Question:
Consider $f(z)=u(x,y)+iv(x,y)$. When $u(x,y)=\frac{x}{x^2+y^2-2y+1}$, find $f(z)$ which can be a regular function.
Answer:
According to Cauchy-Riemann equations, if the function is regular, the following must be held:
\[
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}
\]
Use the first equation.
\begin{eqnarray}
\frac{\partial u}{\partial x} &=& \frac{\partial}{\partial x}\frac{x}{x^2+(y-1)^2} \\
&=& \frac{1\cdot(x^2+(y-1)^2)-x\cdot(2x)}{(x^2+(y-1)^2)^2} \\
&=& \frac{-x^2+(y-1)^2}{(x^2+(y-1)^2)^2} = \frac{\partial v}{\partial y}
\end{eqnarray}
Therefore,
\begin{eqnarray}
v &=& \int \frac{-x^2+(y-1)^2}{(x^2+(y-1)^2)^2}dy \\
&=& \frac{-(y-1)}{x^2+(y-1)^2}+\varphi(x)
\end{eqnarray}
Now, use the second equation.
\begin{eqnarray}
\frac{\partial}{\partial x}v &=& \frac{2x(y-1)}{(x^2+(y-1)^2)^2}+\varphi'(x)
&=& -\frac{\partial}{\partial y}u
\end{eqnarray}
Since $-\frac{\partial u}{\partial y}= \frac{2x(y-1)}{(x^2+(y-1)^2)^2}$,
\begin{equation}
\varphi'(x) = 0 \quad \rightarrow \quad \varphi(x) = C \ \mathrm{(constant)}
\end{equation}
Therefore the function $f(z)$ becomes
\begin{eqnarray}
f(z) &=& \frac{x}{x^2+(y-1)^2}+i \left[ \frac{-(y-1)}{x^2+(y-1)^2}+C \right] \\
&=& \frac{x-iy+i}{x^2+(y-1)^2}+iC \\
\end{eqnarray}
Now, we substitute the following:
\[
x = \frac{z+z'}{2}, \qquad y = \frac{z-z'}{2i}
\]
where $z=x+iy$ and $z'=x-iy$.
\begin{eqnarray}
f(z) &=& \frac{x-iy+i}{x^2+(y-1)^2}+iC \\
&=& \frac{z'+i}{zz'+iz-iz'+1}+iC \\
&=& \frac{z'+i}{(z-i)(z'+i)}+iC \\
\end{eqnarray}
Therefore, the function can be expressed by only $z$.
\begin{equation}
f(z) = \frac{1}{z-i}+iC
\end{equation}
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Showing posts with label complex variables. Show all posts
Showing posts with label complex variables. Show all posts
Thursday, February 11, 2016
Wednesday, February 3, 2016
Momentum space to configuration space
Question:
In quantum theory, the wave function can be convertible either in momentum space or configuration space. We have a wave function in momentum space: $\phi(p) = N/(p+\alpha^2)$. Find the equivalent wave function in configuration space.
Answer:
We can use the Fourier transformation to obtain the wave function in configuration space.
\begin{eqnarray*}
\Phi(x) &=& \frac{1}{\sqrt{2\pi\hbar}}\int^{\infty}_{-\infty}dp \frac{N}{p^2+\alpha^2}e^{ipx/\hbar} \\
&=& \frac{N}{\sqrt{2\pi\hbar}}\left(\int^{0}_{-\infty} \frac{e^{ipx/\hbar}}{p^2+\alpha^2}dp+\int^{\infty}_{0} \frac{e^{ipx/\hbar}}{p^2+\alpha^2}dp\right) \\
&=& \frac{N}{\sqrt{2\pi\hbar}}\left(\int^{\infty}_{0} \frac{e^{-ipx/\hbar}}{p^2+\alpha^2}dp+\int^{\infty}_{0} \frac{e^{ipx/\hbar}}{p^2+\alpha^2}dp\right) \\
&=& \frac{N}{\sqrt{2\pi\hbar}}\int^{\infty}_{0} \frac{e^{ipx/\hbar}+e^{-ipx/\hbar}}{p^2+\alpha^2}dp
\end{eqnarray*}
Use the following formula.
