Consider the function \(f\left ( z\right ) =z^{\frac{1}{n}}\) where \(n\) is a positive integer. The branch point is at \(z=0\) and the branch cut is chosen to be along the positive \(x\) axis. How many sheets are there? What is the range of \(\theta \) corresponding to each sheet?
Solution
Following the example in the class handout, where it showed how to find the number of sheets for \(z^{\frac{1}{2}}\), the same method is used here, which is to keep adding a multiple of \(2\pi \) angles until the same result for the original principal value of the function \(g\left ( z\right ) \) evaluated at \(\theta \) is obtained. This gives the number of sheets.
Let \begin{align} g\left ( z\right ) & =z^{\frac{1}{n}}\nonumber \\ g\left ( r,\theta \right ) & =\left ( re^{i\theta }\right ) ^{\frac{1}{n}}\nonumber \\ g\left ( r,\theta \right ) & =r^{\frac{1}{n}}e^{i\frac{\theta }{n}} \tag{1} \end{align}
In the above, \(\theta \) is called principal argument. And now the idea is to find how many times \(2\pi \) needs to be added to \(\theta \) in order to get back the same value of original of \(g\left ( r,\theta \right ) \) at the starting \(\theta \) that one picks. Adding one time \(2\pi \) to \(\theta \), equation (1) becomes \begin{align*} g\left ( r,\theta +2\pi \right ) & =r^{\frac{1}{n}}e^{i\frac{\left ( \theta +2\pi \right ) }{n}}\\ & =r^{\frac{1}{n}}e^{i\frac{\theta }{n}+i\frac{2\pi }{n}}\\ & =r^{\frac{1}{n}}e^{i\frac{\theta }{n}}e^{i\frac{2\pi }{n}} \end{align*}
And we add another \(2\pi \), or now a total of \(4\pi \) \begin{align*} g\left ( r,\theta +4\pi \right ) & =r^{\frac{1}{n}}e^{i\frac{\left ( \theta +4\pi \right ) }{n}}\\ & =r^{\frac{1}{n}}e^{i\frac{\theta }{n}+i\frac{4\pi }{n}}\\ & =r^{\frac{1}{n}}e^{i\frac{\theta }{n}}e^{i\frac{4\pi }{n}} \end{align*}
And so on. We keep adding \(2\pi \), or a total of \(k\left ( 2\pi \right ) \) such that the last term above, which in term of \(k\) is \(e^{\frac{k\left ( 2\pi \right ) i}{n}}\) simplifies to \(1\) which implies getting back original function value at \(g\left ( r,\theta \right ) \). Hence for \(k\) times we have\begin{align*} g\left ( r,\theta +k\left ( 2\pi \right ) \right ) & =r^{\frac{1}{n}}e^{i\frac{\left ( \theta +k\left ( 2\pi \right ) \right ) }{n}}\\ & =r^{\frac{1}{n}}e^{i\frac{\theta }{n}+i\frac{k\left ( 2\pi \right ) }{n}}\\ & =r^{\frac{1}{n}}e^{i\frac{\theta }{n}}e^{i\frac{k\left ( 2\pi \right ) }{n}} \end{align*}
We see from the above, is that only when \(k=n\), then \(r^{\frac{1}{n}}e^{i\frac{\theta }{n}}e^{i\frac{k\left ( 2\pi \right ) }{n}}=r^{\frac{1}{n}}e^{i\frac{\theta }{n}}e^{2\pi i}\). But \(e^{2\pi i}=1\), therefore it reduces to\begin{align*} g\left ( r,\theta +n\left ( 2\pi \right ) \right ) & =r^{\frac{1}{n}}e^{i\frac{\theta }{n}}\\ & =g\left ( r,\theta \right ) \end{align*}
Which is the original value of the function. Therefore there are \(n\) sheets.
The formula that can also be used to obtain all values for this multivalued function is\[ g\left ( r,\theta \right ) =r^{\frac{1}{n}}e^{i\left ( \frac{\theta }{n}+\frac{2\pi }{n}k\right ) }\qquad k=0,1,\cdots n-1 \] Now to answer the angle \(\theta \) range question. From the above, we see the range of the angle for each sheet is as follows\begin{align*} R_{1} & :0<\theta <2\pi \\ R_{2} & :2\pi <\theta <4\pi \\ R_{3} & :4\pi <\theta <6\pi \\ & \vdots \\ R_{n} & :\left ( n-1\right ) 2\pi <\theta <n\left ( 2\pi \right ) \end{align*}
Sheet \(R_{1}\) is called the principal sheet associated with \(k=0\).
