Lineintegral of absolute value with path $gamma: [0,1]rightarrowmathbb{C},tmapsto i +exp(ipi t)$
up vote
0
down vote
favorite
Calculate:
$int_{gamma} |z|dz$ with $gamma: [0,1]rightarrowmathbb{C},tmapsto i +exp(ipi t)$
I tried calculating it and actually made some progress, where one term vanished when splitting the integral in real and imaginary part, but the substitutions needed to arrive at the value afterwards just seem to complicated (I checked with Wolfram Alpha). Are there any tricks to bring to make this integral more managable ?
complex-analysis complex-integration line-integrals
asked Nov 28 at 9:05
Christian Singer
344113
add a comment |
up vote
0
down vote
favorite
Calculate:
$int_{gamma} |z|dz$ with $gamma: [0,1]rightarrowmathbb{C},tmapsto i +exp(ipi t)$
I tried calculating it and actually made some progress, where one term vanished when splitting the integral in real and imaginary part, but the substitutions needed to arrive at the value afterwards just seem to complicated (I checked with Wolfram Alpha). Are there any tricks to bring to make this integral more managable ?
complex-analysis complex-integration line-integrals
asked Nov 28 at 9:05
Christian Singer
344113
1
Well, the path is just the upper arc of a circle of radius $1$...Can you see it?
– DonAntonio
Nov 28 at 9:16
2
All you need is $2sin (pi t) cos (pi t)=sin (2pi t)$ and $2sin^{2} (pi t)=1-cos (2pi t)$. If you show your work you can get some help. I don't see any complicated expressions here.
– Kavi Rama Murthy
Nov 28 at 9:18
I tought about that (wich would make the value be $1/2cdotpi r^{2}$ but is that a formal argument ? @DonAntonio
– Christian Singer
Nov 28 at 9:18
@Cristian What is a "formal argument"? It's just basic knowledge of complex numbers: $$i+e^{ipi t}=cospi t+i(sinpi t+1)$$ The above is a circle around $;(0,1);$ of radius 1...but only the upper arc as $;0le tlepi;$ ...
– DonAntonio
Nov 28 at 9:42
Well, I started off with using the definition of a line integral so I got $ipi int_{0}^{1}|i+exp(ipi t)|cdot exp(ipi t) dt$ and using the definition of absolute value of a complex number I'm already stuck, I don't see how I could apply those trig identities here. @Kavi Rama Murthy
– Christian Singer
Nov 28 at 10:13
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Calculate:
$int_{gamma} |z|dz$ with $gamma: [0,1]rightarrowmathbb{C},tmapsto i +exp(ipi t)$
I tried calculating it and actually made some progress, where one term vanished when splitting the integral in real and imaginary part, but the substitutions needed to arrive at the value afterwards just seem to complicated (I checked with Wolfram Alpha). Are there any tricks to bring to make this integral more managable ?
complex-analysis complex-integration line-integrals
asked Nov 28 at 9:05
Christian Singer
344113
Calculate:
$int_{gamma} |z|dz$ with $gamma: [0,1]rightarrowmathbb{C},tmapsto i +exp(ipi t)$
I tried calculating it and actually made some progress, where one term vanished when splitting the integral in real and imaginary part, but the substitutions needed to arrive at the value afterwards just seem to complicated (I checked with Wolfram Alpha). Are there any tricks to bring to make this integral more managable ?
complex-analysis complex-integration line-integrals
complex-analysis complex-integration line-integrals
asked Nov 28 at 9:05
Christian Singer
344113
asked Nov 28 at 9:05
Christian Singer
344113
asked Nov 28 at 9:05
Christian Singer
344113
asked Nov 28 at 9:05
Christian Singer
344113
asked Nov 28 at 9:05
Christian Singer
344113
344113
1
Well, the path is just the upper arc of a circle of radius $1$...Can you see it?
– DonAntonio
Nov 28 at 9:16
2
All you need is $2sin (pi t) cos (pi t)=sin (2pi t)$ and $2sin^{2} (pi t)=1-cos (2pi t)$. If you show your work you can get some help. I don't see any complicated expressions here.
