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### Messages - Victor Ivrii

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16
##### Final Exam / Re: FE-P5
« on: December 20, 2018, 05:09:56 AM »
(a) As $|z|=2$ consider $Q(z)=z^3$; then  $|Q(z)|=8$,
$$|P(z)-Q(z)|=|2iz -3-i|\le 4+|3+i|=4+\sqrt{10}<8;$$
therefore $P$ has as many roots in $\{z\colon |z|<2\}$ as $Q(z)$ has, which is $3$ (we count orders).

(c) Then there are no roots in $\{z\colon |z|>2\}$.

(b) Consider $z=w+1$ with $|w|=1$; then $P(w+1)=w^3+3w^2 +5w -i$ and we set $Q(w)=5w$; then $|Q(w)|=5$,
$$|P(w)-Q(w)|=|w^3+3w^2 -i| < 5.$$
Indeed, $|P(w)-Q(w)|\le |w+3|+1< 5$ except $w=1$, and for $w=1$ we have $|w^3+3w^2 -i|=|4-i|=\sqrt{17}<5$. Therefore $P(w+1)$ has as many roots in $\{w\colon |w|<1\}$ as $Q(w)$ has, which is $1$ (we count orders).

Finally, in the domain  $\{z\colon |z-1|>1, |z|<2\}$, there are $3-1=2$ roots.

17
##### Final Exam / FE-P3 official
« on: December 20, 2018, 05:04:59 AM »
$\newcommand{\Res}{\operatorname{Res}}$
(a) Singular points are of $g(z)=\tan(z)$ and $h(z)=z\cot ^2(z)$, that is $z_n= (n+\frac{1}{2})\pi$ and $w_n = \pi n$.

(b) $z_n$ are simple poles and $\Res (f,z_n)= \Res (\tan(z), z_n)= \frac{\sin(z)}{(\cos (z))'}\bigr|_{z=z_n} =-1$.

$w_0$ is a simple pole
$$\Res (f,w_0)= \Res (z\cot^2(z), w_0)= \Res (z\cot (z) \times \cot(z), 0)= \Res (\cot(z),0)= 1$$
because $\lim _{z\to 0} z\cot(z)=1$.

$w_n$ with $n\ne 0$ are double poles and
\begin{align*}
\Res (f,w_n)=&\Res (z\cot^2 (z), w_n)= \Res ((\pi n +w)\cot^2 (w) , 0) =\\
&\pi n\Res (\cot^2 (w) , 0) +\Res (w\cot^2(w),0) =1
\end{align*}
because $\Res (\cot^2 (w) , 0)=0$ (since $\cot^2(w)$ is an even function) and $\Res (w\cot^2(w),0) =1$ we already calculated.

$\infty$ is a not isolated singularity and therefore residue here is not defined.

18
##### Final Exam / FE-P2 official
« on: December 20, 2018, 04:49:22 AM »
(a)--(b) If $|z|=r$ and $\arg{w}=t$, then $z=r e^{it}$ and
\begin{align*}
w= \frac{1}{2}\bigl(re^{it} +r^{-1}e^{-it}\bigr)&= \cosh(s)\cos(t)+i\sinh(s)\sin (t)
\end{align*}
with $s=\ln (r)$ (and $r=e^s$).

Then $w=u+iv$ with $\frac{u^2}{a^2}+\frac{v^2}{b^2}=1$, $a=\cos(s)$, $b=\sinh(s)$ and also
$\frac{u^2}{A^2}-\frac{v^2}{B^2}=1$,
with  $A=\cos (\theta)$, $B=\sin(\theta)$, $A^2+B^2=1$.

(c) Consider, what the circle $\{z\colon |z|=1\}$ is mapped to: $z=e^{it}$, then
$w =\frac{1}{2}(e^{it}+e^{-it})=\cos(t)$ and it runs a line segment $[-1,1]\subset\mathbb{R}$. So $f$ maps the unit disk onto compliment of it $\mathbb{C}^*\setminus [-1,1]$ (and $0$ is mapped to $\infty$).

