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7

$e^{4z}+e^{3z}+e^{2z}+e^z+1=0$ is too hard to solve directly. Instead, solve \begin{equation} e^{5z}-1=(e^z-1)(e^{4z}+e^{3z}+e^{2z}+e^z+1)=0 \end{equation} and remove roots that are not roots of original equation.


5

You can use the lovely trigonometrical identity $$ \lvert \sin{(x+iy)} \rvert^2 = \sin^2{x}+\sinh^2{y} $$ (use the angle-addition formulae, the relationships between the trigonometric and hyperbolic functions, and the Pythagorean identity): since $\sinh^2{y}>0$ unless $y=0$, the only zeros occur when $y=0$ and $x$ is a root of $\sin$, i.e. the usual real ...


4

Let $z = \mathrm{e}^{\mathrm{i} x}$. Then $$ \cos(x) = \dfrac{z + \frac{1}{z}}{2}. $$


4

Observe that the second summand is conjugate of the first: $$ \frac1{\alpha-ix}=\overline{\frac1{\alpha+ix}} $$ thus their sum will be real. In particular $$ \frac1{\alpha+ix}+\frac1{\alpha-ix}=\frac{2\alpha}{\alpha^2+x^2}{} $$ thus the integrand will be $$ \frac{4\alpha^2}{(\alpha^2+x^2)^2}=\frac{4}{\alpha^2}\frac{1}{(1+(x/\alpha)^2)^2} $$ Can you integrate ...


3

The integral is even in $\alpha$, so assume that $\alpha\gt0$. Real Analysis Approach Substitute $x\mapsto\alpha x$, $$ \begin{align} \int_{-\infty}^\infty\left(\frac1{\alpha+ix}+\frac1{\alpha-ix}\right)^2\mathrm{d}x &=\int_{-\infty}^\infty\left(\frac{2\alpha}{\alpha^2+x^2}\right)^2\mathrm{d}x\\ ...


3

$$\lim_{z\to i } \frac{z-1}{z^2 + 1} = \lim_{z\to i } \left|\frac{z-1}{(z-i)(z+i)}\right| \\ =\lim_{z\to i } \frac{|z-1|}{|z-i||z+i|}$$ $$\rm Now,\ as\ z\to i, \ |z-i|\to 0$$ $$\implies f(z) \to \infty$$


2

Hint:$$f(z)=\frac{z^2+2iz-1}{2z^2+iz+1}=\frac{(z+i)^2}{(z+i)(2z-i)}=\frac{z+i}{2z-i},\quad\forall z\ne -i$$


2

Hint: Say $\gamma$ is the unit circle. It would follow that $$\left|\frac1{2\pi i}\int_\gamma(f(z)-\overline z)\,dz\right|\le\frac9{10}.$$But you can easily figure out exactly what that integral is...


2

For any $\epsilon \in (0,\frac{1}{10})$, we have $$\begin{align} \left|\frac{1}{2\pi i}\int_{|z|=1-\epsilon} f(z) dz\right| \ge & \left|\frac{1}{2\pi i}\int_{|z|=1-\epsilon} \bar{z} dz \right| - \left|\frac{1}{2\pi i}\int_{|z|=1-\epsilon} ( f(z) - \bar{z} ) dz \right|\\ \ge & (1-\epsilon)^2 - \frac{1}{2\pi}\int_{|z|=1-\epsilon}|f(z) - \bar{z}| ...


2

If the open subset $G$ were an annulus, then $\mathbb{C}\setminus G$ has two connected components. Suppose $\gamma$ winds around the hole in the annulus. In the unbounded component of $\mathbb{C}\setminus G$ for sure $n(\gamma;w)=0$ since it's part of the unbounded component of $\mathbb{C}\setminus \gamma$, but that's not necessarily true in the bounded ...


2

Let $f(n) = 3n^3+n^2+6^{\log_2 n} = 3n^3+n^2+n^{\log_2 6}$ and $g(n)=n^3$. Show both $f(n) = O(g(n))$ and $g(n) = O(f(n))$. This is exactly showing $f(n) = \Theta(g(n))$.


