:\:\:\frac{{z}−\mathrm{1}}{{z}\sqrt{{z}^{\mathrm{2}} +\mathrm{1}}}\:\mathrm{d}{z}= \\ $$$$\mathrm{Let}'\mathrm{s}\:\mathrm{Substitute}\:{z}=\mathrm{tan}\left({t}\right) \\ $$$$\frac{\mathrm{d}{z}}{\mathrm{d}{t}}=\mathrm{sec}^{\mathrm{2}} \left({t}\right)\:\rightarrow\:\mathrm{d}{z}=\mathrm{sec}^{\mathrm{2}} \left({t}\right)\mathrm{d}{t} \\ $$$$\sqrt{{z}^{\mathrm{2}} +\mathrm{1}}\overset{{z}=\mathrm{tan}\left({t}\right)} {=}\sqrt{\mathrm{tan}^{\mathrm{2}} \left({t}\right)+\mathrm{1}}=\mathrm{sec}\left({t}\right) \\ $$$$\int\:\:\:\:\frac{\mathrm{tan}\left({t}\right)−\mathrm{1}}{\mathrm{tan}\left({t}\right)}\centerdot\mathrm{cos}\left({t}\right)\centerdot\mathrm{sec}^{\mathrm{2}} \left({t}\right)\mathrm{d}{t}=\int\:\:\:\frac{\frac{\mathrm{sin}\left({t}\right)}{\mathrm{cos}\left({t}\right)}−\mathrm{1}}{\frac{\mathrm{sin}\left({t}\right)}{\mathrm{cos}\left({t}\right)}}\centerdot\mathrm{cos}\left({t}\right)\mathrm{sec}^{\mathrm{2}} \left({t}\right)\mathrm{d}{t} \\ $$$$\int\:\:\:\:\:\frac{\mathrm{sin}\left({t}\right)−\mathrm{cos}\left({t}\right)}{\mathrm{sin}\left({t}\right)}\centerdot\frac{\mathrm{1}}{\mathrm{cos}\left({t}\right)}\:\mathrm{d}{t}=\int\:\:\left(\mathrm{1}−\frac{\mathrm{cos}\left({t}\right)}{\mathrm{sin}\left({t}\right)}\right)\frac{\mathrm{1}}{\mathrm{cos}\left({t}\right)}\:\mathrm{d}{t} \\ $$$$\int\:\left(\mathrm{sec}\left({t}\right)−\mathrm{csc}\left({t}\right)\right)\mathrm{d}{t} \\ $$$$\mathrm{E}{hhh}…… \\ $$$$\frac{\mathrm{d}\rho}{\mathrm{d}{t}}=\mathrm{sec}^{\mathrm{2}} \left({t}\right)+\mathrm{sec}\left({t}\right)\mathrm{tan}\left({t}\right)\:\rightarrow\:\mathrm{d}\rho=\mathrm{sec}^{\mathrm{2}} \left({t}\right)+\mathrm{sec}\left({t}\right)\mathrm{tan}\left({t}\right)\mathrm{d}{t} \\ $$$$\left(\because\:\:\frac{\left(\mathrm{sec}\left({t}\right)+\mathrm{tan}\left({t}\right)\right)\mathrm{sec}\left({t}\right)}{\mathrm{sec}\left({t}\right)+\mathrm{tan}\left({t}\right)}=\mathrm{sec}\left({t}\right)\right) \\ $$$$\mathrm{and}\:\rho\overset{\mathrm{substitute}} {=}\mathrm{sec}\left({t}\right)+\mathrm{tan}\left({t}\right) \\ $$$$\frac{\mathrm{d}\rho}{\mathrm{d}{t}}=\mathrm{sec}^{\mathrm{2}} \left({t}\right)+\mathrm{sec}\left({t}\right)\mathrm{tan}\left({t}\right)\:\rightarrow\:\mathrm{d}\rho=\mathrm{sec}^{\mathrm{2}} \left({t}\right)+\mathrm{sec}\left({t}\right)\mathrm{tan}\left({t}\right)\mathrm{d}{t} \\ $$$$\int\:\:\frac{\mathrm{d}\rho}{\rho}=\mathrm{ln}\left(\rho\right)+\mathrm{const}\centerdot\centerdot\centerdot\centerdot\centerdot\centerdot\centerdot\left({A}\right) \\ $$$$\mathrm{and}\:\:−\int\:\mathrm{csc}\left({t}\right)\mathrm{d}{t}=\mathrm{ln}\left(\mathrm{cot}\left({t}\right)+\mathrm{csc}\left({t}\right)\right) \\ $$$$\therefore\:\mathrm{ln}\left(\mathrm{tan}\left({t}\right)+\mathrm{sec}\left({t}\right)\right)+\mathrm{ln}\left(\mathrm{cot}\left({t}\right)+\mathrm{csc}\left({t}\right)\right)+\mathrm{const} \\ $$$${t}=\mathrm{tan}^{−\mathrm{1}} \left({z}\right) \\ $$$$\mathrm{ln}\left(\mathrm{tan}\left(\mathrm{tan}^{−\mathrm{1}} \left({z}\right)\right)+\mathrm{sec}\left(\mathrm{tan}^{−\mathrm{1}} \left({z}\right)\right)\right)+\mathrm{ln}\left(\mathrm{cot}\left(\mathrm{tan}^{−\mathrm{1}} \left({z}\right)\right)+\mathrm{csc}\left(\mathrm{tan}^{−\mathrm{1}} \left({z}\right)\right)\right) \\ $$$$\mathrm{cot}\left(\mathrm{tan}^{−\mathrm{1}} \left({z}\right)\right)=\frac{\mathrm{1}}{{z}}\:\:,\:\mathrm{csc}\left(\mathrm{tan}^{−\mathrm{1}} \left({z}\right)\right)=\frac{\sqrt{{z}^{\mathrm{2}} +\mathrm{1}}}{{z}} \\ $$$$\mathrm{sec}\left(\mathrm{tan}^{−\mathrm{1}} \left({z}\right)\right)=\sqrt{{z}^{\mathrm{2}} +\mathrm{1}}\:,\:\mathrm{tan}\left(\mathrm{tan}^{−\mathrm{1}} \left({z}\right)\right)={z} \\ $$$$\therefore\mathrm{ln}\left(\frac{\mathrm{1}+\sqrt{{z}^{\mathrm{2}} +\mathrm{1}}}{{z}}\right)+\mathrm{ln}\left({z}+\sqrt{{z}^{\mathrm{2}} +\mathrm{1}}\right)=\mathrm{ln}\left(\frac{\left(\sqrt{{z}^{\mathrm{2}} +\mathrm{1}}+\mathrm{1}\right)\left(\sqrt{{z}^{\mathrm{2}} +\mathrm{1}}+{z}\right)}{{z}}\right) \\ $$$$\therefore\mathrm{ln}\left(\mathrm{1}+\sqrt{{z}^{\mathrm{2}} +\mathrm{1}}\right)−\mathrm{ln}\left({z}\right)+\mathrm{sinh}^{−\mathrm{1}} \left({z}\right)+\mathrm{Const}… \\ $$$$\mathrm{Final}!!!! \\ $$$$\therefore\int_{−\mathrm{1}} ^{\:\boldsymbol{{i}}} =−\boldsymbol{{i}}\pi\:\:\: \\ $$$$\boldsymbol{{i}}\centerdot\left(−\boldsymbol{{i}}\pi\right)=\pi \\ $$$$\mathrm{Siuuuuuuu} \\ $$
Commented by Ghisom last updated on 09/Jun/25
complicated...
$$\mathrm{complicated}… \\ $$
Commented by fantastic last updated on 09/Jun/25
wewji getting excited moment
$${wewji}\:{getting}\:{excited}\:{moment} \\ $$
Commented by wewji12 last updated on 09/Jun/25
:(
$$:\left(\right. \\ $$
Answered by Ghisom last updated on 09/Jun/25
∫_i ^(−1) ((z−1)/(z(√(z^2 +1))))dz= [t=z+(√(z^2 +1))] =∫_i ^(−1+(√2)) ((t^2 −2t−1)/(t(t−1)(t+1)))dt= =∫_i ^(−1+(√2)) ((1/t)+(1/(t−1))+(1/(t+1)))dt= =[ln ((t(t+1))/(t−1))]_i ^(−1+(√2)) =πi
$$\underset{\mathrm{i}} {\overset{−\mathrm{1}} {\int}}\frac{{z}−\mathrm{1}}{{z}\sqrt{{z}^{\mathrm{2}} +\mathrm{1}}}{dz}= \\ $$$$\:\:\:\:\:\left[{t}={z}+\sqrt{{z}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=\underset{\mathrm{i}} {\overset{−\mathrm{1}+\sqrt{\mathrm{2}}} {\int}}\frac{{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}}{{t}\left({t}−\mathrm{1}\right)\left({t}+\mathrm{1}\right)}{dt}= \\ $$$$=\underset{\mathrm{i}} {\overset{−\mathrm{1}+\sqrt{\mathrm{2}}} {\int}}\left(\frac{\mathrm{1}}{{t}}+\frac{\mathrm{1}}{{t}−\mathrm{1}}+\frac{\mathrm{1}}{{t}+\mathrm{1}}\right){dt}= \\ $$$$=\left[\mathrm{ln}\:\frac{{t}\left({t}+\mathrm{1}\right)}{{t}−\mathrm{1}}\right]_{\mathrm{i}} ^{−\mathrm{1}+\sqrt{\mathrm{2}}} =\pi\mathrm{i} \\ $$

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