(d) See attachment

(e) $w=\frac{1}{2}(z+z^{-1}) \implies z^2 -2wz +1=0$; it has two roots $z_1$ and $z_2$ s.t.
$z_1z_2=1$ and $z_1+z_2=2w$; exactly one of them for $w\notin [-1,1]$ is in unit disk, and the second one is in $\{z\colon |z|>1\}$.

19
##### Final Exam / FE-P1 official
« on: December 20, 2018, 04:41:00 AM »
Observe that
$$f(z)=\frac{1}{(z+1+i)(z+1-i)}= \frac{1}{2i}\Bigl(\frac{1}{z-z_2}-\frac{1}{z-z_1}\Bigr)$$
has two singular points $z_{1,2} =-1\pm i=\sqrt{2}e^{\pm 5\pi i/4}$.

(a)   So $R=\sqrt{2}$ and
\begin{align*}
f(z)=
&\frac{1}{2i}\Bigl(\frac{1}{z_2-z}-\frac{1}{z_1-z}\Bigr)=
\frac{1}{2i}\Bigl(  \sum_{n=0}^\infty z^n z_2^{-n-1} - \sum_{n=0}^\infty z^n z_1^{-n-1}\Bigr)=\\
&\frac{1}{2i} \sum_{n=0}^\infty z^n \Bigr(z_2^{-n-1} - z_1^{-n-1}\Big)=
\frac{1}{2i} \sum_{n=0}^\infty z^n 2^{-(n+1)/2}\Bigl(e^{5(n+1)\pi i/4} - e^{-5(n+1)\pi i/4}\Bigr)=\\
& \sum_{n=0}^\infty 2^{-(n+1)/2}\sin (5(n+1))\pi /4)z^n
\end{align*}
As $|z|=\sqrt{2}$ terms do not tend to $0$ and the series diverges.

(b) So $R=\sqrt{2}$ and
\begin{align*}
f(z)=
&\frac{1}{2i}\Bigl(\frac{1}{z-z_1}-\frac{1}{z-z_2}\Bigr)=
\frac{1}{2i}\Bigl(  \sum_{n=0}^\infty z^{-n-1} z_1^{n} - \sum_{n=0}^\infty z^{-n-1} z_2^{n}\Bigr)=\\
&\frac{1}{2i} \sum_{n=0}^\infty z^{-n-1} \Bigr(z_1^{n} - z_2^{n}\Big)=
\frac{1}{2i} \sum_{n=0}^\infty z^{-n-1} 2^{n/2}\Bigr(e^{5n\pi i/4} - e^{-5n\pi i/4}\Big)=\\
&\sum_{n=0}^\infty 2^{n/2}\sin (5n\pi /4)z^{-n-1} = \sum_{-\infty}^{-2} 2^{-(n+1)/2}\sin (5z(-n-1)\pi /4)z^{n}
\end{align*}
where we plugged $n=m-1$ with $m=1,2,\ldots$, observed that the term with $m=1$ is $0$ and finally replaced $m$ by $n$.

As $|z|=\sqrt{2}$ terms do not tend to $0$ and the series diverges.

20
##### Final Exam / FE-P6 Official
« on: December 20, 2018, 04:37:41 AM »
There is a single pole of $f(z)=\frac{1}{1+z^n}$ inside $\Gamma$, namely $z=e^{i\pi/n}$, which is a simple pole and the residue is $\frac{1}{(1+z^n)'}\bigr|_{z=e^{i\pi/n}}= \frac{1}{nz^{n-1}}\bigr|_{z=e^{i\pi/n}}=-\frac{1}{n}e^{i\pi/n}$.

Therefore due to the residue theorem $I_R+J_R+K_R= -\frac{2}{n}\pi i \times e^{i\alpha/2}$, where $K_R$ is an integral over an arc and $J_R$ is an integral over the second straight segment.

Then
$$J_R=\int _{R}^0 \frac{e^{i\alpha}\,dt }{1+e^{in\alpha}t^n}=-e^{i\alpha}I_R$$
with $I_R=\int_0^R\frac{dx}{1+x^n}$.

