3 votes
Accepted

Show there is some $c>0$ satisfying $\frac{1}{|\cosh(\pi z)|}<\frac{c}{1+(\Re(z))^2}$

For $z=x + i y$ is $$ \cosh(\pi z) = \cosh(\pi x) \cos(\pi y) + i \sinh(\pi x) \sin(\pi y) $$ and therefore $$ |\cosh(\pi z)| \ge |\operatorname{Re}(\cosh(\pi z)| = \cosh(\pi x)\cdot| \cos(\pi y)| \, ...
Martin R's user avatar
  • 114k
2 votes

Is $\sqrt z$ continuous and differentiable at $z=0$?

Assuming we have selected a branch of $\sqrt{z}$, we have $$(\sqrt{z})^2=z\implies |\sqrt{z}|^2=|z|\implies |\sqrt{z}|=\sqrt{|z|}$$ Therefore, $$\lim_{z\to 0} \left|\sqrt{z}\right|=\lim_{z\to 0} \sqrt{...
Julio Puerta's user avatar
  • 4,931
2 votes

Inequality of Entire functions

You have$$\cos\left(\frac\pi2\right)=1\quad\text{and}\quad\int_\gamma\cos(z)\,\mathrm dz=0$$if $\gamma\colon\left[-\frac\pi2,\frac\pi2\right]\longrightarrow\Bbb C$ is the path defined by $\gamma(t)=t$....
José Carlos Santos's user avatar
2 votes

Derivative of the complex power function (by definition)

If you are allowed to use the chain rule, one gets (at least in the real case): \begin{align*} f(x) := x^{\alpha} = \exp\left(\alpha\ln(x)\right) \Rightarrow f'(x) = \frac{\alpha\exp\left(\alpha\ln(x)\...
Átila Correia's user avatar
1 vote

Evaluate $\int_0^{2 \pi} e^{\sin(e^{i \theta})} \hspace{0.1cm} d \theta$

There is a symmetry around $\theta=\pi$. The real part is even and the imaginary part is odd. So, just remain the integration of the real part tp get $2\pi$ as @Bertrand87 showed in his good answer.
Claude Leibovici's user avatar
1 vote

Evaluate $\int_0^{2 \pi} e^{\sin(e^{i \theta})} \hspace{0.1cm} d \theta$

Let $z=e^{i\theta}$,then $d\theta = \frac{dz}{iz}$, hence the integral is just $$\frac{1}{i}\oint_{|z|=1} \frac{e^{\sin(z)}}{z}dz=(2\pi)\frac{1}{2\pi i}\oint_{|z|=1} \frac{e^{\sin(z)}}{z}dz=(2\pi)(e^{\...
Just a user's user avatar
  • 15.4k
1 vote
Accepted

Evaluate $\int_0^{2 \pi} e^{\sin(e^{i \theta})} \hspace{0.1cm} d \theta$

The correct substitution is: $$ z= e^{i\theta }$$ $$ dz =e^{i\theta}id\theta = zi d\theta \implies d\theta = \frac{dz}{zi} $$ along with the contour $|z|=1$ $$ \sum_{n=0}^\infty \frac{1}{n!} \int_0^{2 ...
Bertrand87's user avatar
  • 2,311
1 vote

Is $\sqrt z$ continuous and differentiable at $z=0$?

Yes the limit exists and it is 0, we can even set aside the branch cut ambiguity since this one doesn’t depend on that. If you write $z = re^{i\theta}$ and we say $f(z)$ is some branch cut of $\sqrt{...
Lee Fisher's user avatar
  • 2,009
1 vote

Prove that $\int_0^{\pi/2} \cos^{p+q-2}(\theta) \cos((p-q)\theta)d\theta = \frac{\pi}{(p+q-1)2^{p+q-1}B(p,q)}$

One solution using mostly real methods: \begin{eqnarray*} \int_0^{\pi/2} \cos^{p+q-2}(\theta) \cos((p-q)\theta)d\theta &=& \frac{1}{2^{p+q-1}}\Re \int_0^{\pi/2} \left(e^{i\theta} + e^{-i\...
Bertrand87's user avatar
  • 2,311

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