How can this equality be proved?
$begingroup$
I confront with a statement:
For any given any complex number $zin B(0,r)$, where $r$ is radius of real number, then we have equations:
$frac{1}{2pi}int_{0}^{2pi}operatorname{Log}|re^{itheta}-z|,dtheta=operatorname{Log}r$
I try to compute it by changing the integral into another form:
$frac{1}{4pi}int_{0}^{2pi} operatorname{Log}(r^2+rho^2-2rrho costheta),dtheta=operatorname{Log}r$
where $rho <r$, but it's still too complicated to compute. Actually, if I can show that $int_{0}^{2pi} operatorname{Log}(1+rho^2-2rho costheta),dtheta=0$, where $rho <1$, then the proof is completed. Hope someone could help, Thanks!
complex-analysis
edited Dec 18 '18 at 12:11
Bernard
123k741117
asked Dec 18 '18 at 12:02
Yuyi ZhangYuyi Zhang
15117
$endgroup$
add a comment |
$begingroup$
I confront with a statement:
For any given any complex number $zin B(0,r)$, where $r$ is radius of real number, then we have equations:
$frac{1}{2pi}int_{0}^{2pi}operatorname{Log}|re^{itheta}-z|,dtheta=operatorname{Log}r$
I try to compute it by changing the integral into another form:
$frac{1}{4pi}int_{0}^{2pi} operatorname{Log}(r^2+rho^2-2rrho costheta),dtheta=operatorname{Log}r$
where $rho <r$, but it's still too complicated to compute. Actually, if I can show that $int_{0}^{2pi} operatorname{Log}(1+rho^2-2rho costheta),dtheta=0$, where $rho <1$, then the proof is completed. Hope someone could help, Thanks!
complex-analysis
edited Dec 18 '18 at 12:11
Bernard
123k741117
asked Dec 18 '18 at 12:02
Yuyi ZhangYuyi Zhang
15117
$endgroup$
$begingroup$
Uhm I would rather try to use the residue theorem: en.wikipedia.org/wiki/Residue_theorem
$endgroup$
– Ant
Dec 18 '18 at 12:14
add a comment |
$begingroup$
I confront with a statement:
For any given any complex number $zin B(0,r)$, where $r$ is radius of real number, then we have equations:
$frac{1}{2pi}int_{0}^{2pi}operatorname{Log}|re^{itheta}-z|,dtheta=operatorname{Log}r$
I try to compute it by changing the integral into another form:
$frac{1}{4pi}int_{0}^{2pi} operatorname{Log}(r^2+rho^2-2rrho costheta),dtheta=operatorname{Log}r$
where $rho <r$, but it's still too complicated to compute. Actually, if I can show that $int_{0}^{2pi} operatorname{Log}(1+rho^2-2rho costheta),dtheta=0$, where $rho <1$, then the proof is completed. Hope someone could help, Thanks!
complex-analysis
edited Dec 18 '18 at 12:11
Bernard
123k741117
asked Dec 18 '18 at 12:02
Yuyi ZhangYuyi Zhang
15117
$endgroup$
I confront with a statement:
For any given any complex number $zin B(0,r)$, where $r$ is radius of real number, then we have equations:
$frac{1}{2pi}int_{0}^{2pi}operatorname{Log}|re^{itheta}-z|,dtheta=operatorname{Log}r$
I try to compute it by changing the integral into another form:
$frac{1}{4pi}int_{0}^{2pi} operatorname{Log}(r^2+rho^2-2rrho costheta),dtheta=operatorname{Log}r$
where $rho <r$, but it's still too complicated to compute. Actually, if I can show that $int_{0}^{2pi} operatorname{Log}(1+rho^2-2rho costheta),dtheta=0$, where $rho <1$, then the proof is completed. Hope someone could help, Thanks!
complex-analysis
complex-analysis
edited Dec 18 '18 at 12:11
Bernard
123k741117
asked Dec 18 '18 at 12:02
Yuyi ZhangYuyi Zhang
15117
edited Dec 18 '18 at 12:11
Bernard
123k741117
asked Dec 18 '18 at 12:02
Yuyi ZhangYuyi Zhang
15117
edited Dec 18 '18 at 12:11
Bernard
123k741117
edited Dec 18 '18 at 12:11
Bernard
123k741117
edited Dec 18 '18 at 12:11
Bernard
123k741117
123k741117
asked Dec 18 '18 at 12:02
Yuyi ZhangYuyi Zhang
15117
asked Dec 18 '18 at 12:02
Yuyi ZhangYuyi Zhang
15117
asked Dec 18 '18 at 12:02
Yuyi ZhangYuyi Zhang
15117
15117
$begingroup$
Uhm I would rather try to use the residue theorem: en.wikipedia.org/wiki/Residue_theorem
$endgroup$
– Ant
Dec 18 '18 at 12:14
add a comment |
$begingroup$
Uhm I would rather try to use the residue theorem: en.wikipedia.org/wiki/Residue_theorem
$endgroup$
– Ant
Dec 18 '18 at 12:14
$begingroup$
Uhm I would rather try to use the residue theorem: en.wikipedia.org/wiki/Residue_theorem
$endgroup$
– Ant
Dec 18 '18 at 12:14
$begingroup$
Uhm I would rather try to use the residue theorem: en.wikipedia.org/wiki/Residue_theorem
$endgroup$
– Ant
Dec 18 '18 at 12:14
add a comment |
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$begingroup$
Uhm I would rather try to use the residue theorem: en.wikipedia.org/wiki/Residue_theorem
$endgroup$
– Ant
Dec 18 '18 at 12:14