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Complex Analysis : Complex Analysis

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Example Questions

Example Question #41 : Complex Analysis

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

Is the above inequality true?

Possible Answers:

The truth of the statement depends on the value of . 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

Yes it is true for the whole complex plane.

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

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

It is true only for the real line.

y=0

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

It is true when:

Z=x+yi=0

Correct answer:

Yes it is true for the whole complex plane.

Explanation:

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

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

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

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

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

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

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

now we do algebra...

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

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

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

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

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

always or greater, or you can reason it this way...

$y^{2}\geqslant0$

$2y^{2}\geqslant0$

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

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

Thus the inequality is true for all complex numbers.

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Example Question #42 : Complex Analysis

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

Possible Answers:

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

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

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

None of the choices

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

Correct answer:

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

Explanation:

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

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Example Question #12 : Elementary Functions

Compute Log(1-i)

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

Converting into polar coordinates

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

This gives us

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

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Example Question #1 : Complex Integration

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

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

Possible Answers:

$\pi i$

$2 \pi i$

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

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

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

Correct answer:

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

Explanation:

Recall Cauchy's integral formula, which states

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

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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Example Question #2 : Complex Integration

Let be the unit circle.

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

Possible Answers:

$-\pi i$

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

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

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

Correct answer:

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

Explanation:

Recall the Taylor expansion of

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

Use this to write

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

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

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

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Example Question #1 : Taylor And Laurent Series

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

Making the appropriate substitutions

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

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

Shifting the index of summation gives us

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

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Example Question #1 : Taylor And Laurent Series

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

A term by term multiplication gives us

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

Combining like terms

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

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Example Question #1 : Residue Theory

Cauchy's Residue Theorem is as follows:

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

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

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

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

in the region |z|=4 .

Possible Answers:

$3 \pi i$

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

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

$\pi i$

$- 2 \pi i$

Correct answer:

$\pi i$

Explanation:

Note, for

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

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

Observe,

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

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

Thus,

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

Therefore, by Cauchy's Residue Theorem,

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

Hence,

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

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Example Question #1 : Residue Theory

Cauchy's Residue Theorem is as follows:

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

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

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

Using Cauchy's Residue Theorem, evaluate the integral of

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

in the region |z|=4

Possible Answers:

$- 2 \pi i$

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

$-e^{2 \pi i}$

$2 \pi i$

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

Correct answer:

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

Explanation:

Note, for

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

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

Observe,

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

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

Thus,

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

Therefore, by Cauchy's Residue Theorem,

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

Hence,

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

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Example Question #1 : Residue Theory

Cauchy's Residue Theorem is as follows:

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

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

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

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

in the region |z|=4 .

Possible Answers:

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

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

$2 \pi i$

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

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

Correct answer:

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

Explanation:

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

Let

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

Then

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

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

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

Observe,

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

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

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

Therefore,

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

Hence, by Cauchy's Residue Theorem,

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

Therefore,

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

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All Complex Analysis Resources

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Complex Analysis : Complex Analysis

Study concepts, example questions & explanations for Complex Analysis

All Complex Analysis Resources

Example Questions

Example Question #41 : Complex Analysis

Example Question #42 : Complex Analysis

Example Question #12 : Elementary Functions

Example Question #1 : Complex Integration

Example Question #2 : Complex Integration

Example Question #1 : Taylor And Laurent Series

Example Question #1 : Taylor And Laurent Series

Example Question #1 : Residue Theory

Example Question #1 : Residue Theory

Example Question #1 : Residue Theory

All Complex Analysis Resources

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