\[ \cos\theta = \frac{e^{i\theta}+e^{-i\theta}}{2} \]
We can rewrite it as
\[ \Phi(x) = 2\frac{N}{\sqrt{2\pi\hbar}}\int^{\infty}_{0} \frac{\cos(px/\hbar)}{p^2+\alpha^2}dp \]
Replace the momentum with the wavenumber; namely, $p=k\hbar$.
\begin{eqnarray*}
\Phi(x) &=& \frac{2}{\hbar}\frac{N}{\sqrt{2\pi\hbar}}\int^{\infty}_{0} \frac{\cos kx}{k^2+\alpha^2/\hbar^2}dk \\
&=& \sqrt{\frac{2N^2}{\pi \hbar^3}}\int^{\infty}_{0} \frac{\cos kx}{k^2+\alpha^2/\hbar^2}dk
\end{eqnarray*}
Let us put $\sqrt{\frac{2N^2}{\pi \hbar^3}}$ as $A$. Now, integrate the above in the complex plane. Consider the path of half circle in the upper half plane. The integral becomes
\[ I = A\oint \frac{e^{ikx}}{k^2+\alpha^2/\hbar^2}dk
= A\left( \int^r_{-r} \frac{e^{ikx}}{k^2+\alpha^2/\hbar^2}dk + \int_{C_r}\frac{e^{izx}}{z^2+\alpha^2/\hbar^2}dz \right) \]
where $z$ is the complex variable of $k$.
The second term will become zero because $1/(z^2+\alpha^2/\hbar^2) = 0$ as $|z|\rightarrow \infty$. Due to Jordan's lemma, when $r \rightarrow \infty$, the integral becomes zero. Namely, only the first term survives.
\[ I = A \int^{\infty}_{-\infty} \frac{e^{ikx}}{k^2+\alpha^2/\hbar^2}dk \]
The pole inside the contour is $i\frac{\alpha}{\hbar}$. Using the residue theorem, we obtain
\[ \mathrm{Res} [f, i\alpha/\hbar] = \lim_{z\rightarrow i\frac{\alpha}{\hbar}}\frac{e^{izx}}{z+i\frac{\alpha}{\hbar}} = \frac{e^{-\frac{\alpha}{\hbar}x}}{2i\frac{\alpha}{\hbar}}\]
Thus
\[ I = A \left(2\pi i \frac{e^{-\frac{\alpha}{\hbar}x}}{2i\frac{\alpha}{\hbar}}\right) = A \frac{\pi e^{-(\alpha/\hbar) x}}{\alpha/\hbar} \]
Take the real part of the integral, $I$.
\begin{eqnarray*}
A \int^{\infty}_{0} \frac{\cos kx}{k^2+\alpha^2/\hbar^2}dk &=& \frac{A}{2}\int^{\infty}_{-\infty} \frac{\cos kx}{k^2+\alpha^2/\hbar^2}dk \\
&=& \frac{A}{2}\frac{\pi e^{-(\alpha/\hbar) x}}{\alpha/\hbar}
\end{eqnarray*}
Therefore, we finally have the wave function in configuration space:
\begin{eqnarray*}
\Phi(x) &=& \sqrt{\frac{2N^2}{\pi \hbar^3}}\frac{1}{2}\frac{\pi e^{-(\alpha/\hbar) x}}{\alpha/\hbar} \\
&=& \frac{N}{\alpha}\sqrt{\frac{\pi}{2\hbar}}e^{-\frac{\alpha}{\hbar}x}
\end{eqnarray*}
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In quantum theory, the wave function can be convertible either in momentum space or configuration space. We have a wave function in momentum space: $\phi(p) = N/(p+\alpha^2)$. Find the equivalent wave function in configuration space.