On the other hand,
$$|K_R|=|\int_0^\alpha \frac{iRe^{it}\,dt}{1+e^{itn}R^n }|\le \frac{R}{R^n-1}\int_0^\alpha\, dt=\frac{\alpha R}{R^n-1}\to 0 \qquad\text{as }\ \ R\to \infty.$$

Then $(1-e^{i\alpha}) I =-\frac{2}{n}\pi i \times e^{i\alpha/2}$ and
\begin{align*}
I =& -\frac{2}{n}\pi i \times  \frac{e^{i\alpha/2}}{1-e^{i\pi \alpha}}=
&&-\frac{2}{n}\pi i \times  \frac{1}{e^{-i\alpha/2}-e^{i\pi \alpha/2}}=\\
&-\frac{2}{n}\pi i \times  \frac{1}{-2i\sin(\alpha/2)}=
&&\frac{\pi}{n\sin(\pi /n)}.
\end{align*}

21
##### Final Exam / FE-P4 official
« on: December 20, 2018, 02:01:42 AM »
$\renewcommand{\Re}{\operatorname{Re}}$
Characteristic equation:
$$\left|\begin{matrix} 1-k & -2\\ 1 & -1-k \end{matrix}\right|= k^2+1=0\implies k_{1,2}=\pm i.$$
Finding eigenvector corresponding to $k_1=i$
$$\begin{pmatrix} 1-i & -2\\ 1 & -1-i \end{pmatrix}\begin{pmatrix} \alpha\\ \beta \end{pmatrix}=0\implies \alpha =(1+i)\beta\implies \mathbf{e}_1 =\begin{pmatrix} 1+i\\ 1 \end{pmatrix}$$
and $\mathbf{e}_2$ is complex conjugate. Then the general solution to the homogeneous equation is
\begin{align}
\begin{pmatrix}x\\y\end{pmatrix}=&\Re \Bigl[(C_1+iC_2) (\cos (t)+i\sin(t))\begin{pmatrix} 1+i\\ 1 \end{pmatrix}\Bigr]\notag\implies \\
&x=  C_1 (\cos(t)-\sin(t))+ C_2( -\cos(t)-\sin(t) )\\
&y=  C_1\cos(t)    -C_2\sin(t)  \,.
\label{eq-4-1}
\end{align}
We are looking for solution to inhomogeneous equation in the same form albeit with variable $C_1,C_2$. Then
\begin{align*}
&\left\{\begin{aligned}
&C'_1 (\cos(t)-\sin(t))+ C'_2( -\cos(t)-\sin(t) )=\sec(t),\\
\end{aligned}\right.\implies\\[4pt]
&\left\{\begin{aligned}
&C'_1\sin(t)+ C'_2\cos(t)=-\sec(t),\\
&C'_1\cos(t)      -\ C'_2\sin(t) =0
\end{aligned}\right.\implies\\
&\left\{\begin{aligned}
&C'_1= -\sec(t)\sin(t)=-\tan(t)\implies C_1=\ln (\cos(t))+c_1\\
&C'_2=-\sec(t)\cos(t)=-1\implies C_2=-t +c_2\,.
\end{aligned}\right.
\end{align*}
Finally
\begin{align*}
&x=  [\ln (\cos(t))+c_1] (\cos(t)-\sin(t))+ [t -c_2]( \cos(t)+\sin(t) )\,\\
&y=[\ln (\cos(t))+c_1]\cos(t)   +[t-c_2]\sin(t) \,.
\end{align*}

22
##### Final Exam / FE-P3 official
« on: December 20, 2018, 01:59:33 AM »
Writing characteristic equation: $L(k):= k^3-2k^2-k+2=0$, with $L(k)=(k-2)k^2 -(k-2)=(k-2)(k^2-1)$; then $k_1=1, k_2=-1, k_3=2$. Then

y^*= C_1e^{t} + C_2 e^{-t}  + C_3 e^{2t}
\label{eq-3-1}

is a general solution to the homogeneous equation.

