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### 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 $\gamma(t)=t$....

### 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)\...
• 15.8k
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.
• 261k
1 vote

• 2,311
1 vote

Yes the limit exists and it is 0, we can even set aside the branch cut ambiguity since this one doesnâ€™t depend on that. If you write $z = re^{i\theta}$ and we say $f(z)$ is some branch cut of $\sqrt{... • 2,009 1 vote ### Prove that$\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\...
• 2,311

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