1

Let $\Sigma$ be a connected compact Riemann surface, and let $f : \Sigma \to \mathbb{C}$ be a holomorphic function. As $|f| : \Sigma \to \mathbb{R}$ is continuous and $\Sigma$ is compact, it attains a maximum value $M$. By continuity, $S := \{p \in \Sigma \mid |f(p)| = M\}$ is closed. Suppose $q \in \Sigma$ is such that $|f(q)| = M$. Let $U$ be an open ...


1

All such functions can be written in the form $$f(z) = \exp(g(z))$$ where $g$ is holomorphic. Such functions certainly satisfy your condition, and the fact that all such functions can be written in this form follows from the existence of the logarithm of non-zero functions.


1

Actually, you are almost done. Consider this: $$\frac{|z-1|}{|z-i||z+i|} \overset{(\text{triangle eq.})}{\geq} \frac{|z-1|}{|z-i|\cdot(|z|+|i|)} = \frac{|z-1|}{|z|+1} \cdot \frac{1}{|z-i|}$$ Let's say you have $K \in \mathbb{R}$. Now you need to choose an appropriate $\delta$ so that $|f(z)| \gt K$ $\forall z \in B_\delta(i)$. Suppose you have a $z \in ...


1

If we know that $\zeta$ is uniformly convergent then $$\lim_s \zeta (s)=\lim_{s}\sum_{n=1}^\infty \frac1{n^s}=\sum_{n=1}^\infty \lim_s \frac1{n^s}=1+0+0+\ldots =1.$$


1

Despite your aversion(?) to integrals, I think they give the quickest and easiest argument. The function $n \mapsto n^{-s}$ is decreasing in $n$ (for $s > 1$), so $$ 1 \le \zeta(s) = \sum_{n=1}^\infty \frac1{n^s} = 1 + \sum_{n=2}^\infty \frac1{n^s} \le 1 + \int_1^\infty \frac{1}{x^s}\,ds = 1 + \frac{1}{s-1}. $$ To see why the inequality between the series ...


1

If you wanted to use CR: (This answer only works for $\phi$ real, but if $\phi=\phi_1+i\phi_2$ with $\phi_1,\phi_2$ real, you can see that it follows from the real case.) Write $z=a+bi$ then $$\frac{\phi(t)}{(t-a)-bi} = \frac{\phi(t)(t-a +bi)}{(t-a)^2+b^2}$$ So $$\int_{-1}^{1} \frac{\phi(t)}{t-z}\,dt = \int_{-1}^1 \frac{\phi(t)(t-a)}{(t-a)^2+b^2}\,dt + ...


1

Let $\gamma$ be a closed curve in $\mathbb{C} \setminus [-1,1]$, not enclosing $[-1,1]$. Then \begin{align} \int_\gamma f(z)\,dz &= \int_\gamma \left( \int_{-1}^1 \frac{\phi(t)}{z-t}\,dt \right) \, dz \\ &= \int_{-1}^1 \left( \int_{\gamma} \frac{\phi(t)}{z-t}\,dz \right) \, dt \\ &= \int_{-1}^1 \phi(t) \left( \int_{\gamma} \frac{1}{z-t}\,dz ...


1

The functions $\eta(z,t) = {\phi(t) \over t-z}$ and ${\partial \eta \over \partial z}$are continuous on $[-1,1]^c \times [-1,1]$, hence by the Leibnitz rule we see that $f$ is differentiable and $f'(z) = \int_{-1,1]} {\partial \eta(z,t) \over \partial z} dt = \int_{-1,1]} {\phi(t) \over (t-z)^2} dt$.


1

For a function $f:\>{\mathbb C}\to{\mathbb C}$ to be entire it should at least be analytic in a full neighborhood of $z=0$. But this is not the case for the function "defined" by the typographical painting $f(z):=z^{1/3}$. Such a function would have to satisfy $\bigl(f(z)\bigr)^3=z$, hence $f(0)=0$, and $|f(z)|=|z|^{1/3}$. This then would imply ...


1

Comparing with $\sum z^n$, you see that the series is absolutely convergent for $|z|<1$. On the other hand, the series fails to converge for $z=1$.



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