Derive the formula \[ \arctan z=\frac{i}{2}\ln \left ( \frac{i+z}{i-z}\right ) \] Solution
Let \(w=\arctan \left ( z\right ) \) hence \begin{align*} z & =\tan \left ( w\right ) \\ z & =\frac{\sin w}{\cos w} \end{align*}
But \(\sin w=\frac{e^{iw}-e^{-iw}}{2i}\) and \(\cos w=\frac{e^{iw}+e^{-iw}}{2}\), hence the above simplifies to\begin{align*} z & =\frac{\frac{e^{iw}-e^{-iw}}{2i}}{\frac{e^{iw}+e^{-iw}}{2}}\\ & =\frac{1}{i}\frac{e^{iw}-e^{-iw}}{e^{iw}+e^{-iw}} \end{align*}
Or\[ iz=\frac{e^{iw}-e^{-iw}}{e^{iw}+e^{-iw}}\] Multiplying the numerator and denominator of the right side by \(e^{iw}\) gives\[ iz=\frac{e^{2iw}-1}{e^{2iw}+1}\] Let \(e^{iw}=x\) then the above is the same as\begin{align*} iz & =\frac{x^{2}-1}{x^{2}+1}\\ iz\left ( x^{2}+1\right ) & =x^{2}-1\\ x^{2}iz+iz & =x^{2}-1\\ x^{2}iz+iz-x^{2}+1 & =0\\ x^{2}\left ( iz-1\right ) +\left ( 1+iz\right ) & =0\\ x^{2} & =\frac{-\left ( 1+iz\right ) }{\left ( iz-1\right ) }\\ & =\frac{\left ( 1+iz\right ) }{\left ( 1-iz\right ) } \end{align*}
Simplifying gives\begin{align*} x^{2} & =\frac{i\left ( -i+z\right ) }{i\left ( -i-z\right ) }\\ & =\frac{\left ( z-i\right ) }{\left ( -i-z\right ) } \end{align*}
Hence \[ x=\pm \left ( \frac{z-i}{-i-z}\right ) ^{\frac{1}{2}}\] But \(x=e^{iw}\), and the above becomes\[ e^{iw}=\pm \left ( \frac{z-i}{-i-z}\right ) ^{\frac{1}{2}}\] We need now to decide which sign to take. Since \(z=\tan \left ( w\right ) \), then when \(w=0\), \(z=0\) because \(\tan \left ( 0\right ) =0\). Putting \(w=0,z=0\) in the above gives\begin{align*} 1 & =\pm \left ( \frac{i}{i}\right ) ^{\frac{1}{2}}\\ & =\pm \left ( 1\right ) ^{\frac{1}{2}}\\ & =\pm 1 \end{align*}
Hence we need to choose the \(+\) sign so both sides is positive. Hence\[ e^{iw}=\left ( \frac{z-i}{-i-z}\right ) ^{\frac{1}{2}}\] Now, taking the natural log of both sides gives\begin{align*} iw & =\ln \left ( \frac{z-i}{-i-z}\right ) ^{\frac{1}{2}}\\ iw & =\frac{1}{2}\ln \left ( \frac{z-i}{-i-z}\right ) \\ w & =\frac{1}{2i}\ln \left ( \frac{z-i}{-i-z}\right ) \\ & =\frac{-i}{2}\ln \left ( \frac{z-i}{-i-z}\right ) \\ & =\frac{i}{2}\ln \left ( \left ( \frac{z-i}{-i-z}\right ) ^{-1}\right ) \\ & =\frac{i}{2}\ln \left ( \frac{-i-z}{z-i}\right ) \\ & =\frac{i}{2}\ln \left ( \frac{-\left ( z+i\right ) }{-\left ( i-z\right ) }\right ) \\ & =\frac{i}{2}\ln \left ( \frac{z+i}{i-z}\right ) \end{align*}
But \(w=\arctan \left ( z\right ) \), hence the final result is\[ \arctan \left ( z\right ) =\frac{i}{2}\ln \left ( \frac{i+z}{i-z}\right ) \]
Using the formula for \(\arctan z\) from the previous problem, find the real functions \(u\left ( x,y\right ) \) and \(v\left ( x,y\right ) \) in the expression \(\arctan z=u\left ( x,y\right ) +iv\left ( x,y\right ) \)
Solution