– Kavi Rama Murthy
Nov 28 at 9:18
I tought about that (wich would make the value be $1/2cdotpi r^{2}$ but is that a formal argument ? @DonAntonio
– Christian Singer
Nov 28 at 9:18
@Cristian What is a "formal argument"? It's just basic knowledge of complex numbers: $$i+e^{ipi t}=cospi t+i(sinpi t+1)$$ The above is a circle around $;(0,1);$ of radius 1...but only the upper arc as $;0le tlepi;$ ...
– DonAntonio
Nov 28 at 9:42
Well, I started off with using the definition of a line integral so I got $ipi int_{0}^{1}|i+exp(ipi t)|cdot exp(ipi t) dt$ and using the definition of absolute value of a complex number I'm already stuck, I don't see how I could apply those trig identities here. @Kavi Rama Murthy
– Christian Singer
Nov 28 at 10:13
add a comment |
1
Well, the path is just the upper arc of a circle of radius $1$...Can you see it?
– DonAntonio
Nov 28 at 9:16
2
All you need is $2sin (pi t) cos (pi t)=sin (2pi t)$ and $2sin^{2} (pi t)=1-cos (2pi t)$. If you show your work you can get some help. I don't see any complicated expressions here.
– Kavi Rama Murthy
Nov 28 at 9:18
I tought about that (wich would make the value be $1/2cdotpi r^{2}$ but is that a formal argument ? @DonAntonio
– Christian Singer
Nov 28 at 9:18
@Cristian What is a "formal argument"? It's just basic knowledge of complex numbers: $$i+e^{ipi t}=cospi t+i(sinpi t+1)$$ The above is a circle around $;(0,1);$ of radius 1...but only the upper arc as $;0le tlepi;$ ...
– DonAntonio
Nov 28 at 9:42
Well, I started off with using the definition of a line integral so I got $ipi int_{0}^{1}|i+exp(ipi t)|cdot exp(ipi t) dt$ and using the definition of absolute value of a complex number I'm already stuck, I don't see how I could apply those trig identities here. @Kavi Rama Murthy
– Christian Singer
Nov 28 at 10:13
1
1
Well, the path is just the upper arc of a circle of radius $1$...Can you see it?
– DonAntonio
Nov 28 at 9:16
Well, the path is just the upper arc of a circle of radius $1$...Can you see it?
– DonAntonio
Nov 28 at 9:16
2
2
All you need is $2sin (pi t) cos (pi t)=sin (2pi t)$ and $2sin^{2} (pi t)=1-cos (2pi t)$. If you show your work you can get some help. I don't see any complicated expressions here.
– Kavi Rama Murthy
Nov 28 at 9:18
All you need is $2sin (pi t) cos (pi t)=sin (2pi t)$ and $2sin^{2} (pi t)=1-cos (2pi t)$. If you show your work you can get some help. I don't see any complicated expressions here.
– Kavi Rama Murthy
Nov 28 at 9:18
I tought about that (wich would make the value be $1/2cdotpi r^{2}$ but is that a formal argument ? @DonAntonio
– Christian Singer
Nov 28 at 9:18
I tought about that (wich would make the value be $1/2cdotpi r^{2}$ but is that a formal argument ? @DonAntonio
– Christian Singer
Nov 28 at 9:18
@Cristian What is a "formal argument"? It's just basic knowledge of complex numbers: $$i+e^{ipi t}=cospi t+i(sinpi t+1)$$ The above is a circle around $;(0,1);$ of radius 1...but only the upper arc as $;0le tlepi;$ ...
– DonAntonio
Nov 28 at 9:42
@Cristian What is a "formal argument"? It's just basic knowledge of complex numbers: $$i+e^{ipi t}=cospi t+i(sinpi t+1)$$ The above is a circle around $;(0,1);$ of radius 1...but only the upper arc as $;0le tlepi;$ ...