We are looking for solution to the inhomogeneous equation as (\ref{eq-3-1}) with unknown functions $C_1,C_2,C_3$ s.t.
\begin{align*}
&\left\{\begin{aligned}
&C_1' e^{t} + C_2' e^{-t}+ C_3' e^{2t}=0,\\
&C_1' e^{t} - C_2' e^{-t} + 2C_3' e^{2t}=0,\\
&C_1' e^{t} + C_2' e^{-t}+4 C_3' e^{2t}=\frac{12e^{2t}}{e^t+1};
\end{aligned}\right.\implies\\[4pt]
&\left\{\begin{aligned}
2&C_1' e^{t} + 3C_3' e^{2t}=0,\\
&C_1' e^{t} - C_2' e^{-t} + 2C_3' e^{2t}=0,\\
2&C_1' e^{t} +  6 C_3' e^{2t}=\frac{12e^{2t}}{e^t+1};
\end{aligned}\right.\implies\\[4pt]
&C_1=-\int \frac{6e^{t}\,dt}{e^t+1}= -6 \ln (e^t+1) +c_1,\\[4pt]
&C_2= \int \frac{2e^{3t}\,dt}{e^t+1}= \int \Bigl[2e^{2t}- 2 e^{t}\Bigr] \,dt +\int  \frac{2e^{t}\,dt}{e^t+1}
=e^{2t} -2e^{t} +2\ln (e^t+1)+c_2,\\[4pt]
&C_3=\int \frac{4\,dt}{e^t+1}=\int \frac{4e^{-t}\,dt}{1+e^{-t}}= -4\ln (1+e^{-t})+c_3= -4\ln (e^t+1)+4t +c_3.
\end{align*}
Then
\begin{align*}
y= &\bigl[-6 \ln (e^t+1) +c_1\bigr]e^{t}+\bigl[e^{2t} -2e^{t} +2\ln (e^t+1)+c_2\bigr]e^{-t}+
\bigl[-4\ln (e^t+1)+4t +c_3\bigr]e^{2t}=\\
&-6  e^{t}\ln (e^t+1)+ 2e^{-t}\ln (e^t+1)
-4e^{2t}\ln (e^t+1)+4te^{2t}-2 +
c_1e^{t} + c_2 e^{-t}  + c_3 e^{2t}
\end{align*}
with $c_1:= c_1+1$ in the last transition.

23
##### Final Exam / FE-P2 official
« on: December 20, 2018, 01:53:44 AM »
Writing characteristic equation: $L(k):= k^3-3k^2+4k-2=0$. Obviously, one root $k_1=1$; then $k_2+k_3= 2$, $k_1k_2=2$ and they satisfy $k^2-2k+2=0\implies k_{1,2}= 1\pm i$. Then

y^*= C_1e^t + C_2 e^t \cos(t) + C_3 e^t\sin(t)
\label{eq-2-1}

is a general solution to the homogeneous equation.

Solving inhomogeneous equations with RHE $f_1=10e^{t}$, $f_2=10 e^{-t}$, $f_3=\cos(t)$:
\begin{align*}
&y_{p1}= At e^t,\\
&y_{p2}=Be^{-t},\\
&y_{p3}= C\cos(t)+D\sin(t).
\end{align*}
Here $A L'(k)|_{k=1} = A(3k^2-6k+4)|_{k=1}=10\implies A=10$,
$BL(-1) =-10 B=10\implies B=-1$ and
$$(C+iD)L(i)= (A+iB) (1+3i)=20\implies C+iD= \frac{20}{1+3i}=\frac{20(1-3i)}{10}= 2-6i\implies C=2, D=6.$$
Then
\begin{align*}
&y_{p1}= 10t e^t,\\
&y_{p2}=-e^{-t},\\
&y_{p3}= 2\cos(t)+6\sin(t).
\end{align*}
Finally
$$y= \underbracket{10t e^{t}}_{y_{p1}}\underbracket{-e^{-t}}_{y_{p2}}+ \underbracket{2\cos(t)+6\sin(t)}_{y_{p3}} + \underbracket{C_1e^t + C_2 e^t \cos(t) + C_3 e^t\sin(t)}_{y^*}.$$