Let \[ \frac{i}{2}\ln \left ( \frac{i+z}{i-z}\right ) =u+iv \] where \(u\equiv u\left ( x,y\right ) ,v\equiv v\left ( x,y\right ) \) are the real and imaginary parts of \(\arctan \left ( z\right ) \). Therefore\begin{align} \frac{i}{2}\ln \left ( \frac{i+z}{i-z}\right ) & =\frac{i}{2}\left ( \ln \left \vert \frac{i+z}{i-z}\right \vert +i\left ( \arg \left ( \frac{i+z}{i-z}\right ) +2n\pi \right ) \right ) \qquad n=0,\pm 1,\pm 2,\cdots \nonumber \\ & =\frac{i}{2}\ln \left \vert \frac{i+z}{i-z}\right \vert -\frac{1}{2}\left ( \arg \left ( \frac{i+z}{i-z}\right ) +2n\pi \right ) \tag{1} \end{align}
Where \(\arg \left ( \frac{i+z}{i-z}\right ) \) is the principal argument. But since\(\ z=x+iy\) then we see that\begin{align} \left \vert \frac{i+z}{i-z}\right \vert & =\left \vert \frac{i+\left ( x+iy\right ) }{i-\left ( x+iy\right ) }\right \vert \nonumber \\ & =\left \vert \frac{i+x+iy}{i-x-iy}\right \vert \nonumber \\ & =\left \vert \frac{x+i\left ( 1+y\right ) }{-x+i\left ( 1-y\right ) }\right \vert \nonumber \\ & =\frac{\sqrt{x^{2}+\left ( 1+y\right ) ^{2}}}{\sqrt{x^{2}+\left ( 1-y\right ) ^{2}}}\nonumber \\ & =\sqrt{\frac{x^{2}+\left ( 1+y\right ) ^{2}}{x^{2}+\left ( 1-y\right ) ^{2}}} \tag{2} \end{align}
And the principal argument is \begin{align*} \arg \left ( \frac{i+z}{i-z}\right ) & =\arg \left ( i+z\right ) -\arg \left ( i-z\right ) \\ & =\arg \left ( i\left ( 1-iz\right ) \right ) -\arg \left ( i\left ( 1+iz\right ) \right ) \\ & =\arg i+\arg \left ( 1-iz\right ) -\arg i+\arg \left ( 1+iz\right ) \\ & =\arg \left ( 1-iz\right ) +\arg \left ( 1+iz\right ) \end{align*}
Letting \(z=x+iy\) in the above results in\begin{align} \arg \left ( \frac{i+z}{i-z}\right ) & =\arg \left ( 1-i\left ( x+iy\right ) \right ) -\arg \left ( 1+i\left ( x+iy\right ) \right ) \nonumber \\ & =\arg \left ( 1-ix+y\right ) -\arg \left ( 1+ix-y\right ) \nonumber \\ & =\arg \left ( \left ( 1+y\right ) -ix\right ) -\arg \left ( \left ( 1-y\right ) +ix\right ) \nonumber \\ & =\arctan \left ( \frac{-x}{1+y}\right ) -\arctan \left ( \frac{x}{1-y}\right ) \tag{3} \end{align}
Substituting (2,3) into (1) gives\begin{align*} \frac{i}{2}\ln \left ( \frac{i+z}{i-z}\right ) & =\frac{i}{2}\left ( \ln \sqrt{\frac{x^{2}+\left ( 1+y\right ) ^{2}}{x^{2}+\left ( 1-y\right ) ^{2}}}+i\left ( \arctan \left ( \frac{-x}{1+y}\right ) -\arctan \left ( \frac{x}{1-y}\right ) +2n\pi \right ) \right ) \qquad n=0,\pm 1,\pm 2,\cdots \\ & =\frac{i}{4}\ln \left ( \frac{x^{2}+\left ( 1+y\right ) ^{2}}{x^{2}+\left ( 1-y\right ) ^{2}}\right ) -\frac{1}{2}\left ( \arctan \left ( \frac{-x}{1+y}\right ) -\arctan \left ( \frac{x}{1-y}\right ) +2n\pi \right ) \end{align*}
Setting the above equal to \(u+iv\) shows that the real part and the imaginary parts are\begin{align*} u & =-\frac{1}{2}\left ( \arctan \left ( \frac{-x}{1+y}\right ) -\arctan \left ( \frac{x}{1-y}\right ) +2n\pi \right ) \qquad n=0,\pm 1,\pm 2,\cdots \\ v & =\frac{1}{4}\ln \left ( \frac{x^{2}+\left ( y+1\right ) ^{2}}{x^{2}+\left ( 1-y\right ) ^{2}}\right ) \end{align*}
Therefore\begin{align*} \arctan \left ( z\right ) & =\frac{i}{2}\ln \left ( \frac{i+z}{i-z}\right ) \\ & =u+iv \end{align*}
Where \(u,v\) are given above. We see that \(\arctan \left ( z\right ) \) is multivalued as it depends on the value of \(n\).