– DonAntonio
Nov 28 at 9:42
Well, I started off with using the definition of a line integral so I got $ipi int_{0}^{1}|i+exp(ipi t)|cdot exp(ipi t) dt$ and using the definition of absolute value of a complex number I'm already stuck, I don't see how I could apply those trig identities here. @Kavi Rama Murthy
– Christian Singer
Nov 28 at 10:13
Well, I started off with using the definition of a line integral so I got $ipi int_{0}^{1}|i+exp(ipi t)|cdot exp(ipi t) dt$ and using the definition of absolute value of a complex number I'm already stuck, I don't see how I could apply those trig identities here. @Kavi Rama Murthy
– Christian Singer
Nov 28 at 10:13
add a comment |
1 Answer
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Well, writing down using the definitions:
begin{align}int_gamma |z|dz & = ipiint_0^1|i+e^{ipi t}|e^{ipi t}dt \& = ipiint_0^1 left(cos(pi t)^2+(1+sin(pi t))^2right)^{1/2}(cos(pi t)+isin(pi t))dt .end{align}
Now, we compute the real and imaginary part of that integral in two separated integrals:
begin{align} int_0^1left(cos(pi t)^2+(1+sin(pi t))^2right)^{1/2}cos(pi t)dt & =int_0^1 left(1-sin^2(pi t)+(1+sin(pi t))^2right)^{1/2}cos(pi t)dt \
& = sqrt{2}int_0^1 (1+sin(pi t))^{1/2}cos(pi t)dt \ & = frac{2sqrt{2}}{3pi}left(1+sin(pi t)right)^{3/2}Bigr|_0^1 \ & = 0end{align}
since $int (1+sin(pi t))^{1/2}cos(pi t)=frac{1}{pi}int u^{1/2} du$ using $u=1+sin(pi t)$.
Now, the imaginary part: It's harder to calculate, but using several times chain rule, you will get:
$$int (1+sin(pi t))^{1/2}sin(pi t)=-frac{sqrt{sin(pi x)+1}(-4sin^3(frac{pi x}{2})+3cos(frac{pi x}{2})+cos(frac{3pi x}{2}))}{3pi(sin(frac{pi x}{2})+cos(frac{pi x}{2}))},$$
and replacing:
$$int_0^1 (1+sin(pi t))^{1/2}sin(pi t)=frac{8}{3pi}.$$
So
$$int_gamma |z|dz= ifrac{8}{3pi}.$$
answered Nov 28 at 16:57
José Alejandro Aburto Araneda
41019
1
Didn't you forget a square root when calculating the absolute value ?
– Christian Singer
Nov 28 at 17:03
Yes, indeed. I'll correct now
– José Alejandro Aburto Araneda
Nov 28 at 17:04
add a comment |
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1 Answer
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Well, writing down using the definitions:
begin{align}int_gamma |z|dz & = ipiint_0^1|i+e^{ipi t}|e^{ipi t}dt \& = ipiint_0^1 left(cos(pi t)^2+(1+sin(pi t))^2right)^{1/2}(cos(pi t)+isin(pi t))dt .end{align}
Now, we compute the real and imaginary part of that integral in two separated integrals:
begin{align} int_0^1left(cos(pi t)^2+(1+sin(pi t))^2right)^{1/2}cos(pi t)dt & =int_0^1 left(1-sin^2(pi t)+(1+sin(pi t))^2right)^{1/2}cos(pi t)dt \
& = sqrt{2}int_0^1 (1+sin(pi t))^{1/2}cos(pi t)dt \ & = frac{2sqrt{2}}{3pi}left(1+sin(pi t)right)^{3/2}Bigr|_0^1 \ & = 0end{align}
since $int (1+sin(pi t))^{1/2}cos(pi t)=frac{1}{pi}int u^{1/2} du$ using $u=1+sin(pi t)$.
Now, the imaginary part: It's harder to calculate, but using several times chain rule, you will get:
$$int (1+sin(pi t))^{1/2}sin(pi t)=-frac{sqrt{sin(pi x)+1}(-4sin^3(frac{pi x}{2})+3cos(frac{pi x}{2})+cos(frac{3pi x}{2}))}{3pi(sin(frac{pi x}{2})+cos(frac{pi x}{2}))},$$
and replacing:
$$int_0^1 (1+sin(pi t))^{1/2}sin(pi t)=frac{8}{3pi}.$$
So
$$int_gamma |z|dz= ifrac{8}{3pi}.$$
answered Nov 28 at 16:57
José Alejandro Aburto Araneda
41019
1
Didn't you forget a square root when calculating the absolute value ?
