Plotting odd complex functions w/o computer help












1














I am evaluating a complex integral that utilises the Cauchy Integral Formula and its properties.



In the book I'm reading, they give examples of evaluating integrals using CIT by graphing them, which really does help to see if points of a function is analytic in a certain domain.



For example, in evaluating this integral - how would I be able to plot the function shown in this integral, where C is a circle $|z|=1$ traversed once counter clockwise. $$int_{C}frac{z+i}{z^3+2z^2}$$



Thanks.






complex-analysis complex-numbers







share|cite|improve this question






















  • You may be able to find ways of visualizing complex mappings but in general there is no way to plot the graph of a function from $mathbb{C}to mathbb{C}$ analogously to graphs of functions from $mathbb{R} to mathbb{R}$ (or up to 3-space). The point $(z, f(z))$ is in some sense a point of $mathbb{R}^4$ which our brains find difficult to visualize...
    – LoveTooNap29
    5 hours ago










  • I understand. This is something that I will come to terms with. I'm used to plotting basic real functions. Thanks for that. So in a case like this, how would one go about solving it? I understand that the integrand fails to be analytic point z = 0, but where to from there?
    – Dr.Doofus
    4 hours ago










  • Indeed, it would be relatively straightforward to apply residue theorem (although probably tedious) but it appears your only given tool is Cauchy’s integral formula, correct? Just want to make sure before I offer a sketch of a solution.
    – LoveTooNap29
    4 hours ago










  • Yes, sir. It's how they want us to go about it.
    – Dr.Doofus
    4 hours ago
















1














I am evaluating a complex integral that utilises the Cauchy Integral Formula and its properties.



In the book I'm reading, they give examples of evaluating integrals using CIT by graphing them, which really does help to see if points of a function is analytic in a certain domain.



For example, in evaluating this integral - how would I be able to plot the function shown in this integral, where C is a circle $|z|=1$ traversed once counter clockwise. $$int_{C}frac{z+i}{z^3+2z^2}$$



Thanks.






complex-analysis complex-numbers







share|cite|improve this question






















  • You may be able to find ways of visualizing complex mappings but in general there is no way to plot the graph of a function from $mathbb{C}to mathbb{C}$ analogously to graphs of functions from $mathbb{R} to mathbb{R}$ (or up to 3-space). The point $(z, f(z))$ is in some sense a point of $mathbb{R}^4$ which our brains find difficult to visualize...
    – LoveTooNap29
    5 hours ago










  • I understand. This is something that I will come to terms with. I'm used to plotting basic real functions. Thanks for that. So in a case like this, how would one go about solving it? I understand that the integrand fails to be analytic point z = 0, but where to from there?
    – Dr.Doofus
    4 hours ago










  • Indeed, it would be relatively straightforward to apply residue theorem (although probably tedious) but it appears your only given tool is Cauchy’s integral formula, correct? Just want to make sure before I offer a sketch of a solution.
    – LoveTooNap29
    4 hours ago










  • Yes, sir. It's how they want us to go about it.
    – Dr.Doofus
    4 hours ago














1












1








1







I am evaluating a complex integral that utilises the Cauchy Integral Formula and its properties.



In the book I'm reading, they give examples of evaluating integrals using CIT by graphing them, which really does help to see if points of a function is analytic in a certain domain.



For example, in evaluating this integral - how would I be able to plot the function shown in this integral, where C is a circle $|z|=1$ traversed once counter clockwise. $$int_{C}frac{z+i}{z^3+2z^2}$$



Thanks.






complex-analysis complex-numbers







share|cite|improve this question













I am evaluating a complex integral that utilises the Cauchy Integral Formula and its properties.



In the book I'm reading, they give examples of evaluating integrals using CIT by graphing them, which really does help to see if points of a function is analytic in a certain domain.



