Hot answers tagged

5

If $f(z)$ and $\overline{f(z)}$ are analytic, then so is $h(z):=f(z)+\overline{f(z)}$, but the latter is real-valued (since $w+\overline{w}=2\textrm{Re}(w)$ ). Thus, the imaginary part of $h$ is zero, and this, by the Cauchy Riemann equations implies that the real part of $h$ is constant, so $h$ is a real constant (on each connected component of its ...


4

You can apply Estimation lemma. Since $$ \left|\int_{\Gamma_2} \frac{1}{1+z^2}dz\right| \le \frac{\pi R}{R^2 -1} $$ for large $R$, $$ \lim_{R\to\infty}\left|\int_{\Gamma_2} \frac{1}{1+z^2}dz\right|=0. $$ Then you can get what you want.


2

If $A = \int_a^b f(t)dt$, notice that, since $\theta$ is the principal argument of $A$, $$A = |A| e^{i \theta}$$ Which gives: $$|A| = Ae^{-i\theta} = \left(\int_a^b f(t)dt \right) e^{-i\theta} = \int_a^b e^{-i\theta} f(t)dt$$ Hence $\int_a^b e^{-i\theta} f(t) dt = |A| \in \Bbb R$


2

$\sin z=-1\implies\cos z=0\implies e^{iz}=-i=\cos\dfrac\pi2-i\sin\dfrac\pi2=e^{-i\pi/2}$ $$z=2n\pi-\dfrac\pi2$$


2

If you have some theorem, proposition, or lemma that the sum, difference, quotient, and product of analytic functions are analytic (except at points where the denominator is zero), then $\frac{z^2}{z-3}$ is analytic because the function $z\mapsto z$ is.


2

You've picked the wrong relation between the sine and the exponential. You want $\mathrm{e}^{\mathrm{i} a z} = \cos( a z) + \mathrm{i} \sin (a z )$. Do you see that you then only want to know the imaginary part of the integral with $\mathrm{e}^{\mathrm{i} a z}$? Edit to respond to question in comment below: \begin{align} \int_{-\infty}^\infty ...


1

You can use the exponential form. $z=2(1+ \sqrt3 i)^{1/2}=2\sqrt2(\frac{1}{2}+ \frac{\sqrt3}{2} i)^{1/2}=2\sqrt2(e^{\frac{i\pi}{3}})^{\frac{1}{2}}=2\sqrt2(e^{\frac{i\pi}{6}})=2\sqrt2(\frac{\sqrt3}{2}+i\frac{1}{2})=\sqrt2\sqrt3+i\sqrt2=\sqrt6+i\sqrt2$


2

This is similar to the difference between integral and Cauchy principal value For example you know that $$\int _ {\mathbb R} x dx$$ does not exists, but $$\lim_{R \to \infty} \int_{-R}^R x dx = 0$$ which is the Cauchy principal value. The main point is that $\int_\mathbb R$ is not defined as a limit $\lim_{R \to \infty} \int_{-R}^R$.. For the riemann ...


1

Both parts look fine. Maybe the second could be formulated somewhat simpler. But let's start with part (a): Since $\varphi$ is analytic at $a$, we can find a series representation as Taylor series \begin{align*} \varphi(z)=\sum_{k=0}^{\infty}\frac{\varphi^{(k)}(a)}{k!}(z-a)^k\tag{1} \end{align*} We obtain \begin{align*} ...


1

With total cheat, use the change of variable $w=\dfrac{1+i}{\sqrt2}z$. The equation becomes $$w^4-1=0$$ with the obvious solutions $$w=\pm1,\pm i.$$



Only top voted, non community-wiki answers of a minimum length are eligible