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Complex Analysis : Complex Analysis

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Example Questions

Example Question #11 : Elementary Functions

$|e^{Z^{2}}|\leq\ e^{|Z|^{2}}$ $\displaystyle |e^{Z^{2}}|\leq\ e^{|Z|^{2}}$

Is the above inequality true?

Possible Answers:

The truth of the statement depends on the value of $\displaystyle Z$. In other words it is true for some restricted domain of the complex plane.

It is true when:

$\displaystyle Z=x+yi=0$

The truth of the statement depends on the value of $\displaystyle Z$. In other words it is true for some restricted domain of the complex plane.

It is true only for the real line.

y=0 $\displaystyle y=0$

No it is not true. The statement is false for the entire complex plane.

The truth of the statement depends on the value of $\displaystyle Z$. In other words it is true for some restricted domain of the complex plane.

It is true only for the pure imaginary line.

x=0 $\displaystyle x=0$

Yes it is true for the whole complex plane.

Correct answer:

Yes it is true for the whole complex plane.

Explanation:

$|e^{Z^{2}}|=|e^{(x+yi)^{2}}|=|e^{(x^{2}-y^{2})+(2xy)i}|$ $\displaystyle |e^{Z^{2}}|=|e^{(x+yi)^{2}}|=|e^{(x^{2}-y^{2})+(2xy)i}|$

$|e^{(x^{2}-y^{2})+(2xy)i}|=|e^{x^2-y^2}|*|e^{2xyi}|=e^{x^2-y^2}*1=e^{x^2-y^2}$ $\displaystyle |e^{(x^{2}-y^{2})+(2xy)i}|=|e^{x^2-y^2}|*|e^{2xyi}|=e^{x^2-y^2}*1=e^{x^2-y^2}$

above are the steps to get the magnitude of the left side of the inequality.

$e^{|Z|^{2}}=e^{|x+yi|^{2}}=e^{(\sqrt{x^2+y^2})^{2}}=e^{x^2+y^2}$ $\displaystyle e^{|Z|^{2}}=e^{|x+yi|^{2}}=e^{(\sqrt{x^2+y^2})^{2}}=e^{x^2+y^2}$

above are the steps to get the magnitude of the right side of the inequality.

Thus $|e^{Z^{2}}|\leq\ e^{|Z|^{2}}$ $\displaystyle |e^{Z^{2}}|\leq\ e^{|Z|^{2}}$ becomes...

$e^{x^2-y^2}\leq e^{x^2+y^2}$ $\displaystyle e^{x^2-y^2}\leq e^{x^2+y^2}$

now we do algebra...

$e^{x^{2}}/e^{y^{2}}\leq e^{x^{2}}*e^{y^{2}}$ $\displaystyle e^{x^{2}}/e^{y^{2}}\leq e^{x^{2}}*e^{y^{2}}$

$e^{x^{2}}\leq e^{x^{2}}*e^{y^{2}}*e^{y^{2}}$ $\displaystyle e^{x^{2}}\leq e^{x^{2}}*e^{y^{2}}*e^{y^{2}}$

$1\leq e^{y^{2}}*e^{y^{2}}$ $\displaystyle 1\leq e^{y^{2}}*e^{y^{2}}$

$1\leq e^{2y^{2}}$ $\displaystyle 1\leq e^{2y^{2}}$

with this last inequality you can graph the right hand side and see that it is

always $\displaystyle 1$ or greater, or you can reason it this way...

$y^{2}\geqslant0$ $\displaystyle y^{2}\geqslant0$

$2y^{2}\geqslant0$ $\displaystyle 2y^{2}\geqslant0$

$e^{2y^{2}}\geq e^{0}$ $\displaystyle e^{2y^{2}}\geq e^{0}$

$e^{2y^{2}}\geq 1$ $\displaystyle e^{2y^{2}}\geq 1$

Thus the inequality is true for all complex numbers.