For example, in evaluating this integral - how would I be able to plot the function shown in this integral, where C is a circle $|z|=1$ traversed once counter clockwise. $$int_{C}frac{z+i}{z^3+2z^2}$$



Thanks.






complex-analysis complex-numbers




complex-analysis complex-numbers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 5 hours ago









Dr.Doofus

9210




9210












  • You may be able to find ways of visualizing complex mappings but in general there is no way to plot the graph of a function from $mathbb{C}to mathbb{C}$ analogously to graphs of functions from $mathbb{R} to mathbb{R}$ (or up to 3-space). The point $(z, f(z))$ is in some sense a point of $mathbb{R}^4$ which our brains find difficult to visualize...
    – LoveTooNap29
    5 hours ago










  • I understand. This is something that I will come to terms with. I'm used to plotting basic real functions. Thanks for that. So in a case like this, how would one go about solving it? I understand that the integrand fails to be analytic point z = 0, but where to from there?
    – Dr.Doofus
    4 hours ago










  • Indeed, it would be relatively straightforward to apply residue theorem (although probably tedious) but it appears your only given tool is Cauchy’s integral formula, correct? Just want to make sure before I offer a sketch of a solution.
    – LoveTooNap29
    4 hours ago










  • Yes, sir. It's how they want us to go about it.
    – Dr.Doofus
    4 hours ago


















  • You may be able to find ways of visualizing complex mappings but in general there is no way to plot the graph of a function from $mathbb{C}to mathbb{C}$ analogously to graphs of functions from $mathbb{R} to mathbb{R}$ (or up to 3-space). The point $(z, f(z))$ is in some sense a point of $mathbb{R}^4$ which our brains find difficult to visualize...
    – LoveTooNap29
    5 hours ago










  • I understand. This is something that I will come to terms with. I'm used to plotting basic real functions. Thanks for that. So in a case like this, how would one go about solving it? I understand that the integrand fails to be analytic point z = 0, but where to from there?
    – Dr.Doofus
    4 hours ago










  • Indeed, it would be relatively straightforward to apply residue theorem (although probably tedious) but it appears your only given tool is Cauchy’s integral formula, correct? Just want to make sure before I offer a sketch of a solution.
    – LoveTooNap29
    4 hours ago










  • Yes, sir. It's how they want us to go about it.
    – Dr.Doofus
    4 hours ago
















You may be able to find ways of visualizing complex mappings but in general there is no way to plot the graph of a function from $mathbb{C}to mathbb{C}$ analogously to graphs of functions from $mathbb{R} to mathbb{R}$ (or up to 3-space). The point $(z, f(z))$ is in some sense a point of $mathbb{R}^4$ which our brains find difficult to visualize...
– LoveTooNap29
5 hours ago




You may be able to find ways of visualizing complex mappings but in general there is no way to plot the graph of a function from $mathbb{C}to mathbb{C}$ analogously to graphs of functions from $mathbb{R} to mathbb{R}$ (or up to 3-space). The point $(z, f(z))$ is in some sense a point of $mathbb{R}^4$ which our brains find difficult to visualize...
– LoveTooNap29
5 hours ago












I understand. This is something that I will come to terms with. I'm used to plotting basic real functions. Thanks for that. So in a case like this, how would one go about solving it? I understand that the integrand fails to be analytic point z = 0, but where to from there?
– Dr.Doofus
4 hours ago




I understand. This is something that I will come to terms with. I'm used to plotting basic real functions. Thanks for that. So in a case like this, how would one go about solving it? I understand that the integrand fails to be analytic point z = 0, but where to from there?
– Dr.Doofus
4 hours ago












Indeed, it would be relatively straightforward to apply residue theorem (although probably tedious) but it appears your only given tool is Cauchy’s integral formula, correct? Just want to make sure before I offer a sketch of a solution.
– LoveTooNap29
4 hours ago




Indeed, it would be relatively straightforward to apply residue theorem (although probably tedious) but it appears your only given tool is Cauchy’s integral formula, correct? Just want to make sure before I offer a sketch of a solution.
– LoveTooNap29
4 hours ago












Yes, sir. It's how they want us to go about it.
– Dr.Doofus
4 hours ago




Yes, sir. It's how they want us to go about it.
– Dr.Doofus
4 hours ago










2 Answers
2






active

oldest

votes


















2














The idea is that you only need to know where the function is singular



enter image description here



In your case it is at $z= 0$ and $z = -2$, but the second point is outside the region of integration, so we do not care about it



$$
oint_C frac{z + i}{z^2(z + 2)} {rm d}z= oint_C frac{f(z)}{z^2}{rm d}z
$$



where the function



$$
f(z) = frac{z + i}{z + 2}
$$



is well behaved in the region $|z| leq 1$. The problem is at $z = 0$, where you have second order singularity, so you can use Cauchy's integral form



$$
left.frac{{rm d}^n f}{{rm d}z^n}right|_{z = 0} = frac{n!}{2pi i}oint_C frac{f(z)}{z^{n + 1}}{rm d}z
$$



with $n = 1$ which yields



$$
oint_Cfrac{z + i}{z^2(z + 2)} = 2 pi i frac{{rm d}}{{rm d}z}left(frac{z + i}{z + 2}right)_{z = 0} = 2pi i left(frac{1}{2} - frac{i}{4} right)
$$






share|cite|improve this answer





















  • This is great. Thank you. Better than the provided solutions!
    – Dr.Doofus
    4 hours ago