Answer:
We can use the Fourier transformation to obtain the wave function in configuration space.
\begin{eqnarray*}
\Phi(x) &=& \frac{1}{\sqrt{2\pi\hbar}}\int^{\infty}_{-\infty}dp \frac{N}{p^2+\alpha^2}e^{ipx/\hbar} \\
&=& \frac{N}{\sqrt{2\pi\hbar}}\left(\int^{0}_{-\infty} \frac{e^{ipx/\hbar}}{p^2+\alpha^2}dp+\int^{\infty}_{0} \frac{e^{ipx/\hbar}}{p^2+\alpha^2}dp\right) \\
&=& \frac{N}{\sqrt{2\pi\hbar}}\left(\int^{\infty}_{0} \frac{e^{-ipx/\hbar}}{p^2+\alpha^2}dp+\int^{\infty}_{0} \frac{e^{ipx/\hbar}}{p^2+\alpha^2}dp\right) \\
&=& \frac{N}{\sqrt{2\pi\hbar}}\int^{\infty}_{0} \frac{e^{ipx/\hbar}+e^{-ipx/\hbar}}{p^2+\alpha^2}dp
\end{eqnarray*}
Use the following formula.
\[ \cos\theta = \frac{e^{i\theta}+e^{-i\theta}}{2} \]
We can rewrite it as
\[ \Phi(x) = 2\frac{N}{\sqrt{2\pi\hbar}}\int^{\infty}_{0} \frac{\cos(px/\hbar)}{p^2+\alpha^2}dp \]
Replace the momentum with the wavenumber; namely, $p=k\hbar$.
\begin{eqnarray*}
\Phi(x) &=& \frac{2}{\hbar}\frac{N}{\sqrt{2\pi\hbar}}\int^{\infty}_{0} \frac{\cos kx}{k^2+\alpha^2/\hbar^2}dk \\
&=& \sqrt{\frac{2N^2}{\pi \hbar^3}}\int^{\infty}_{0} \frac{\cos kx}{k^2+\alpha^2/\hbar^2}dk
\end{eqnarray*}
Let us put $\sqrt{\frac{2N^2}{\pi \hbar^3}}$ as $A$. Now, integrate the above in the complex plane. Consider the path of half circle in the upper half plane. The integral becomes
\[ I = A\oint \frac{e^{ikx}}{k^2+\alpha^2/\hbar^2}dk
= A\left( \int^r_{-r} \frac{e^{ikx}}{k^2+\alpha^2/\hbar^2}dk + \int_{C_r}\frac{e^{izx}}{z^2+\alpha^2/\hbar^2}dz \right) \]
where $z$ is the complex variable of $k$.
The second term will become zero because $1/(z^2+\alpha^2/\hbar^2) = 0$ as $|z|\rightarrow \infty$. Due to Jordan's lemma, when $r \rightarrow \infty$, the integral becomes zero. Namely, only the first term survives.
\[ I = A \int^{\infty}_{-\infty} \frac{e^{ikx}}{k^2+\alpha^2/\hbar^2}dk \]
The pole inside the contour is $i\frac{\alpha}{\hbar}$. Using the residue theorem, we obtain
\[ \mathrm{Res} [f, i\alpha/\hbar] = \lim_{z\rightarrow i\frac{\alpha}{\hbar}}\frac{e^{izx}}{z+i\frac{\alpha}{\hbar}} = \frac{e^{-\frac{\alpha}{\hbar}x}}{2i\frac{\alpha}{\hbar}}\]
Thus
\[ I = A \left(2\pi i \frac{e^{-\frac{\alpha}{\hbar}x}}{2i\frac{\alpha}{\hbar}}\right) = A \frac{\pi e^{-(\alpha/\hbar) x}}{\alpha/\hbar} \]
Take the real part of the integral, $I$.
