How to evaluate a complex integral? [closed]

Show that |$int_{C} frac{e^z}{bar z + 1} dz$| ≤ $2pi e^2$ where C is the circle |z-1| = 1.

I think C can be parametrized by z = $1 + e^{it}$ and $z' = ie^{it}dt$

with 0 ≤ t ≤ $2 pi$

How do i have to continue this proof?

asked Nov 29 at 12:13

Peter van de Berg

198

closed as off-topic by Brahadeesh, Did, Alexander Gruber♦ Nov 30 at 3:18

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Show that |$int_{C} frac{e^z}{bar z + 1} dz$| ≤ $2pi e^2$ where C is the circle |z-1| = 1.

I think C can be parametrized by z = $1 + e^{it}$ and $z' = ie^{it}dt$

with 0 ≤ t ≤ $2 pi$

How do i have to continue this proof?

asked Nov 29 at 12:13

Peter van de Berg

198

closed as off-topic by Brahadeesh, Did, Alexander Gruber♦ Nov 30 at 3:18

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, Did, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment |

Show that |$int_{C} frac{e^z}{bar z + 1} dz$| ≤ $2pi e^2$ where C is the circle |z-1| = 1.

I think C can be parametrized by z = $1 + e^{it}$ and $z' = ie^{it}dt$

with 0 ≤ t ≤ $2 pi$

How do i have to continue this proof?

asked Nov 29 at 12:13

Peter van de Berg

198

Show that |$int_{C} frac{e^z}{bar z + 1} dz$| ≤ $2pi e^2$ where C is the circle |z-1| = 1.

I think C can be parametrized by z = $1 + e^{it}$ and $z' = ie^{it}dt$

with 0 ≤ t ≤ $2 pi$

How do i have to continue this proof?

complex-analysis complex-integration

asked Nov 29 at 12:13

Peter van de Berg

198

asked Nov 29 at 12:13

Peter van de Berg

198

asked Nov 29 at 12:13

Peter van de Berg

198

asked Nov 29 at 12:13

Peter van de Berg

198

asked Nov 29 at 12:13

Peter van de Berg

198

closed as off-topic by Brahadeesh, Did, Alexander Gruber♦ Nov 30 at 3:18

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, Did, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Brahadeesh, Did, Alexander Gruber♦ Nov 30 at 3:18

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, Did, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment |

1 Answer
1

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oldest

votes

By the ML inequality, we have that

$$Bigg|int_Gamma f(z)mathrm{d}zBigg| leq max_{z in Gamma}f(z) cdot l(Gamma)$$

where $l(Gamma)$ is the arc length of $Gamma$.

Thus, for your given integral, it is :

$$Bigg|int_C frac{e^z}{bar{z}+1}mathrm{d}zBigg| leq max_{z in Gamma}Bigg{frac{e^z}{bar{z}+1}Bigg}cdot l(Gamma)$$

But $C$ is the circle $|z-1| = 1$, which means that it has radius $r=1$ and thus its length is $l(Gamma) = 2pi$. Can you use complex algebra now and make a case for the maximum of the integrand function ?

answered Nov 29 at 12:26

Rebellos

14.3k31244

Thanks, i already see what i have to do!
– Peter van de Berg
Nov 29 at 12:30

@PetervandeBerg Glad I could help !
– Rebellos
Nov 29 at 12:33

add a comment |

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

By the ML inequality, we have that

$$Bigg|int_Gamma f(z)mathrm{d}zBigg| leq max_{z in Gamma}f(z) cdot l(Gamma)$$

where $l(Gamma)$ is the arc length of $Gamma$.

Thus, for your given integral, it is :

$$Bigg|int_C frac{e^z}{bar{z}+1}mathrm{d}zBigg| leq max_{z in Gamma}Bigg{frac{e^z}{bar{z}+1}Bigg}cdot l(Gamma)$$

answered Nov 29 at 12:26

Rebellos

14.3k31244

Thanks, i already see what i have to do!
– Peter van de Berg
Nov 29 at 12:30

@PetervandeBerg Glad I could help !
– Rebellos
Nov 29 at 12:33

add a comment |

By the ML inequality, we have that

$$Bigg|int_Gamma f(z)mathrm{d}zBigg| leq max_{z in Gamma}f(z) cdot l(Gamma)$$

where $l(Gamma)$ is the arc length of $Gamma$.

Thus, for your given integral, it is :

$$Bigg|int_C frac{e^z}{bar{z}+1}mathrm{d}zBigg| leq max_{z in Gamma}Bigg{frac{e^z}{bar{z}+1}Bigg}cdot l(Gamma)$$

answered Nov 29 at 12:26

Rebellos

14.3k31244

Thanks, i already see what i have to do!
– Peter van de Berg
Nov 29 at 12:30

@PetervandeBerg Glad I could help !
– Rebellos
Nov 29 at 12:33

add a comment |

By the ML inequality, we have that

$$Bigg|int_Gamma f(z)mathrm{d}zBigg| leq max_{z in Gamma}f(z) cdot l(Gamma)$$

where $l(Gamma)$ is the arc length of $Gamma$.

Thus, for your given integral, it is :

$$Bigg|int_C frac{e^z}{bar{z}+1}mathrm{d}zBigg| leq max_{z in Gamma}Bigg{frac{e^z}{bar{z}+1}Bigg}cdot l(Gamma)$$

answered Nov 29 at 12:26

Rebellos

14.3k31244

By the ML inequality, we have that

$$Bigg|int_Gamma f(z)mathrm{d}zBigg| leq max_{z in Gamma}f(z) cdot l(Gamma)$$

where $l(Gamma)$ is the arc length of $Gamma$.

Thus, for your given integral, it is :

$$Bigg|int_C frac{e^z}{bar{z}+1}mathrm{d}zBigg| leq max_{z in Gamma}Bigg{frac{e^z}{bar{z}+1}Bigg}cdot l(Gamma)$$

answered Nov 29 at 12:26

Rebellos

14.3k31244

answered Nov 29 at 12:26

Rebellos

14.3k31244

answered Nov 29 at 12:26

Rebellos

14.3k31244

answered Nov 29 at 12:26

Rebellos

14.3k31244

Thanks, i already see what i have to do!
– Peter van de Berg
Nov 29 at 12:30

@PetervandeBerg Glad I could help !
– Rebellos
Nov 29 at 12:33

add a comment |

Thanks, i already see what i have to do!
– Peter van de Berg
Nov 29 at 12:30

@PetervandeBerg Glad I could help !
– Rebellos
Nov 29 at 12:33

Thanks, i already see what i have to do!
– Peter van de Berg
Nov 29 at 12:30

@PetervandeBerg Glad I could help !
– Rebellos
Nov 29 at 12:33

add a comment |

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