  • Ah it looks like you beat me to it! Excellent visual too. Good to at least confirm by someone else's work that I can still recall this material correctly.
    – LoveTooNap29
    4 hours ago



















2














Sketch:
Note that the pole at $z=0$ has order $2$. Use the slightly generalized Cauchy formula for derivatives,
$$f^{(n)}(a)=frac{n!}{2pi i}oint frac{f(z)}{(z-a)^{n+1}} mathrm{d}z,$$
Write the original integrand as $g(z)$ and identify $f(z)$ from $g(z)=f(z)/z^2$ in your case for $n=1$ and $a=0$. Then compute the first derivative of $f$ at $a=0$ to compute the original integral (remember to multiply this by $2pi i$). This yields $pi (frac12 +i)$.



Be sure to check all hypotheses of the Cauchy integral formula though to justify this rigorously, though (that $f$ is holomorphic inside the curve is required, not $g$, etc). Comment if you want some steps expanded, but preferably with where you got stuck also, or if you spot any typos, etc. It has been some time since I've gotten my hands dirty with contour integrals with only Cauchy's integral formula, however, the example on wikipedia is rather illustrative, though the poles there are only of order 1, unlike here.






share|cite|improve this answer





















  • Thank you. I genuinely appreciate it.
    – Dr.Doofus
    4 hours ago










  • Sorry, didn't know you were working on an answer as well (+1)
    – caverac
    4 hours ago











Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














The idea is that you only need to know where the function is singular



enter image description here



In your case it is at $z= 0$ and $z = -2$, but the second point is outside the region of integration, so we do not care about it



$$
oint_C frac{z + i}{z^2(z + 2)} {rm d}z= oint_C frac{f(z)}{z^2}{rm d}z
$$



where the function



$$
f(z) = frac{z + i}{z + 2}
$$



is well behaved in the region $|z| leq 1$. The problem is at $z = 0$, where you have second order singularity, so you can use Cauchy's integral form



$$
left.frac{{rm d}^n f}{{rm d}z^n}right|_{z = 0} = frac{n!}{2pi i}oint_C frac{f(z)}{z^{n + 1}}{rm d}z
$$



with $n = 1$ which yields



$$
oint_Cfrac{z + i}{z^2(z + 2)} = 2 pi i frac{{rm d}}{{rm d}z}left(frac{z + i}{z + 2}right)_{z = 0} = 2pi i left(frac{1}{2} - frac{i}{4} right)
$$






share|cite|improve this answer





















  • This is great. Thank you. Better than the provided solutions!
    – Dr.Doofus
    4 hours ago










  • Ah it looks like you beat me to it! Excellent visual too. Good to at least confirm by someone else's work that I can still recall this material correctly.
    – LoveTooNap29
    4 hours ago
















2














The idea is that you only need to know where the function is singular



enter image description here



In your case it is at $z= 0$ and $z = -2$, but the second point is outside the region of integration, so we do not care about it



$$
oint_C frac{z + i}{z^2(z + 2)} {rm d}z= oint_C frac{f(z)}{z^2}{rm d}z
$$



where the function



$$
f(z) = frac{z + i}{z + 2}
$$



is well behaved in the region $|z| leq 1$. The problem is at $z = 0$, where you have second order singularity, so you can use Cauchy's integral form



$$
left.frac{{rm d}^n f}{{rm d}z^n}right|_{z = 0} = frac{n!}{2pi i}oint_C frac{f(z)}{z^{n + 1}}{rm d}z
$$



with $n = 1$ which yields



$$
oint_Cfrac{z + i}{z^2(z + 2)} = 2 pi i frac{{rm d}}{{rm d}z}left(frac{z + i}{z + 2}right)_{z = 0} = 2pi i left(frac{1}{2} - frac{i}{4} right)
$$






share|cite|improve this answer





















  • This is great. Thank you. Better than the provided solutions!
    – Dr.Doofus
    4 hours ago










