Complex Integration & Cauchy's Theorem

Solution:

Parameterize: $z(\theta) = e^{i\theta}$ for $0 \leq \theta \leq 2\pi$, so $dz = ie^{i\theta}d\theta$

$\oint_{|z|=1} \frac{dz}{z} = \int_0^{2\pi} \frac{ie^{i\theta}d\theta}{e^{i\theta}} = \int_0^{2\pi} id\theta = 2\pi i$

Note: This is non-zero because $1/z$ is not analytic at $z = 0$, which is inside the curve.

Solution:

Parameterize: $z(t) = t(1+i)$ for $0 \leq t \leq 1$, so $dz = (1+i)dt$

$\int_C z^2 dz = \int_0^1 [t(1+i)]^2 (1+i)dt = (1+i)^3 \int_0^1 t^2 dt = \frac{(1+i)^3}{3} = \frac{-2+2i}{3}$

Solution:

Since $z^2$ is analytic everywhere (it's an entire function) and the unit disk is simply connected, by Cauchy's theorem, the integral around any closed curve is zero.

Alternatively, parameterizing $z = e^{i\theta}$: $\oint_{|z|=1} z^2 dz = \int_0^{2\pi} e^{2i\theta} \cdot ie^{i\theta}d\theta = i\int_0^{2\pi} e^{3i\theta}d\theta = 0$

Solution:

The function has poles at $z=0$ and $z=1$, both inside $|z|=3$.

By deformation, we can split this into integrals around small circles around each pole, but since the function is not analytic at these points, we'll need Cauchy's integral formula (covered in the next course) or residue theory to evaluate this.

For now, we note that Cauchy's theorem does not apply here because $f$ is not analytic at $z=0$ and $z=1$.

Solution:

In the right half-plane, we can take the principal branch of the logarithm. An antiderivative of $1/z$ is $\ln z$ (principal branch).

Therefore: $\int_C \frac{1}{z}dz = \ln i - \ln 1 = \ln i = i\frac{\pi}{2}$

Solution:

Parameterize: $z = e^{i\theta}$ for $0 \leq \theta \leq 2\pi$, so $dz = ie^{i\theta}d\theta$

$\int_C \frac{dz}{z} = \int_0^{2\pi} \frac{ie^{i\theta}}{e^{i\theta}}d\theta = i\int_0^{2\pi}d\theta = 2\pi i$

Note: This is not zero because $1/z$ has a singularity at $z = 0$, so it's not analytic in a simply connected domain containing $C$.

Solution:

Note that $e^{\cos\theta + i\sin\theta} = e^{e^{i\theta}}$. On the unit circle $z = e^{i\theta}$:

$\text{Re}(e^z) = e^{\cos\theta}\cos(\sin\theta)$ when $z = e^{i\theta}$

$\int_0^{2\pi} e^{\cos\theta}\cos(\sin\theta)d\theta = \text{Re}\left(\oint_{|z|=1} \frac{e^z}{iz}dz\right)$

Since $e^z$ is entire and $1/z$ has residue 1 at $z = 0$ (we'll learn this in residue theory), the integral equals $2\pi$.

Solution:

If $f$ is analytic, then $f'$ is also analytic (a key result we'll prove later). By Cauchy's theorem, the integral is zero.

Alternatively, if $\gamma$ starts and ends at $z_0$, then $\oint_\gamma f'(z)dz = f(z_0) - f(z_0) = 0$ by the fundamental theorem of calculus for complex functions.

Solution:

Parameterize: $z(t) = t(1+i)$ for $0 \leq t \leq 1$, so $\overline{z(t)} = t(1-i)$, $dz = (1+i)dt$

$\int_C \overline{z}dz = \int_0^1 t(1-i)(1+i)dt = \int_0^1 2tdt = 1$

Note: $\overline{z}$ is not analytic, so the value depends on the path.

Solution:

On the unit circle, $|z|^2 = 1$, so $|z|^2 = 1$

Parameterize: $z = e^{i\theta}$, $dz = ie^{i\theta}d\theta$

$\int_C |z|^2dz = \int_0^{2\pi} 1 \cdot ie^{i\theta}d\theta = i\int_0^{2\pi}e^{i\theta}d\theta = 0$

Solution:

Split: $\int_C \frac{z^2 + 1}{z}dz = \int_C zdz + \int_C \frac{1}{z}dz$

Since $z$ is analytic, $\int_C zdz = 0$ by Cauchy's theorem.

From Example 4.6, $\int_C \frac{1}{z}dz = 2\pi i$

Therefore, the integral equals $2\pi i$.

Solution:

By deformation theorem, we can shrink $C$ to a small circle around the origin. On a small circle, $e^z \approx 1$, so the integral approximates $\oint \frac{1}{z}dz = 2\pi i$.

More precisely, we can use Cauchy's integral formula (covered in the next course) to get the exact result.

Solution:

On the unit circle $z = e^{i\theta}$, we have $\text{Im}(e^z) = e^{\cos\theta}\sin(\sin\theta)$.

The integral is the imaginary part of $\oint_{|z|=1} \frac{e^z}{iz}dz$, which equals 0 (as we'll see with residue theory, or by symmetry).

Solution:

Using the mean value property (from Cauchy's integral formula), $f(0) = \frac{1}{2\pi} \int_0^{2\pi} f(Re^{i\theta})d\theta$

Taking absolute values: $|f(0)| \leq \frac{1}{2\pi} \int_0^{2\pi} |f(Re^{i\theta})|d\theta \leq \frac{1}{2\pi} \int_0^{2\pi} Md\theta = M$

Solution:

The function has poles at $z = \pm i$, both inside $C$.