\begin{eqnarray*}
A \int^{\infty}_{0} \frac{\cos kx}{k^2+\alpha^2/\hbar^2}dk &=& \frac{A}{2}\int^{\infty}_{-\infty} \frac{\cos kx}{k^2+\alpha^2/\hbar^2}dk \\
&=& \frac{A}{2}\frac{\pi e^{-(\alpha/\hbar) x}}{\alpha/\hbar}
\end{eqnarray*}
Therefore, we finally have the wave function in configuration space:
\begin{eqnarray*}
\Phi(x) &=& \sqrt{\frac{2N^2}{\pi \hbar^3}}\frac{1}{2}\frac{\pi e^{-(\alpha/\hbar) x}}{\alpha/\hbar} \\
&=& \frac{N}{\alpha}\sqrt{\frac{\pi}{2\hbar}}e^{-\frac{\alpha}{\hbar}x}
\end{eqnarray*}
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Tuesday, February 2, 2016
Trigonometric functions and complex variables
Question:
Define $z=x+iy$. Prove the following equation
\[ \frac{\sin 2x + i\sinh 2y}{\cos 2x + \cosh 2y} = \tan z \]
Answer:
There are following relationships:
\begin{eqnarray*}
\sinh x &=& -i\sin ix \\
\cosh x &=& \cos ix \\
\sin A + \sin B &=& 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \\
\cos A + \cos B &=& 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
\end{eqnarray*}
Thus,
\begin{eqnarray*}
\frac{\sin 2x + i\sinh 2y}{\cos 2x + \cosh 2y} &=& \frac{\sin 2x + i(-i\sin i2y)}{\cos 2x + \cos i2y}\\
&=& \frac{\sin 2x + \sin i2y}{\cos 2x + \cos i2y} \\
&=& \frac{2\sin\frac{2x+i2y}{2}\cos\frac{2x-i2y}{2}}{2\cos\frac{2x+i2y}{2}\cos\frac{2x-i2y}{2}} \\
&=& \frac{2\sin(x+iy)\cos(x-iy)}{2\cos(x+iy)\cos(x-iy)} \\
&=& \frac{\sin(x+iy)}{\cos(x+iy)} \\
&=& \tan(x+iy) = \tan z
\end{eqnarray*}
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Define $z=x+iy$. Prove the following equation
\[ \frac{\sin 2x + i\sinh 2y}{\cos 2x + \cosh 2y} = \tan z \]
Answer:
There are following relationships:
\begin{eqnarray*}
\sinh x &=& -i\sin ix \\
\cosh x &=& \cos ix \\
\sin A + \sin B &=& 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \\
\cos A + \cos B &=& 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
\end{eqnarray*}
Thus,
\begin{eqnarray*}
\frac{\sin 2x + i\sinh 2y}{\cos 2x + \cosh 2y} &=& \frac{\sin 2x + i(-i\sin i2y)}{\cos 2x + \cos i2y}\\
&=& \frac{\sin 2x + \sin i2y}{\cos 2x + \cos i2y} \\
&=& \frac{2\sin\frac{2x+i2y}{2}\cos\frac{2x-i2y}{2}}{2\cos\frac{2x+i2y}{2}\cos\frac{2x-i2y}{2}} \\
&=& \frac{2\sin(x+iy)\cos(x-iy)}{2\cos(x+iy)\cos(x-iy)} \\
&=& \frac{\sin(x+iy)}{\cos(x+iy)} \\
&=& \tan(x+iy) = \tan z
\end{eqnarray*}
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Saturday, January 30, 2016
De Moivre's theorem and application
Question:
By using De Moivre's theorem, simplify the following expressions:
(1) $\left(\frac{1+\sqrt{3}i}{2}\right)^{10}$
(2) $\left(\frac{\sqrt{3}+i}{1+i}\right)^{6}$
Answer:
De Moivre's theorem shows
\[ (\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta \]
(1) First, we obtain the magnitude.