  • Ah it looks like you beat me to it! Excellent visual too. Good to at least confirm by someone else's work that I can still recall this material correctly.
    – LoveTooNap29
    4 hours ago














2












2








2






The idea is that you only need to know where the function is singular



enter image description here



In your case it is at $z= 0$ and $z = -2$, but the second point is outside the region of integration, so we do not care about it



$$
oint_C frac{z + i}{z^2(z + 2)} {rm d}z= oint_C frac{f(z)}{z^2}{rm d}z
$$



where the function



$$
f(z) = frac{z + i}{z + 2}
$$



is well behaved in the region $|z| leq 1$. The problem is at $z = 0$, where you have second order singularity, so you can use Cauchy's integral form



$$
left.frac{{rm d}^n f}{{rm d}z^n}right|_{z = 0} = frac{n!}{2pi i}oint_C frac{f(z)}{z^{n + 1}}{rm d}z
$$



with $n = 1$ which yields



$$
oint_Cfrac{z + i}{z^2(z + 2)} = 2 pi i frac{{rm d}}{{rm d}z}left(frac{z + i}{z + 2}right)_{z = 0} = 2pi i left(frac{1}{2} - frac{i}{4} right)
$$






share|cite|improve this answer












The idea is that you only need to know where the function is singular



enter image description here



In your case it is at $z= 0$ and $z = -2$, but the second point is outside the region of integration, so we do not care about it



$$
oint_C frac{z + i}{z^2(z + 2)} {rm d}z= oint_C frac{f(z)}{z^2}{rm d}z
$$



where the function



$$
f(z) = frac{z + i}{z + 2}
$$



is well behaved in the region $|z| leq 1$. The problem is at $z = 0$, where you have second order singularity, so you can use Cauchy's integral form



$$
left.frac{{rm d}^n f}{{rm d}z^n}right|_{z = 0} = frac{n!}{2pi i}oint_C frac{f(z)}{z^{n + 1}}{rm d}z
$$



with $n = 1$ which yields



$$
oint_Cfrac{z + i}{z^2(z + 2)} = 2 pi i frac{{rm d}}{{rm d}z}left(frac{z + i}{z + 2}right)_{z = 0} = 2pi i left(frac{1}{2} - frac{i}{4} right)
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 4 hours ago









caverac

13.9k21130




13.9k21130












  • This is great. Thank you. Better than the provided solutions!
    – Dr.Doofus
    4 hours ago










  • Ah it looks like you beat me to it! Excellent visual too. Good to at least confirm by someone else's work that I can still recall this material correctly.
    – LoveTooNap29
    4 hours ago


















  • This is great. Thank you. Better than the provided solutions!
    – Dr.Doofus
    4 hours ago










  • Ah it looks like you beat me to it! Excellent visual too. Good to at least confirm by someone else's work that I can still recall this material correctly.
    – LoveTooNap29
    4 hours ago
















This is great. Thank you. Better than the provided solutions!
– Dr.Doofus
4 hours ago




This is great. Thank you. Better than the provided solutions!
– Dr.Doofus
4 hours ago












Ah it looks like you beat me to it! Excellent visual too. Good to at least confirm by someone else's work that I can still recall this material correctly.
– LoveTooNap29
4 hours ago




Ah it looks like you beat me to it! Excellent visual too. Good to at least confirm by someone else's work that I can still recall this material correctly.
– LoveTooNap29
4 hours ago











2














Sketch:
Note that the pole at $z=0$ has order $2$. Use the slightly generalized Cauchy formula for derivatives,
$$f^{(n)}(a)=frac{n!}{2pi i}oint frac{f(z)}{(z-a)^{n+1}} mathrm{d}z,$$
Write the original integrand as $g(z)$ and identify $f(z)$ from $g(z)=f(z)/z^2$ in your case for $n=1$ and $a=0$. Then compute the first derivative of $f$ at $a=0$ to compute the original integral (remember to multiply this by $2pi i$). This yields $pi (frac12 +i)$.