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Example Question #41 : Complex Analysis

Compute $e^{\left(\frac{2+\pi i}{4}\right)}$ $\displaystyle e^{\left(\frac{2+\pi i}{4}\right)}$

Possible Answers:

None of the choices

$\frac{\sqrt{2}}{2}(1+i)$ $\displaystyle \frac{\sqrt{2}}{2}(1+i)$

$\sqrt{e}(1+i)$ $\displaystyle \sqrt{e}(1+i)$

$\sqrt{2e}(1+i)$ $\displaystyle \sqrt{2e}(1+i)$

$\sqrt{\frac{e}{2}} (1+i)$ $\displaystyle \sqrt{\frac{e}{2}} (1+i)$

Correct answer:

$\sqrt{\frac{e}{2}} (1+i)$ $\displaystyle \sqrt{\frac{e}{2}} (1+i)$

Explanation:

$e^{\frac{2+\pi i}{4}} = e^{\frac{2}{4}}*e^{\frac{\pi i}{4}} = \sqrt{e} * \frac{\sqrt{2}}{2}(1+i) = \sqrt{\frac{e}{2}}(1+i)$ $\displaystyle e^{\frac{2+\pi i}{4}} = e^{\frac{2}{4}}*e^{\frac{\pi i}{4}} = \sqrt{e} * \frac{\sqrt{2}}{2}(1+i) = \sqrt{\frac{e}{2}}(1+i)$

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Example Question #13 : Elementary Functions

Compute $\displaystyle Log(1-i)$

Possible Answers:

$\frac{1}{2}\ln2 + \frac{\pi}{4}i$ $\displaystyle \frac{1}{2}\ln2 + \frac{\pi}{4}i$

$\frac{1}{2}\ln2 - \frac{\pi}{4}i$ $\displaystyle \frac{1}{2}\ln2 - \frac{\pi}{4}i$

$-\frac{i}{2}\ln2 + \frac{\pi}{4}$ $\displaystyle -\frac{i}{2}\ln2 + \frac{\pi}{4}$

$-\frac{i}{2}\ln2 - \frac{\pi}{4}$ $\displaystyle -\frac{i}{2}\ln2 - \frac{\pi}{4}$

$\sqrt{2} - \frac{\pi}{4}i$ $\displaystyle \sqrt{2} - \frac{\pi}{4}i$

Correct answer:

$\frac{1}{2}\ln2 - \frac{\pi}{4}i$ $\displaystyle \frac{1}{2}\ln2 - \frac{\pi}{4}i$

Explanation:

$w = \Log(1-i)$ $\displaystyle w = \Log(1-i)$ where $\displaystyle w$ is the complex number such that $\displaystyle e^w = 1-i$

Converting into polar coordinates

$e^w = 1-i \implies e^xe^{iy} = \sqrt{2}e^{-\frac{\pi}{4}i} \implies x = \frac{1}{2}\ln 2 \text{ and } y = -\frac{\pi}{4}$ $\displaystyle e^w = 1-i \implies e^xe^{iy} = \sqrt{2}e^{-\frac{\pi}{4}i} \implies x = \frac{1}{2}\ln 2 \text{ and } y = -\frac{\pi}{4}$

This gives us

$w = x+yi = \frac{1}{2}\ln2 - \frac{\pi}{4}i$ $\displaystyle w = x+yi = \frac{1}{2}\ln2 - \frac{\pi}{4}i$

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Example Question #1 : Complex Integration

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

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

Possible Answers:

$\frac{\pi i }{2}$ $\displaystyle \frac{\pi i }{2}$

$\frac{\pi i }{4}$ $\displaystyle \frac{\pi i }{4}$

$2 \pi i$ $\displaystyle 2 \pi i$

$\pi i$ $\displaystyle \pi i$

$\frac{\pi i}{8}$ $\displaystyle \frac{\pi i}{8}$

Correct answer:

$\frac{\pi i }{4}$ $\displaystyle \frac{\pi i }{4}$

Explanation:

Recall Cauchy's integral formula, which states

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

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

Plugging in gives us

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

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Example Question #2 : Complex Integration

Let $\displaystyle C$be the unit circle.