\[ \left| \frac{1+\sqrt{3}i}{2} \right| = \sqrt{\left(\frac{1}{2}\right)^2+ \left(\frac{\sqrt{3}}{2}\right)^2} = 1 \]
The argument $\theta$ ranges from 0 to 360 degrees. Comparing with $\cos\theta + i\sin\theta$, we can have
\[ \cos\theta = \frac{1}{2}, \sin\theta = \frac{\sqrt{3}}{2} \]
Therefore, $\theta = 60^o$. Use the theorem.
\[ (\cos 60^o + i\sin 60^o)^{10} = \cos(10\times60^o)+i\sin(10\times60^o) \\
= \cos 600^o + i\sin 600^o \\
= \cos 240^o + i\sin 240^o \\
= -\frac{1}{2}-\frac{\sqrt{3}}{2}i \]
(2)
Let us separate the numerator and denominator.
\[ \frac{(\sqrt{3}+i)^6}{(1+i)^6} \]
For the numerator, the magnitude and argument are
\[ |\sqrt{3}+i| = 2 \\
\cos\theta_1 = \frac{\sqrt{3}}{2}, \sin\theta_1 = \frac{1}{2} \rightarrow \theta_1 = 30^o \]
Thus, $\sqrt{3}+i = 2(\cos 30^o + i\sin 30^o)$.
For the denominator, we have
\[ |1+i| = \sqrt{2} \\
\cos\theta_2 = \frac{1}{\sqrt{2}}, \sin\theta_2 = \frac{1}{\sqrt{2}} \rightarrow \theta_2 = 45^o \]
Thus, $1+i = \sqrt{2}(\cos 45^o + i\sin 45^o)$.
We have
\[ \frac{\sqrt{3}+i}{1+i}=\frac{2(\cos 30^o + i\sin 30^o)}{\sqrt{2}(\cos 45^o + i\sin 45^o)} \\
= \sqrt{2}[\cos(-15^o)+i\sin(-15^o)] \]
Use the theorem.
\[ (\sqrt{2}[\cos(-15^o)+i\sin(-15^o)])^6 = 2^3[\cos(-6\times 15^o) + i\sin(-6\times 15^o)] \\
= 8[\cos(-90^o) + i\sin(-90^o)] \\
=-8i \]
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By using De Moivre's theorem, simplify the following expressions:
(1) $\left(\frac{1+\sqrt{3}i}{2}\right)^{10}$
(2) $\left(\frac{\sqrt{3}+i}{1+i}\right)^{6}$
Answer:
De Moivre's theorem shows
\[ (\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta \]
(1) First, we obtain the magnitude.
\[ \left| \frac{1+\sqrt{3}i}{2} \right| = \sqrt{\left(\frac{1}{2}\right)^2+ \left(\frac{\sqrt{3}}{2}\right)^2} = 1 \]
The argument $\theta$ ranges from 0 to 360 degrees. Comparing with $\cos\theta + i\sin\theta$, we can have
\[ \cos\theta = \frac{1}{2}, \sin\theta = \frac{\sqrt{3}}{2} \]
Therefore, $\theta = 60^o$. Use the theorem.
\[ (\cos 60^o + i\sin 60^o)^{10} = \cos(10\times60^o)+i\sin(10\times60^o) \\
= \cos 600^o + i\sin 600^o \\
= \cos 240^o + i\sin 240^o \\
= -\frac{1}{2}-\frac{\sqrt{3}}{2}i \]
(2)
Let us separate the numerator and denominator.
\[ \frac{(\sqrt{3}+i)^6}{(1+i)^6} \]
For the numerator, the magnitude and argument are
\[ |\sqrt{3}+i| = 2 \\
\cos\theta_1 = \frac{\sqrt{3}}{2}, \sin\theta_1 = \frac{1}{2} \rightarrow \theta_1 = 30^o \]
Thus, $\sqrt{3}+i = 2(\cos 30^o + i\sin 30^o)$.