– Christian Singer
Nov 28 at 17:03
Yes, indeed. I'll correct now
– José Alejandro Aburto Araneda
Nov 28 at 17:04
add a comment |
up vote
1
down vote
Well, writing down using the definitions:
begin{align}int_gamma |z|dz & = ipiint_0^1|i+e^{ipi t}|e^{ipi t}dt \& = ipiint_0^1 left(cos(pi t)^2+(1+sin(pi t))^2right)^{1/2}(cos(pi t)+isin(pi t))dt .end{align}
Now, we compute the real and imaginary part of that integral in two separated integrals:
begin{align} int_0^1left(cos(pi t)^2+(1+sin(pi t))^2right)^{1/2}cos(pi t)dt & =int_0^1 left(1-sin^2(pi t)+(1+sin(pi t))^2right)^{1/2}cos(pi t)dt \
& = sqrt{2}int_0^1 (1+sin(pi t))^{1/2}cos(pi t)dt \ & = frac{2sqrt{2}}{3pi}left(1+sin(pi t)right)^{3/2}Bigr|_0^1 \ & = 0end{align}
since $int (1+sin(pi t))^{1/2}cos(pi t)=frac{1}{pi}int u^{1/2} du$ using $u=1+sin(pi t)$.
Now, the imaginary part: It's harder to calculate, but using several times chain rule, you will get:
$$int (1+sin(pi t))^{1/2}sin(pi t)=-frac{sqrt{sin(pi x)+1}(-4sin^3(frac{pi x}{2})+3cos(frac{pi x}{2})+cos(frac{3pi x}{2}))}{3pi(sin(frac{pi x}{2})+cos(frac{pi x}{2}))},$$
and replacing:
$$int_0^1 (1+sin(pi t))^{1/2}sin(pi t)=frac{8}{3pi}.$$
So
$$int_gamma |z|dz= ifrac{8}{3pi}.$$
answered Nov 28 at 16:57
José Alejandro Aburto Araneda
41019
1
Didn't you forget a square root when calculating the absolute value ?
– Christian Singer
Nov 28 at 17:03
Yes, indeed. I'll correct now
– José Alejandro Aburto Araneda
Nov 28 at 17:04
add a comment |
up vote
1
down vote
up vote
1
down vote
Well, writing down using the definitions:
begin{align}int_gamma |z|dz & = ipiint_0^1|i+e^{ipi t}|e^{ipi t}dt \& = ipiint_0^1 left(cos(pi t)^2+(1+sin(pi t))^2right)^{1/2}(cos(pi t)+isin(pi t))dt .end{align}
Now, we compute the real and imaginary part of that integral in two separated integrals:
begin{align} int_0^1left(cos(pi t)^2+(1+sin(pi t))^2right)^{1/2}cos(pi t)dt & =int_0^1 left(1-sin^2(pi t)+(1+sin(pi t))^2right)^{1/2}cos(pi t)dt \
& = sqrt{2}int_0^1 (1+sin(pi t))^{1/2}cos(pi t)dt \ & = frac{2sqrt{2}}{3pi}left(1+sin(pi t)right)^{3/2}Bigr|_0^1 \ & = 0end{align}
since $int (1+sin(pi t))^{1/2}cos(pi t)=frac{1}{pi}int u^{1/2} du$ using $u=1+sin(pi t)$.