Be sure to check all hypotheses of the Cauchy integral formula though to justify this rigorously, though (that $f$ is holomorphic inside the curve is required, not $g$, etc). Comment if you want some steps expanded, but preferably with where you got stuck also, or if you spot any typos, etc. It has been some time since I've gotten my hands dirty with contour integrals with only Cauchy's integral formula, however, the example on wikipedia is rather illustrative, though the poles there are only of order 1, unlike here.






share|cite|improve this answer





















  • Thank you. I genuinely appreciate it.
    – Dr.Doofus
    4 hours ago










  • Sorry, didn't know you were working on an answer as well (+1)
    – caverac
    4 hours ago
















2














Sketch:
Note that the pole at $z=0$ has order $2$. Use the slightly generalized Cauchy formula for derivatives,
$$f^{(n)}(a)=frac{n!}{2pi i}oint frac{f(z)}{(z-a)^{n+1}} mathrm{d}z,$$
Write the original integrand as $g(z)$ and identify $f(z)$ from $g(z)=f(z)/z^2$ in your case for $n=1$ and $a=0$. Then compute the first derivative of $f$ at $a=0$ to compute the original integral (remember to multiply this by $2pi i$). This yields $pi (frac12 +i)$.



Be sure to check all hypotheses of the Cauchy integral formula though to justify this rigorously, though (that $f$ is holomorphic inside the curve is required, not $g$, etc). Comment if you want some steps expanded, but preferably with where you got stuck also, or if you spot any typos, etc. It has been some time since I've gotten my hands dirty with contour integrals with only Cauchy's integral formula, however, the example on wikipedia is rather illustrative, though the poles there are only of order 1, unlike here.






share|cite|improve this answer





















  • Thank you. I genuinely appreciate it.
    – Dr.Doofus
    4 hours ago










  • Sorry, didn't know you were working on an answer as well (+1)
    – caverac
    4 hours ago














2












2








2






Sketch:
Note that the pole at $z=0$ has order $2$. Use the slightly generalized Cauchy formula for derivatives,
$$f^{(n)}(a)=frac{n!}{2pi i}oint frac{f(z)}{(z-a)^{n+1}} mathrm{d}z,$$
Write the original integrand as $g(z)$ and identify $f(z)$ from $g(z)=f(z)/z^2$ in your case for $n=1$ and $a=0$. Then compute the first derivative of $f$ at $a=0$ to compute the original integral (remember to multiply this by $2pi i$). This yields $pi (frac12 +i)$.



Be sure to check all hypotheses of the Cauchy integral formula though to justify this rigorously, though (that $f$ is holomorphic inside the curve is required, not $g$, etc). Comment if you want some steps expanded, but preferably with where you got stuck also, or if you spot any typos, etc. It has been some time since I've gotten my hands dirty with contour integrals with only Cauchy's integral formula, however, the example on wikipedia is rather illustrative, though the poles there are only of order 1, unlike here.






share|cite|improve this answer












Sketch:
Note that the pole at $z=0$ has order $2$. Use the slightly generalized Cauchy formula for derivatives,
$$f^{(n)}(a)=frac{n!}{2pi i}oint frac{f(z)}{(z-a)^{n+1}} mathrm{d}z,$$
Write the original integrand as $g(z)$ and identify $f(z)$ from $g(z)=f(z)/z^2$ in your case for $n=1$ and $a=0$. Then compute the first derivative of $f$ at $a=0$ to compute the original integral (remember to multiply this by $2pi i$). This yields $pi (frac12 +i)$.



Be sure to check all hypotheses of the Cauchy integral formula though to justify this rigorously, though (that $f$ is holomorphic inside the curve is required, not $g$, etc). Comment if you want some steps expanded, but preferably with where you got stuck also, or if you spot any typos, etc. It has been some time since I've gotten my hands dirty with contour integrals with only Cauchy's integral formula, however, the example on wikipedia is rather illustrative, though the poles there are only of order 1, unlike here.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 4 hours ago









LoveTooNap29

1,0241613




1,0241613












  • Thank you. I genuinely appreciate it.
    – Dr.Doofus
    4 hours ago










  • Sorry, didn't know you were working on an answer as well (+1)
    – caverac
    4 hours ago


















  • Thank you. I genuinely appreciate it.
    – Dr.Doofus
    4 hours ago










  • Sorry, didn't know you were working on an answer as well (+1)
    – caverac
    4 hours ago
















Thank you. I genuinely appreciate it.
– Dr.Doofus
4 hours ago




Thank you. I genuinely appreciate it.
– Dr.Doofus
4 hours ago












Sorry, didn't know you were working on an answer as well (+1)
– caverac
4 hours ago




