Complex Integration - Complex Analysis

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Question

Let be the region of the complex plane enclosed by $x = \pm 2 $\text{ and }$ y = \pm 2$

Compute $\int_C $\frac{\cos $z}{z(z^2$+8)}$ dz$

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Answer

Recall Cauchy's integral formula, which states

$f(z) = $\frac{1}{2\pi i}$\int_C$\frac{f(s)}{s-z}$ds$

In this case, we have $f(s) = $$\frac{cos(s)}{s^2$ + 8}$$

Plugging in gives us

$\int_C $\frac{\frac{\cos $z}{z^2$+8}$}{z} dz = 2\pi i f(0) = 2\pi i $\frac{1}{8}$ = $\frac{\pi i }{4}$$

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Question

Let be the unit circle.

Compute $\int_C $z^2$ sin\left($\frac{1}{z}$\right) dz$

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Answer

Recall the Taylor expansion of

$\sin(z) = z - $$\frac{z^3$}{3!}$ + $$\frac{z^5$}{5!}$ - $$\frac{z^7$}{7!}$ + ...$

Use this to write

The coefficient of $\frac{1}{z}$ is the residue, so by the residue theorem, the value of the integral is

$-$\frac{\pi i}{3}$$

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Question

Let be the region of the complex plane enclosed by $x = \pm 2 $\text{ and }$ y = \pm 2$

Compute $\int_C $\frac{\cos $z}{z(z^2$+8)}$ dz$

Tap to reveal answer

Answer

Recall Cauchy's integral formula, which states

$f(z) = $\frac{1}{2\pi i}$\int_C$\frac{f(s)}{s-z}$ds$

In this case, we have $f(s) = $$\frac{cos(s)}{s^2$ + 8}$$

Plugging in gives us

$\int_C $\frac{\frac{\cos $z}{z^2$+8}$}{z} dz = 2\pi i f(0) = 2\pi i $\frac{1}{8}$ = $\frac{\pi i }{4}$$

← Didn't Know|Knew It →

Question

Let be the unit circle.

Compute $\int_C $z^2$ sin\left($\frac{1}{z}$\right) dz$

Tap to reveal answer

Answer

Recall the Taylor expansion of

$\sin(z) = z - $$\frac{z^3$}{3!}$ + $$\frac{z^5$}{5!}$ - $$\frac{z^7$}{7!}$ + ...$

Use this to write

The coefficient of $\frac{1}{z}$ is the residue, so by the residue theorem, the value of the integral is

$-$\frac{\pi i}{3}$$

← Didn't Know|Knew It →

Question

Let be the region of the complex plane enclosed by $x = \pm 2 $\text{ and }$ y = \pm 2$

Compute $\int_C $\frac{\cos $z}{z(z^2$+8)}$ dz$

Tap to reveal answer

Answer

Recall Cauchy's integral formula, which states

$f(z) = $\frac{1}{2\pi i}$\int_C$\frac{f(s)}{s-z}$ds$

In this case, we have $f(s) = $$\frac{cos(s)}{s^2$ + 8}$$

Plugging in gives us

$\int_C $\frac{\frac{\cos $z}{z^2$+8}$}{z} dz = 2\pi i f(0) = 2\pi i $\frac{1}{8}$ = $\frac{\pi i }{4}$$

← Didn't Know|Knew It →

Question

Let be the unit circle.

Compute $\int_C $z^2$ sin\left($\frac{1}{z}$\right) dz$

Tap to reveal answer

Answer

Recall the Taylor expansion of

$\sin(z) = z - $$\frac{z^3$}{3!}$ + $$\frac{z^5$}{5!}$ - $$\frac{z^7$}{7!}$ + ...$

Use this to write

The coefficient of $\frac{1}{z}$ is the residue, so by the residue theorem, the value of the integral is

$-$\frac{\pi i}{3}$$

← Didn't Know|Knew It →

Question

Let be the region of the complex plane enclosed by $x = \pm 2 $\text{ and }$ y = \pm 2$

Compute $\int_C $\frac{\cos $z}{z(z^2$+8)}$ dz$

Tap to reveal answer

Answer

Recall Cauchy's integral formula, which states

$f(z) = $\frac{1}{2\pi i}$\int_C$\frac{f(s)}{s-z}$ds$

In this case, we have $f(s) = $$\frac{cos(s)}{s^2$ + 8}$$

Plugging in gives us

$\int_C $\frac{\frac{\cos $z}{z^2$+8}$}{z} dz = 2\pi i f(0) = 2\pi i $\frac{1}{8}$ = $\frac{\pi i }{4}$$

← Didn't Know|Knew It →

Question

Let be the unit circle.