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

Possible Answers:

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

$-\frac{\pi i}{6}$ $\displaystyle -\frac{\pi i}{6}$

$-\frac{\pi i}{2}$ $\displaystyle -\frac{\pi i}{2}$

$-\pi i$ $\displaystyle -\pi i$

$\displaystyle 0$

Correct answer:

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

Explanation:

Recall the Taylor expansion of

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

Use this to write

$z^2\sin\left(\frac{1}{z}\right) = z - \frac{1}{z3!} + \frac{1}{z^35!} - \frac{1}{z^57!} + ...$ $\displaystyle z^2\sin\left(\frac{1}{z}\right) = z - \frac{1}{z3!} + \frac{1}{z^35!} - \frac{1}{z^57!} + ...$

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

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

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Example Question #1 : Taylor And Laurent Series

Find the Taylor Series expansion of $f(z) = z^2e^{3z}$ $\displaystyle f(z) = z^2e^{3z}$

Possible Answers:

$z^2e^{3z} = \sum_{n=0}^\infty \frac{3^{n}z^{n}}{n!}$ $\displaystyle z^2e^{3z} = \sum_{n=0}^\infty \frac{3^{n}z^{n}}{n!}$

$z^2e^{3z} = \sum_{n=0}^\infty \frac{3^{n-2}z^{n}}{(n-2)!}$ $\displaystyle z^2e^{3z} = \sum_{n=0}^\infty \frac{3^{n-2}z^{n}}{(n-2)!}$

$z^2e^{3z} = \sum_{n=2}^\infty \frac{3^{n-2}z^{n}}{(n-2)!}$ $\displaystyle z^2e^{3z} = \sum_{n=2}^\infty \frac{3^{n-2}z^{n}}{(n-2)!}$

$z^2e^{3z} = \sum_{n=2}^\infty \frac{3^{n}z^{n}}{(n-2)!}$ $\displaystyle z^2e^{3z} = \sum_{n=2}^\infty \frac{3^{n}z^{n}}{(n-2)!}$

$z^2e^{3z} = \sum_{n=2}^\infty \frac{3^{n-2}z^{n}}{n!}$ $\displaystyle z^2e^{3z} = \sum_{n=2}^\infty \frac{3^{n-2}z^{n}}{n!}$

Correct answer:

$z^2e^{3z} = \sum_{n=2}^\infty \frac{3^{n-2}z^{n}}{(n-2)!}$ $\displaystyle z^2e^{3z} = \sum_{n=2}^\infty \frac{3^{n-2}z^{n}}{(n-2)!}$

Explanation:

It is well known (can be shown by the definition) that that

$e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$ $\displaystyle e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$

Making the appropriate substitutions

$e^{3z} = \sum_{n=0}^\infty \frac{3^nz^n}{n!}$ $\displaystyle e^{3z} = \sum_{n=0}^\infty \frac{3^nz^n}{n!}$

$z^2e^{3z} = \sum_{n=0}^\infty \frac{3^nz^{n+2}}{n!}$ $\displaystyle z^2e^{3z} = \sum_{n=0}^\infty \frac{3^nz^{n+2}}{n!}$

Shifting the index of summation gives us

$z^2e^{3z} = \sum_{n=2}^\infty \frac{3^{n-2}z^{n}}{(n-2)!}$ $\displaystyle z^2e^{3z} = \sum_{n=2}^\infty \frac{3^{n-2}z^{n}}{(n-2)!}$

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Example Question #2 : Taylor And Laurent Series

Find the first three terms of the Taylor Series of $\frac{e^z}{1+z} \text{ where }|z|<1$ $\displaystyle \frac{e^z}{1+z} \text{ where }|z|< 1$

Possible Answers:

$\frac{e^z}{1+z} = 1 + z^2 - z^3 + ....$ $\displaystyle \frac{e^z}{1+z} = 1 + z^2 - z^3 + ....$