For the denominator, we have
\[ |1+i| = \sqrt{2} \\
\cos\theta_2 = \frac{1}{\sqrt{2}}, \sin\theta_2 = \frac{1}{\sqrt{2}} \rightarrow \theta_2 = 45^o \]
Thus, $1+i = \sqrt{2}(\cos 45^o + i\sin 45^o)$.
We have
\[ \frac{\sqrt{3}+i}{1+i}=\frac{2(\cos 30^o + i\sin 30^o)}{\sqrt{2}(\cos 45^o + i\sin 45^o)} \\
= \sqrt{2}[\cos(-15^o)+i\sin(-15^o)] \]
Use the theorem.
\[ (\sqrt{2}[\cos(-15^o)+i\sin(-15^o)])^6 = 2^3[\cos(-6\times 15^o) + i\sin(-6\times 15^o)] \\
= 8[\cos(-90^o) + i\sin(-90^o)] \\
=-8i \]
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Thursday, January 28, 2016
Residue theorem and integral of a complex-valued function
Question:
Find the value of
\[ \int_C \frac{dz}{z^2+2iz-4} \]
The contour, $C$, is a circle that has center $z=1$ and radius $\sqrt{2}$. It is directed positively.
Answer:
The integrand can be reduced in two factors.
\[ I = \int_C \frac{dz}{z^2+2iz-4} = \int_C \frac{dz}{(z+(\sqrt{3}+i))(z-(\sqrt{3}-i))} \]
The value when the denominator becomes zero is the singular point in the contour. $z=-(\sqrt{3}+i)$ and $z=\sqrt{3}-i$ can be the points, but only $z=\sqrt{3}-i$ is the pole inside contour.
Calculate the residue.
\[ \mathrm{Res}(f(z):\sqrt{3}-i) = \lim_{z\rightarrow \sqrt{3}-i}(z-(\sqrt{3}-i)f(z) \\
= \lim_{z\rightarrow \sqrt{3}-i}(z-(\sqrt{3}-i)\frac{1}{(z+(\sqrt{3}+i))(z-(\sqrt{3}-i))} \\
= \lim_{z\rightarrow \sqrt{3}-i}\frac{1}{z+(\sqrt{3}+i)} \\
= \frac{1}{2\sqrt{3}}\]
Then, using the residue theorem, we have the value of integral:
\[ I = 2\pi i \times \frac{1}{2\sqrt{3}} = \frac{\sqrt{3}\pi i}{3} \]
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Find the value of
\[ \int_C \frac{dz}{z^2+2iz-4} \]
The contour, $C$, is a circle that has center $z=1$ and radius $\sqrt{2}$. It is directed positively.
Answer:
The integrand can be reduced in two factors.
\[ I = \int_C \frac{dz}{z^2+2iz-4} = \int_C \frac{dz}{(z+(\sqrt{3}+i))(z-(\sqrt{3}-i))} \]
The value when the denominator becomes zero is the singular point in the contour. $z=-(\sqrt{3}+i)$ and $z=\sqrt{3}-i$ can be the points, but only $z=\sqrt{3}-i$ is the pole inside contour.
Calculate the residue.
\[ \mathrm{Res}(f(z):\sqrt{3}-i) = \lim_{z\rightarrow \sqrt{3}-i}(z-(\sqrt{3}-i)f(z) \\
= \lim_{z\rightarrow \sqrt{3}-i}(z-(\sqrt{3}-i)\frac{1}{(z+(\sqrt{3}+i))(z-(\sqrt{3}-i))} \\
= \lim_{z\rightarrow \sqrt{3}-i}\frac{1}{z+(\sqrt{3}+i)} \\
= \frac{1}{2\sqrt{3}}\]
Then, using the residue theorem, we have the value of integral:
\[ I = 2\pi i \times \frac{1}{2\sqrt{3}} = \frac{\sqrt{3}\pi i}{3} \]
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