Now, the imaginary part: It's harder to calculate, but using several times chain rule, you will get:
$$int (1+sin(pi t))^{1/2}sin(pi t)=-frac{sqrt{sin(pi x)+1}(-4sin^3(frac{pi x}{2})+3cos(frac{pi x}{2})+cos(frac{3pi x}{2}))}{3pi(sin(frac{pi x}{2})+cos(frac{pi x}{2}))},$$
and replacing:
$$int_0^1 (1+sin(pi t))^{1/2}sin(pi t)=frac{8}{3pi}.$$
So
$$int_gamma |z|dz= ifrac{8}{3pi}.$$
answered Nov 28 at 16:57
José Alejandro Aburto Araneda
41019
Well, writing down using the definitions:
begin{align}int_gamma |z|dz & = ipiint_0^1|i+e^{ipi t}|e^{ipi t}dt \& = ipiint_0^1 left(cos(pi t)^2+(1+sin(pi t))^2right)^{1/2}(cos(pi t)+isin(pi t))dt .end{align}
Now, we compute the real and imaginary part of that integral in two separated integrals:
begin{align} int_0^1left(cos(pi t)^2+(1+sin(pi t))^2right)^{1/2}cos(pi t)dt & =int_0^1 left(1-sin^2(pi t)+(1+sin(pi t))^2right)^{1/2}cos(pi t)dt \
& = sqrt{2}int_0^1 (1+sin(pi t))^{1/2}cos(pi t)dt \ & = frac{2sqrt{2}}{3pi}left(1+sin(pi t)right)^{3/2}Bigr|_0^1 \ & = 0end{align}
since $int (1+sin(pi t))^{1/2}cos(pi t)=frac{1}{pi}int u^{1/2} du$ using $u=1+sin(pi t)$.
Now, the imaginary part: It's harder to calculate, but using several times chain rule, you will get:
$$int (1+sin(pi t))^{1/2}sin(pi t)=-frac{sqrt{sin(pi x)+1}(-4sin^3(frac{pi x}{2})+3cos(frac{pi x}{2})+cos(frac{3pi x}{2}))}{3pi(sin(frac{pi x}{2})+cos(frac{pi x}{2}))},$$
and replacing:
$$int_0^1 (1+sin(pi t))^{1/2}sin(pi t)=frac{8}{3pi}.$$
So
$$int_gamma |z|dz= ifrac{8}{3pi}.$$
answered Nov 28 at 16:57
José Alejandro Aburto Araneda
41019
edited Dec 5 at 0:28
answered Nov 28 at 16:57
José Alejandro Aburto Araneda
41019
answered Nov 28 at 16:57
José Alejandro Aburto Araneda
41019
answered Nov 28 at 16:57
José Alejandro Aburto Araneda
41019
41019
1
Didn't you forget a square root when calculating the absolute value ?
– Christian Singer
Nov 28 at 17:03
Yes, indeed. I'll correct now
– José Alejandro Aburto Araneda
Nov 28 at 17:04
add a comment |
1
Didn't you forget a square root when calculating the absolute value ?
– Christian Singer
Nov 28 at 17:03
Yes, indeed. I'll correct now
– José Alejandro Aburto Araneda
Nov 28 at 17:04
1
1
Didn't you forget a square root when calculating the absolute value ?
– Christian Singer
Nov 28 at 17:03
Didn't you forget a square root when calculating the absolute value ?
– Christian Singer
Nov 28 at 17:03
Yes, indeed. I'll correct now
– José Alejandro Aburto Araneda
Nov 28 at 17:04
Yes, indeed. I'll correct now
– José Alejandro Aburto Araneda
Nov 28 at 17:04
add a comment |
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1
Well, the path is just the upper arc of a circle of radius $1$...Can you see it?
– DonAntonio
Nov 28 at 9:16
2
All you need is $2sin (pi t) cos (pi t)=sin (2pi t)$ and $2sin^{2} (pi t)=1-cos (2pi t)$. If you show your work you can get some help. I don't see any complicated expressions here.
– Kavi Rama Murthy
Nov 28 at 9:18
I tought about that (wich would make the value be $1/2cdotpi r^{2}$ but is that a formal argument ? @DonAntonio
– Christian Singer
Nov 28 at 9:18
@Cristian What is a "formal argument"? It's just basic knowledge of complex numbers: $$i+e^{ipi t}=cospi t+i(sinpi t+1)$$ The above is a circle around $;(0,1);$ of radius 1...but only the upper arc as $;0le tlepi;$ ...
– DonAntonio
Nov 28 at 9:42
Well, I started off with using the definition of a line integral so I got $ipi int_{0}^{1}|i+exp(ipi t)|cdot exp(ipi t) dt$ and using the definition of absolute value of a complex number I'm already stuck, I don't see how I could apply those trig identities here. @Kavi Rama Murthy
– Christian Singer
Nov 28 at 10:13