$\frac{e^z}{1+z} = 1 + \frac{1}{2}z - \frac{1}{3}z^2 + ....$ $\displaystyle \frac{e^z}{1+z} = 1 + \frac{1}{2}z - \frac{1}{3}z^2 + ....$

$\frac{e^z}{1+z} = 1 + z - z^2 + ....$ $\displaystyle \frac{e^z}{1+z} = 1 + z - z^2 + ....$

$\frac{e^z}{1+z} = 1 + \frac{1}{2}z^2 - \frac{1}{3}z^3 + ....$ $\displaystyle \frac{e^z}{1+z} = 1 + \frac{1}{2}z^2 - \frac{1}{3}z^3 + ....$

$\frac{e^z}{1+z} = 1 + \frac{1}{3}z^2 - \frac{1}{5}z^3 + ....$ $\displaystyle \frac{e^z}{1+z} = 1 + \frac{1}{3}z^2 - \frac{1}{5}z^3 + ....$

Correct answer:

$\frac{e^z}{1+z} = 1 + \frac{1}{2}z^2 - \frac{1}{3}z^3 + ....$ $\displaystyle \frac{e^z}{1+z} = 1 + \frac{1}{2}z^2 - \frac{1}{3}z^3 + ....$

Explanation:

Taylor Series expansion of the numerator and the denominator seperately gives us

$e^z = 1+z+\frac{z^2}{2} + \frac{z^3}{6} + .....$ $\displaystyle e^z = 1+z+\frac{z^2}{2} + \frac{z^3}{6} + .....$ $\frac{1}{1+z} = \frac{1}{1-(-z)} = 1 - z + z^2 - z^3 + ....$ $\displaystyle \frac{1}{1+z} = \frac{1}{1-(-z)} = 1 - z + z^2 - z^3 + ....$

A term by term multiplication gives us

$(1+z+\frac{1}{2}z^2+\frac{1}{6}z^3+...) + (-z-z^2-\frac{1}{2}z^3-\frac{1}{6}z^4 + ...) + (z^2+z^3+\frac{1}{2}z^4+\frac{1}{6}z^5 + ...) + (-z^3-z^4 -\frac{1}{2}z^5-\frac{1}{6}z^6 + ...)$ $\displaystyle (1+z+\frac{1}{2}z^2+\frac{1}{6}z^3+...) + (-z-z^2-\frac{1}{2}z^3-\frac{1}{6}z^4 + ...) + (z^2+z^3+\frac{1}{2}z^4+\frac{1}{6}z^5 + ...) + (-z^3-z^4 -\frac{1}{2}z^5-\frac{1}{6}z^6 + ...)$

Combining like terms

$\frac{e^z}{1+z} = 1 + \frac{1}{2}z^2 - \frac{1}{3}z^3 + ....$ $\displaystyle \frac{e^z}{1+z} = 1 + \frac{1}{2}z^2 - \frac{1}{3}z^3 + ....$

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Example Question #1 : Residue Theory

Cauchy's Residue Theorem is as follows:

Let $\displaystyle C$ be a simple closed contour, described positively. If a function $\displaystyle f$ is analytic inside $\displaystyle C$ except for a finite number of singular points $z_{k}$ $\displaystyle z_{k}$ inside $\displaystyle C$, then

$\displaystyle\int_{C} f(z) dz = 2 \pi i \displaystyle\sum_{k=1}^{n} \text{Res}_{z=z_{k}} f(z)$ $\displaystyle \displaystyle\int_{C} f(z) dz = 2 \pi i \displaystyle\sum_{k=1}^{n} \text{Res}_{z=z_{k}} f(z)$

Brown, J. W., & Churchill, R. V. (2009). Complex variables and applications. Boston, MA: McGraw-Hill Higher Education.

Use Cauchy's Residue Theorem to evaluate the integral of

$f(z) = \frac{e^{-z}}{z^3}$ $\displaystyle f(z) = \frac{e^{-z}}{z^3}$

in the region |z|=4 $\displaystyle |z|=4$.