24
##### Final Exam / FE-P1 official
« on: December 20, 2018, 01:49:48 AM »
As $M=2x\sin(y) +1$, $N= 4x^2\cos(y) + 3x\cot(y)+10 \sin(y)\cos(y)$ we have
$$M_y-N_x= 2x\cos(y)-[8x\cos(y)+3\cot(y)]=-6x \cos(y) -3\frac{\cos(y)}{\sin(y)} \implies \frac{N_x-M_y}{M}= 3\cot(y).$$
Therefore we are looking for an integrating factor $\mu(y)$, satisfying
$$\frac{\mu'}{\mu}= 3\cot(y)\implies \ln (\mu)=3\int \cot(y)\,dy=3\ln (\sin(y))$$
(we take constant equal $0$). Then $\mu =\sin^3(y)$.

After multiplication we get
$$\bigl[2x\sin^4(y) +\sin^3(y)\bigr]\,dx + \bigl[4x^2\sin^3(y)\cos(y) + 3x\sin^2(y)\cos(y)+10 \sin^4(y)\cos(y)\bigr]\,dy=0\,.$$
Looking for $H(x,y)$ satisfying
\begin{align}
&H_x = 2x\sin^4(y) +\sin^3(y)\,,\label{eq-1-1}\\
&H_y=4x^2\sin^3(y)\cos(y) + 3x\sin^2(y)\cos(y)+10 \sin^4(y)\cos(y)\,.
\label{eq-1-2}

\end{align}
(\ref{eq-1-1}) implies
$$H(x,y)= \int \bigl[2x\sin^4(y) +\sin^3(y)\bigr]\,dx= x^2\sin^4(y)+ x\sin^3(y) +h(y).$$
Then
$$H_y= 4x^2\sin^3(y)\cos(y)+ 3x\sin^2\cos(y) +h'(y)$$
and comparing with (\ref{eq-1-2}) we see that
$$h'(y)=10\sin^4(y)\cos(y)\implies h(y)=\int 10\sin^4(y)\cos(y)\,dy= 2\sin^5(y);$$
we pick-up constant $0$.

Finally
$$H(x,y)= x^2\sin^4(y)+ x\sin^3(y) +2\sin^5(y) =C$$
is a solution to the original problem.

25
##### Final Exam / Re: FE-P6
« on: December 18, 2018, 01:43:13 PM »
Need to simplify to a real number

26
##### Final Exam / Re: FE-P6
« on: December 18, 2018, 12:29:26 PM »
$k=?$

Need to simplify to a real number

27
##### Final Exam / FE-P6
« on: December 18, 2018, 06:22:02 AM »
Calculate for real $n>1$
$$I:= \int_0^\infty\frac{dx}{1+x^n}.$$

Hint:  Consider
$$\int_\gamma \frac{dz}{1+z^n}$$
with with an arc of radius $R\to \infty$ and an angle $\alpha=\frac{2\pi}{n}$. Express the integral over the second straight segment through integral over the first one.

28
##### Final Exam / FE-P5
« on: December 18, 2018, 06:18:35 AM »
Consider $P(z)= z^3 +2z -3-i$ and, using the argument theorem and RouchÃ©'s theorem calculate the number of its roots in each of the following domains:

(a)  $\{z\colon |z-1|<1\}$;

(b)  $\{z\colon |z-1|>1, |z|<2\}$,

(c) $\{z\colon |z|>2\}$.

29
##### Final Exam / FE-P4
« on: December 18, 2018, 06:17:32 AM »
(a) Find the MÃ¶bius transformation (fractional-linear transformation) $f(z)$ mapping the unit disk $\{z\colon |z|<1\}$ onto itself, such that $f(0)=\frac{1}{2}$ and $f(1)=-1$.

(b) Find the fixed points of $f$ (points s.t. $f(z)=z$)

(c) Find the stretch ($|f'(z)|$) and the rotation angle ($\arg(f'(z))$) of $f$ at $z$.

30
##### Final Exam / FE-P3
« on: December 18, 2018, 06:14:31 AM »
Find all singular points, classify them, and find residues at these points of
$$f(z)= \tan (z) + z\cot^2(z);$$
infinity included.

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