Compute $\int_C $z^2$ sin\left($\frac{1}{z}$\right) dz$

Tap to reveal answer

Answer

Recall the Taylor expansion of

$\sin(z) = z - $$\frac{z^3$}{3!}$ + $$\frac{z^5$}{5!}$ - $$\frac{z^7$}{7!}$ + ...$

Use this to write

The coefficient of $\frac{1}{z}$ is the residue, so by the residue theorem, the value of the integral is

$-$\frac{\pi i}{3}$$

← Didn't Know|Knew It →

Knew It

Complex Integration - Complex Analysis

Let be the region of the complex plane enclosed by Compute

Recall Cauchy's integral formula, which states In this case, we have Plugging in gives us

Let be the unit circle. Compute

Recall the Taylor expansion of Use this to write The coefficient of is the residue, so by the residue theorem, the value of the integral is

Let be the region of the complex plane enclosed by Compute

Recall Cauchy's integral formula, which states In this case, we have Plugging in gives us

Let be the unit circle. Compute

Recall the Taylor expansion of Use this to write The coefficient of is the residue, so by the residue theorem, the value of the integral is

Let be the region of the complex plane enclosed by Compute

Recall Cauchy's integral formula, which states In this case, we have Plugging in gives us

Let be the unit circle. Compute

Recall the Taylor expansion of Use this to write The coefficient of is the residue, so by the residue theorem, the value of the integral is

Let be the region of the complex plane enclosed by Compute

Recall Cauchy's integral formula, which states In this case, we have Plugging in gives us

Let be the unit circle. Compute

Recall the Taylor expansion of Use this to write The coefficient of is the residue, so by the residue theorem, the value of the integral is

Let be the region of the complex plane enclosed by $x = \pm 2 $\text{ and }$ y = \pm 2$

Compute $\int_C $\frac{\cos $z}{z(z^2$+8)}$ dz$

Recall Cauchy's integral formula, which states

$f(z) = $\frac{1}{2\pi i}$\int_C$\frac{f(s)}{s-z}$ds$

In this case, we have $f(s) = $$\frac{cos(s)}{s^2$ + 8}$$

Plugging in gives us

$\int_C $\frac{\frac{\cos $z}{z^2$+8}$}{z} dz = 2\pi i f(0) = 2\pi i $\frac{1}{8}$ = $\frac{\pi i }{4}$$

Let be the unit circle.

Compute $\int_C $z^2$ sin\left($\frac{1}{z}$\right) dz$

Recall the Taylor expansion of

$\sin(z) = z - $$\frac{z^3$}{3!}$ + $$\frac{z^5$}{5!}$ - $$\frac{z^7$}{7!}$ + ...$

Use this to write

The coefficient of $\frac{1}{z}$ is the residue, so by the residue theorem, the value of the integral is

$-$\frac{\pi i}{3}$$

Let be the region of the complex plane enclosed by $x = \pm 2 $\text{ and }$ y = \pm 2$

Compute $\int_C $\frac{\cos $z}{z(z^2$+8)}$ dz$

Recall Cauchy's integral formula, which states

$f(z) = $\frac{1}{2\pi i}$\int_C$\frac{f(s)}{s-z}$ds$

In this case, we have $f(s) = $$\frac{cos(s)}{s^2$ + 8}$$

Plugging in gives us

$\int_C $\frac{\frac{\cos $z}{z^2$+8}$}{z} dz = 2\pi i f(0) = 2\pi i $\frac{1}{8}$ = $\frac{\pi i }{4}$$

Let be the unit circle.

Compute $\int_C $z^2$ sin\left($\frac{1}{z}$\right) dz$

Recall the Taylor expansion of

$\sin(z) = z - $$\frac{z^3$}{3!}$ + $$\frac{z^5$}{5!}$ - $$\frac{z^7$}{7!}$ + ...$

Use this to write

The coefficient of $\frac{1}{z}$ is the residue, so by the residue theorem, the value of the integral is

$-$\frac{\pi i}{3}$$

Let be the region of the complex plane enclosed by $x = \pm 2 $\text{ and }$ y = \pm 2$

Compute $\int_C $\frac{\cos $z}{z(z^2$+8)}$ dz$

Recall Cauchy's integral formula, which states

$f(z) = $\frac{1}{2\pi i}$\int_C$\frac{f(s)}{s-z}$ds$

In this case, we have $f(s) = $$\frac{cos(s)}{s^2$ + 8}$$

Plugging in gives us

$\int_C $\frac{\frac{\cos $z}{z^2$+8}$}{z} dz = 2\pi i f(0) = 2\pi i $\frac{1}{8}$ = $\frac{\pi i }{4}$$

Let be the unit circle.

Compute $\int_C $z^2$ sin\left($\frac{1}{z}$\right) dz$

Recall the Taylor expansion of

$\sin(z) = z - $$\frac{z^3$}{3!}$ + $$\frac{z^5$}{5!}$ - $$\frac{z^7$}{7!}$ + ...$

Use this to write

The coefficient of $\frac{1}{z}$ is the residue, so by the residue theorem, the value of the integral is

$-$\frac{\pi i}{3}$$

Let be the region of the complex plane enclosed by $x = \pm 2 $\text{ and }$ y = \pm 2$

Compute $\int_C $\frac{\cos $z}{z(z^2$+8)}$ dz$

Recall Cauchy's integral formula, which states

$f(z) = $\frac{1}{2\pi i}$\int_C$\frac{f(s)}{s-z}$ds$

In this case, we have $f(s) = $$\frac{cos(s)}{s^2$ + 8}$$

Plugging in gives us

$\int_C $\frac{\frac{\cos $z}{z^2$+8}$}{z} dz = 2\pi i f(0) = 2\pi i $\frac{1}{8}$ = $\frac{\pi i }{4}$$

Let be the unit circle.

Compute $\int_C $z^2$ sin\left($\frac{1}{z}$\right) dz$

Recall the Taylor expansion of

$\sin(z) = z - $$\frac{z^3$}{3!}$ + $$\frac{z^5$}{5!}$ - $$\frac{z^7$}{7!}$ + ...$

Use this to write

The coefficient of $\frac{1}{z}$ is the residue, so by the residue theorem, the value of the integral is

$-$\frac{\pi i}{3}$$