Possible Answers:

$\pi i$ $\displaystyle \pi i$

$3 \pi i$ $\displaystyle 3 \pi i$

$- 2 \pi i$ $\displaystyle - 2 \pi i$

$- \frac{\pi i }{2}$ $\displaystyle - \frac{\pi i }{2}$

$- \frac{\pi i }{4}$ $\displaystyle - \frac{\pi i }{4}$

Correct answer:

$\pi i$ $\displaystyle \pi i$

Explanation:

Note, for

$f(z) = \frac{e^{-z}}{z^3}$ $\displaystyle f(z) = \frac{e^{-z}}{z^3}$

a singularity exists where z= 0 $\displaystyle z= 0$. Thus, since where z=0 $\displaystyle z=0$ is the only singularity for f(z) $\displaystyle f(z)$ inside |z|=3 $\displaystyle |z|=3$, we seek to evaluate the residue for z=0 $\displaystyle z=0$.

Observe,

$f(z) \newline = \frac{e^{-z}}{z^3} \newline = \frac{1}{z^3} \displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k z^k}{k!} \newline = \displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k z^{k-3}}{k!}$ $\displaystyle f(z) \newline = \frac{e^{-z}}{z^3} \newline = \frac{1}{z^3} \displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k z^k}{k!} \newline = \displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k z^{k-3}}{k!}$

The coefficient of $\frac{1}{z}$ $\displaystyle \frac{1}{z}$ is $\frac{(-1)^{2}}{2!} = \frac{1}{2}$ $\displaystyle \frac{(-1)^{2}}{2!} = \frac{1}{2}$.

Thus,

$\text{Res}_{z=0} [f(z)] = \text{Res}_{z=0}[\frac{e^{-z}}{z^3}] = \frac{1}{2}$ $\displaystyle \text{Res}_{z=0} [f(z)] = \text{Res}_{z=0}[\frac{e^{-z}}{z^3}] = \frac{1}{2}$.

Therefore, by Cauchy's Residue Theorem,

$\displaystyle \int_{|z|=3} f(z) dz = \int_{|z|=3} \frac{e^{-z}}{z^2}dz = 2 \pi i \text{Res}_{z=0}[\frac{e^{-z}}{z^2}] = 2 \pi i (\frac{1}{2}) = \pi i$ $\displaystyle \displaystyle \int_{|z|=3} f(z) dz = \int_{|z|=3} \frac{e^{-z}}{z^2}dz = 2 \pi i \text{Res}_{z=0}[\frac{e^{-z}}{z^2}] = 2 \pi i (\frac{1}{2}) = \pi i$

Hence,

$\int_{|z|=3} \frac{e^{-z}}{z^2} dz = \pi i$ $\displaystyle \int_{|z|=3} \frac{e^{-z}}{z^2} dz = \pi i$

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Example Question #2 : Residue Theory

Cauchy's Residue Theorem is as follows:

Brown, J. W., & Churchill, R. V. (2009). Complex variables and applications. Boston, MA: McGraw-Hill Higher Education.

Using Cauchy's Residue Theorem, evaluate the integral of

$f(z) = \frac{e^{-z}}{(z-1)^3}$ $\displaystyle f(z) = \frac{e^{-z}}{(z-1)^3}$

in the region |z|=4 $\displaystyle |z|=4$

Possible Answers:

$2 \pi i$ $\displaystyle 2 \pi i$

$\frac{ \pi i}{e}$ $\displaystyle \frac{ \pi i}{e}$

$- 2 \pi i(1 + e)$ $\displaystyle - 2 \pi i(1 + e)$

$- 2 \pi i$ $\displaystyle - 2 \pi i$

$-e^{2 \pi i}$ $\displaystyle -e^{2 \pi i}$

Correct answer:

$\frac{ \pi i}{e}$ $\displaystyle \frac{ \pi i}{e}$

Explanation:

Note, for

$f(z) = \frac{e^{-z}}{(z-1)^3}$ $\displaystyle f(z) = \frac{e^{-z}}{(z-1)^3}$

a singularity exists where z= 1 $\displaystyle z= 1$. Thus, since where z=1 $\displaystyle z=1$ is the only singularity for f(z) $\displaystyle f(z)$ inside |z|=4 $\displaystyle |z|=4$, we seek to evaluate the residue for z=1 $\displaystyle z=1$.

