Questions tagged [complex-analysis]

For questions mainly about theory of complex analytic/holomorphic functions of one complex variable. Use [tag:complex-numbers] instead for questions about complex numbers. Use [tag:several-complex-variables] instead for questions about holomorphic functions of more than one complex variables.

Complex analysis is a branch of mathematical analysis that investigates functions of one or several complex variables. Typical topics include Cauchy's integral formula, singularities, poles, holomorphic and meromorphic functions, Laurent and Taylor series, maximum modulus principle, isolated zeros principle, etc.

Independent variable and dependent variable of complex functions are both complex numbers, and may be separated into real (denoted by $\Re$) and imaginary parts (denoted by $\Im$) : $z = x +iy$ and $w = f(z) =u(x,y)+iv(x,y)$ where $x, y \in \mathbb{R}$, $u(x,y)$ and $v(x,y)$ being real-valued functions.

A fundamental result of complex analysis is the Cauchy-Riemann equations, which give the conditions for a function to have a complex derivative, which is a complex generalization of the (real) derivative. If the complex derivative is defined at each point in an open connected subset of $\mathbb C$, the function is called analytic, or holomorphic. If the complex derivative is defined at each point in $\mathbb C$, the function is called entire.

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Finding Residue

I am having difficulty with with calculating the residue for $$\text{res}[\frac{\exp(\frac{1}{z})}{z^{2}-16,},z=0]$$ I was able to calculate the residues when $z=4$ and $z=-4$. However Im not sure how to approach this part of the question.
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Exam problem on poles and singularity of a complex function

I am solving the following Exam question: Question Choose the correct options from the following Consider the function $\displaystyle{% \,\mathrm{F}\left(\, z\,\right) = \int_{1}^{2}{\mathrm{d}x \over \left(x - z\right)^{2}} \,,\quad \Im\left(\,…
Srijan
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Show that for any polynomial $p(z)$ there is a $z$ with $|z|=1$ such that $|p(z)-1/z|\geq 1$.

I'm having a bit of trouble on another problem, and I'm not sure where to start: Show that for any polynomial $p(z)$ there is a $z$ with $|z|=1$ such that $|p(z)-1/z|\geq 1$. Could anybody get me started with a tip or two? Thanks in advance.
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Find the region where an integral is defined

How to determine where $$f(z)=\int_0^\infty \frac{e^{tz}}{1+t^2} \, dt$$ is defined and holomorphic using Morera's and Fubini's theorem?
abby
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Number of roots of $z^6-5z^2+8z+2$ in closed unit disk

This problem, of course, screams Rouche's Theorem. Unfortunately, the coefficients were chosen in such a way that getting the strict inequality necessary for application of the theorem is not easy. Since we have a disk of radius less than or equal…
Curious
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Suppose that $f$ is an analytic function from the Riemann sphere to the Riemann sphere, must f be a rational function?

Then Riemann sphere is defined by charts $(\mathbb C,Id_{\mathbb C})$ and $(\mathbb C-\{0\}\cup\{\infty\},\phi)$, $\phi(z) = \frac{1}{z},$ if $z \neq 0$, $\phi(z) = 0$ if $z = \infty$. I was told that if $f$ is an analytic function from the Riemann…
z.z
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A Riemann surface for the function $f(z) = z^{1/p} + z^{1/q}$

When considering the Riemann surface for the function $f(z) = z^{1/p} + z^{1/q}$, where $p,q$ are positive integers, I always thought that it consists of $pq$ branches. However, then I analyzed the function $$f(z) = z^{1/2} + z^{1/4}.$$ It has a…
user54031
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A specific consequence of Cauchy's integral formula

If f is holomorphic in an open subset $G \subset \mathbb{C}$, and if $f'(a)\neq0$ for some $a \in G$, then there exists $r>0$ such that \begin{eqnarray}|f'(z)-f'(a)|<|f'(a)|,\end{eqnarray} for $z \in D(a,r)$ ($D$ for 'disk' with centre $a$, radius…
Fuente
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Showing an analytic function on the unit disk is nonzero in a certain neighborhood

Suppose $f(z)$ is analytic for $|z|\le 1$ and $f(0) = a_0 \ne 0$. If $M = \max_{|z|=1} |f(z)|$, then show $f(z)\ne 0$ for all $z$ with $|z| < \frac{|a_0|}{|a_0|+M} =:r$. I know we can write $f(z) = a_0 + z^kg(z)$, some $k\ge 1$ and $g$ analytic and…
Curious
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Why does my textbook not define complex square and square root functions at the origin?

For example, the square function is defined as $w=f(z)=z^2=r^2e^{i2\theta}$, where r>0 and $-\pi<\theta\leq \pi$. Why is the origin always excluded? Also, why do we use the branch cut on the negative real axis for the square root function when $f$…
SihOASHoihd
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Prove that if |f(z)| is constant then f(z) is also constant (f holomorphic)

I tried proving that, with the given assumptions, $f'(0)$ is $0$, that way I could say that $f(z)$ is constant, since I have already proved that if the derivative is $0$ then the function is constant. I wrote the following $$ \left|\frac{f(z_1) -…
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What was used in this proof that there can not be a holomorphic square root defined on an annulus?

Given an annulus around $0$ with the radii $a$ and $b$, can there be a holomorphic square root of $z$? The proof given is : No, because assuming it existed $$ g(z)^2 = z $$ it would follow that : $$ 2g(z)g'(z)=1$$ and then $$\int _ \gamma…
bakabakabaka
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$f(z)$ has a zero of order $k$ iff $1/f(z)$ has a pole of order $k$

If $f(z)$ is analytic at $z_0$, show that $f(z)$ has a zero of order $k$ at $z_0$ if and only if $\dfrac 1 {f(z)}$ has a pole of order $k$ at $z_0$. I solved it but I'm not sure about my solution. ($\Rightarrow$) Since $f(z)$ is analytic at…
Gobi
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If $f$ and $g$ are branches of $z^a$ and $z^b$ respectively show that $fg$ is a branch of $z^{a+b}$

Suppose $f: G\rightarrow \mathbb{C}$ is a branch of $z^a$. Then $f(z) = e^{ah_1(z)}$ for all $z\in G$. Likewise if $g:G\rightarrow \mathbb{C}$ is a branch of $z^b$ then $g(z) = e^{bh_2(z)}$ for all $z\in G$. Here $h_1$ and $h_2$ are branches of the…
fosho
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Evaluating this complicated integral using complex analysis

I am trying to evaluate this integral: $$\boxed{\int_{-\pi}^\pi\sin(2\cos\theta)\cos((2m+1)\theta)\,d\theta}$$ where $m\in\mathbb{N}$, using complex analysis methods. What I have done is to find the residue of $\sin(z+\frac 1z)$: Since…
yoyostein
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