Sorry, didn't know you were working on an answer as well (+1)
– caverac
4 hours ago


















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CARDNET

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CARDNET は日本カードネットワークが運営するクレジット決済ネットワーク。日本カードネットワークの略称としてもCARDNETが使用されている。 事件・事故 2017年4月15日午前11時8分からクレジットカードの決済がしづらくなる障害が発生した。障害は同日午後5時18分に復旧した。原因は、複数あるCARDNETセンターの拠点のうち1つで、2重化してあるL3スイッチの片方が故障し、1系統にトラフィックが集中したことによる輻輳が発生したため。対処としては、故障したL3スイッチ交換とシステム再起動が行われた [1] 。 出典 ^ 金子寛人 ( 2017年4月17日 ). “CARDNETのクレジット決済に6時間強障害、原因はL3スイッチ故障”. 日経コンピュータ. 2018年10月18日 閲覧。 この項目は、コンピュータに関連した 書きかけの項目 です。この項目を加筆・訂正などしてくださる協力者を求めています（PJ:コンピュータ/P:コンピュータ）。 この項目は、経済に関連した 書きかけの項目 です。この項目を加筆・訂正などしてくださる協力者を求めています（ポータル 経済学、プロジェクト 経済）。 This page is only for reference, If you need detailed information, please check here
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Boot-repair Failure: Unable to locate package grub-common:i386

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0 I am on a system dual-booting Windows and Debian (usually using the latter), and a recent Windows update appears to have wreaked havoc on the Debian boot system. I managed to boot Windows again, but not Linux. Following online suggestions, I created a Live USB of Ubuntu, and booted from that. Then I have installed and run the boot-repair utility. The utility instructs me to type several lines including sudo chroot "/mnt/boot-sav/sda6" apt-get purge -y grub*-common grub-common:i386 To which I receive the error E: Unable to locate package grub-common:i386 I am given to understand the issue may have to do with having UEFI vs BIOS (I believe I have UEFI), but do not understand this well. In any case I have subsequently run the boot-repair BootInfo Summary utility which created the following file h...
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濃尾地震

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濃尾地震 『岐阜市街大地震之図』 歌川国利画 本震 発生日 1891年（明治24年）10月28日 発生時刻 6時38分50秒（JST） 震央 日本 岐阜県本巣郡西根尾村（現・本巣市） 北緯35度35分 東経136度20分（ 地図 ） 規模    マグニチュード（M）8.0 最大震度    震度7： 注1 福井県今立郡鯖江町、愛知県葉栗郡大田島村、東春日井郡勝川村 地震の種類 直下型地震 被害 死傷者数 死者7,273人、負傷者17,175人 注1：当時の震度階級では「激烈」 プロジェクト:地球科学 プロジェクト:災害 テンプレートを表示 濃尾地震 （のうびじしん）は、1891年（明治24年）10月28日に濃尾地方で発生した、日本史上最大の内陸地殻内地震。「 美濃・尾張地震 （みの・おわりじしん）」とも呼ばれている。辛卯の年に発生したことから 辛卯震災 と呼んでいる報告書もある。 目次 1 概要 1.1 震源断層 2 被害 3 各地の震度 4 前兆現象 5 報道 6 学術的な意義 7 地震防災 8 脚注 9 関連項目 10 外部リンク 概要 濃尾地震発生当時の根尾谷断層 濃尾地震を引き起こした根尾谷断層 写真中央を斜めに走る段差が根尾谷断層 濃尾地震は、1891年10月28日6時38分50秒に発生した。震源は、岐阜県本巣郡西根尾村（現・本巣市）、北緯35度35分、東経136度20分付近。河角廣 (1951) は岐阜市付近（北緯35.6度、東経136.6度）に震央を仮定し規模 M K = 7.0 を与え [1] 、マグニチュードは M = 8.4 に換算されているが、明治・大正期の地震については0.5程度大きく見積もられているとされる [2] 。また、震央距離と震度との関係など当時のデータから後にM8.0 [3] とも推定される。アメリカ地質調査所 (USGS) でも8.0としている [4] 。「根尾谷断層帯」が活動した典型的な内陸地殻内地震（いわゆる直下型地震）であり、これは記録が残っている日本の内陸...
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