Observe,

$f(z) \newline = \frac{e^{-z}}{(z-1)^3} \newline = \frac{e^{-u-1}}{u^3} \newline = \frac{ 1}{e} * \frac{1}{u^3} \displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k u^k}{k!} \newline = \frac{1}{e}\displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k u^{k-3}}{k!} \newline = \frac{1}{e}\displaystyle \sum_{k=0}^{\infty} \frac{-1^{k}(z-1)^{k-3}}{k!}$ $\displaystyle f(z) \newline = \frac{e^{-z}}{(z-1)^3} \newline = \frac{e^{-u-1}}{u^3} \newline = \frac{ 1}{e} * \frac{1}{u^3} \displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k u^k}{k!} \newline = \frac{1}{e}\displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k u^{k-3}}{k!} \newline = \frac{1}{e}\displaystyle \sum_{k=0}^{\infty} \frac{-1^{k}(z-1)^{k-3}}{k!}$

The coefficient of $\frac{1}{z-1}$ $\displaystyle \frac{1}{z-1}$ is $\frac{1}{e} * \frac{(-1)^{2}}{2!} = \frac{1}{2e}$ $\displaystyle \frac{1}{e} * \frac{(-1)^{2}}{2!} = \frac{1}{2e}$.

Thus,

$\text{Res}_{z=0} [f(z)] = \text{Res}_{z=0}[\frac{e^{-z}}{(z-1)^2}] = \frac{1}{2e}$ $\displaystyle \text{Res}_{z=0} [f(z)] = \text{Res}_{z=0}[\frac{e^{-z}}{(z-1)^2}] = \frac{1}{2e}$.

Therefore, by Cauchy's Residue Theorem,

$\displaystyle \int_{|z|=3} f(z) dz = \int_{|z|=3} \frac{e^{-z}}{(z-1)^2}dz = 2 \pi i \text{Res}_{z=1}[\frac{e^{-z}}{(z-1)^2}] = 2 \pi i (\frac{1}{2e}) = \frac{ \pi i}{e}$ $\displaystyle \displaystyle \int_{|z|=3} f(z) dz = \int_{|z|=3} \frac{e^{-z}}{(z-1)^2}dz = 2 \pi i \text{Res}_{z=1}[\frac{e^{-z}}{(z-1)^2}] = 2 \pi i (\frac{1}{2e}) = \frac{ \pi i}{e}$

Hence,

$\int_{|z|=3} \frac{e^{-z}}{(z-1)^2} dz = \frac{ \pi i}{e}$ $\displaystyle \int_{|z|=3} \frac{e^{-z}}{(z-1)^2} dz = \frac{ \pi i}{e}$

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Example Question #3 : Residue Theory

Cauchy's Residue Theorem is as follows:

Brown, J. W., & Churchill, R. V. (2009). Complex variables and applications. Boston, MA: McGraw-Hill Higher Education.

Use Cauchy's Residue Theorem to evaluate the integral of

$f(z) = z^3 e^{\frac{1}{z}}$ $\displaystyle f(z) = z^3 e^{\frac{1}{z}}$

in the region |z|=4 $\displaystyle |z|=4$.

Possible Answers:

$2 \pi i$ $\displaystyle 2 \pi i$

$\frac{\pi i}{4}$ $\displaystyle \frac{\pi i}{4}$

$- \frac{\pi i }{6}$ $\displaystyle - \frac{\pi i }{6}$

$\frac{\pi i}{12}$ $\displaystyle \frac{\pi i}{12}$

$\frac{\pi i }{24}$ $\displaystyle \frac{\pi i }{24}$

Correct answer:

$\frac{\pi i}{12}$ $\displaystyle \frac{\pi i}{12}$

Explanation:

Note, there is one singularity for f(z) $\displaystyle f(z)$ where z = 0 $\displaystyle z = 0$.