Using deformation, we can split into integrals around small circles. However, we'll need residue theory (covered later) to evaluate this properly. The result is 0 by the residue theorem since the residues at $\pm i$ cancel out.

Solution:

Since $z^2$ is entire, the integral is path-independent. An antiderivative is $z^3/3$.

$\int_0^{1+i} z^2dz = \left[\frac{z^3}{3}\right]_0^{1+i} = \frac{(1+i)^3}{3} = \frac{-2+2i}{3}$

Solution:

Since $e^z$ is entire, we can use an antiderivative: $e^z$

$\int_0^{\pi + i\pi} e^z dz = e^z\Big|_0^{\pi + i\pi} = e^{\pi + i\pi} - e^0 = e^{\pi}e^{i\pi} - 1 = -e^{\pi} - 1$

Solution:

Parameterize: $z = e^{i\theta}$, $dz = ie^{i\theta}d\theta$

$\oint_{|z|=1} \frac{dz}{z^2} = \int_0^{2\pi} \frac{ie^{i\theta}}{e^{i2\theta}}d\theta = i\int_0^{2\pi} e^{-i\theta}d\theta = i\left[\frac{e^{-i\theta}}{-i}\right]_0^{2\pi} = 0$

Solution:

Consider the closed curve $\gamma = \gamma_1 - \gamma_2$ (going along $\gamma_1$ then backwards along $\gamma_2$).

By Cauchy's theorem: $\oint_\gamma f(z)dz = 0$

But $\oint_\gamma f(z)dz = \int_{\gamma_1} f(z)dz - \int_{\gamma_2} f(z)dz = 0$

Therefore: $\int_{\gamma_1} f(z)dz = \int_{\gamma_2} f(z)dz$

Solution:

The function has poles at $z = \pm 2i$. The circle $|z - i| = 1$ is centered at $i$ with radius 1, so it contains $z = 2i$ but not $z = -2i$.

Using residue theory (to be covered), the integral equals $2\pi i$ times the residue at $z = 2i$.

Solution:

This is exactly Cauchy's theorem. Since $f$ is analytic in a simply connected domain containing $C$ and its interior, the integral is zero.

Solution:

This follows from the residue theorem (covered later). If $f$ is analytic inside $\gamma$, then all residues are zero, so the integral is zero.

Solution:

Since $e^z$ is entire, the integral is path-independent. An antiderivative is $e^z$.

$\int_0^{i\pi} e^z dz = e^{i\pi} - e^0 = -1 - 1 = -2$

Solution:

On $C$, we have $|z-1| = 2$. Using the triangle inequality:

$|z+1| = |(z-1) + 2| \leq |z-1| + 2 = 2 + 2 = 4$

Therefore: $\left|\frac{z+1}{z-1}\right| \leq \frac{4}{2} = 2$ on $C$

$\left|\oint_C \frac{z+1}{z-1}dz\right| \leq \oint_C \left|\frac{z+1}{z-1}\right||dz| \leq 2 \oint_C |dz| = 2 \cdot 4\pi = 8\pi$

Solution:

Parameterize: $z = e^{i\theta}$, $dz = ie^{i\theta}d\theta$

$\oint_{|z|=1} \frac{dz}{z^n} = \int_0^{2\pi} \frac{ie^{i\theta}}{e^{in\theta}}d\theta = i\int_0^{2\pi} e^{i(1-n)\theta}d\theta$

This equals $2\pi i$ if $n = 1$, and 0 otherwise.

Solution:

If $C$ doesn't contain the origin, then $1/z^n$ is analytic inside $C$ (for $n \geq 0$) or the origin is outside $C$ (for $n < 0$).

By Cauchy's theorem, if the function is analytic inside $C$, the integral is zero. For $n \neq 1$, an antiderivative exists, so the integral over a closed curve is zero.

Solution:

Standard parameterization: $z(t) = a\cos t + ib\sin t$ for $t \in [0, 2\pi]$

Then $dz = (-a\sin t + ib\cos t) \, dt$

The integral:

Expanding: $= \int_0^{2\pi} (-a^2\cos t\sin t + iab\cos^2 t - iab\sin^2 t - b^2\sin t\cos t) \, dt$

$= \int_0^{2\pi} [-(a^2+b^2)\sin t\cos t + iab(\cos^2 t - \sin^2 t)] \, dt$

Using $\sin t\cos t = \frac{1}{2}\sin(2t)$ and $\cos^2 t - \sin^2 t = \cos(2t)$:

$= -\frac{a^2+b^2}{2}\int_0^{2\pi} \sin(2t) \, dt + iab\int_0^{2\pi} \cos(2t) \, dt = 0$

Alternatively, since $f(z) = z$ is entire (analytic everywhere), by Cauchy's theorem the integral over any closed curve is zero.

Solution:

Parameterize using $x = t$, so $y = t^2$ for $t \in [0, 1]$

Then $z(t) = t + it^2$, and $dz = (1 + 2it) \, dt$

The integral:

Expanding $(t + it^2)^2 = t^2 + 2it^3 - t^4 = t^2 - t^4 + 2it^3$

$= \int_0^1 (t^2 - t^4 + 2it^3)(1 + 2it) \, dt$

$= \int_0^1 [(t^2 - t^4) + 2it^3 + 2it(t^2 - t^4) - 4t^4] \, dt$

$= \int_0^1 [t^2 - 5t^4 + 2it^3 + 2it^3 - 2it^5] \, dt$

$= \int_0^1 (t^2 - 5t^4) \, dt + i\int_0^1 (4t^3 - 2t^5) \, dt$

$= \left[\frac{t^3}{3} - t^5\right]_0^1 + i\left[t^4 - \frac{t^6}{3}\right]_0^1 = -\frac{2}{3} + i\cdot\frac{2}{3} = -\frac{2}{3}(1 - i)$