Let

$u = \frac{1}{z}$ $\displaystyle u = \frac{1}{z}$

Then

$z = \frac{1}{u}$ $\displaystyle z = \frac{1}{u}$

$f(z) = z^3 e^{\frac{1}{z}} = \frac{1}{u^3} e^{u}$ $\displaystyle f(z) = z^3 e^{\frac{1}{z}} = \frac{1}{u^3} e^{u}$.

Therefore, there is one singularity for f(z) $\displaystyle f(z)$ where u = 0 $\displaystyle u = 0$. Hence, we seek to compute the residue for f(z) $\displaystyle f(z)$ where $\displaystyle u = 0 = z$

Observe,

$f(z) \newline = z^{3} e^{\frac{1}{z}} \newline = \frac{1}{u^3} e^u \newline = \frac{1}{u^3} \displaystyle \sum_{k=0}^{\infty} \frac{u^k}{k!} \newline = \displaystyle \sum_{k=0}^{\infty} \frac{u^{k-3}}{k!} \newline = \displaystyle \sum_{k=0}^{\infty} \frac{z^{3-k}}{k!}$ $\displaystyle f(z) \newline = z^{3} e^{\frac{1}{z}} \newline = \frac{1}{u^3} e^u \newline = \frac{1}{u^3} \displaystyle \sum_{k=0}^{\infty} \frac{u^k}{k!} \newline = \displaystyle \sum_{k=0}^{\infty} \frac{u^{k-3}}{k!} \newline = \displaystyle \sum_{k=0}^{\infty} \frac{z^{3-k}}{k!}$

So, when k = 4 $\displaystyle k = 4$, $z^{3-k} = z^{-1} = \frac{1}{z}$ $\displaystyle z^{3-k} = z^{-1} = \frac{1}{z}$.

Thus, the coefficient of $\frac{1}{z}$ $\displaystyle \frac{1}{z}$ is $\frac{1}{4!}$ $\displaystyle \frac{1}{4!}$.

Therefore,

$\text{Res}_{z=0} [f(z)] = \text{Res}_{z=0} [z^2 e^{\frac{1}{z}}] = \frac{1}{4!}$ $\displaystyle \text{Res}_{z=0} [f(z)] = \text{Res}_{z=0} [z^2 e^{\frac{1}{z}}] = \frac{1}{4!}$

Hence, by Cauchy's Residue Theorem,

$\displaystyle \int_{|z|=3} z^{2} e^{\frac{1}{z}} dz = 2 \pi i \text{Res}_{z=0}[z^2 e^{\frac{1}{z}}] = 2 \pi i * \frac{1}{4!} = \frac{\pi i}{12}$ $\displaystyle \displaystyle \int_{|z|=3} z^{2} e^{\frac{1}{z}} dz = 2 \pi i \text{Res}_{z=0}[z^2 e^{\frac{1}{z}}] = 2 \pi i * \frac{1}{4!} = \frac{\pi i}{12}$

Therefore,

$\displaystyle \int_{|z|=3} z^{2} e^{\frac{1}{z}} dz = \frac{\pi i}{12}$ $\displaystyle \displaystyle \int_{|z|=3} z^{2} e^{\frac{1}{z}} dz = \frac{\pi i}{12}$

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Complex Analysis : Complex Analysis

Study concepts, example questions & explanations for Complex Analysis

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Example Questions

Example Question #11 : Elementary Functions

Example Question #41 : Complex Analysis

Example Question #13 : Elementary Functions

Example Question #1 : Complex Integration

Example Question #2 : Complex Integration

Example Question #1 : Taylor And Laurent Series

Example Question #2 : Taylor And Laurent Series

Example Question #1 : Residue Theory

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Example Question #3 : Residue Theory

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