For illustration of \(u\left ( x,y\right ) \) and \(v\left ( x,y\right ) \), the following is a plot of the above found solution showing the real part \(u\left ( x,y\right ) \) for \(n=0\) (principal sheet)
And the following shows \(u\left ( x,y\right ) \) with both \(n=0\) and \(n=1\) on the same plot showing two sheets
And the following plot shows the imaginary part \(v\left ( x,y\right ) \)
In the domain \(r>0,0<\theta <2\pi \). show that the function \(u=\ln r\) is harmonic and find its conjugate. Do this in both Cartesian and polar coordinates.
A function \(u\left ( x,y\right ) \) is harmonic if it satisfies the Laplace PDE \(u_{xx}+u_{yy}=0\). Since \[ r=\sqrt{x^{2}+y^{2}}\] Then\begin{align*} u & =\ln r\\ & =\ln \sqrt{x^{2}+y^{2}}\\ & =\frac{1}{2}\ln \left ( x^{2}+y^{2}\right ) \end{align*}
We now need to calculate \(u_{xx}\) and \(u_{yy}\). \begin{align*} u_{x} & =\frac{1}{2}\frac{\partial }{\partial x}\ln \left ( x^{2}+y^{2}\right ) \\ & =\frac{1}{2}\frac{2x}{x^{2}+y^{2}}\\ & =\frac{x}{x^{2}+y^{2}} \end{align*}
And\[ u_{xx}=\frac{\partial }{\partial x}\frac{x}{x^{2}+y^{2}}\] Applying the integration rule \(\frac{\partial }{\partial x}\frac{f\left ( x\right ) }{g\left ( x\right ) }=\frac{f^{\prime }g-fg}{g^{2}}\) to the above, where \(f=x\) and \(g=x^{2}+y^{2}\) results in\begin{align} u_{xx} & =\frac{x^{2}+y^{2}-x\left ( 2x\right ) }{\left ( x^{2}+y^{2}\right ) ^{2}}\nonumber \\ & =\frac{x^{2}+y^{2}-2x^{2}}{\left ( x^{2}+y^{2}\right ) ^{2}}\nonumber \\ & =\frac{y^{2}-x^{2}}{\left ( x^{2}+y^{2}\right ) ^{2}} \tag{1} \end{align}
Similarly\begin{align*} u_{y} & =\frac{1}{2}\frac{\partial }{\partial y}\ln \left ( x^{2}+y^{2}\right ) \\ & =\frac{1}{2}\frac{2y}{x^{2}+y^{2}}\\ & =\frac{y}{x^{2}+y^{2}} \end{align*}
Applying the integration rule \(\frac{\partial }{\partial y}\frac{f\left ( y\right ) }{g\left ( y\right ) }=\frac{f^{\prime }g-fg}{g^{2}}\) to the above, where \(f=y\) and \(g=x^{2}+y^{2}\) results in\begin{align} u_{yy} & =\frac{x^{2}+y^{2}-y\left ( 2y\right ) }{\left ( x^{2}+y^{2}\right ) ^{2}}\nonumber \\ & =\frac{x^{2}+y^{2}-2y^{2}}{\left ( x^{2}+y^{2}\right ) ^{2}}\nonumber \\ & =\frac{x^{2}-y^{2}}{\left ( x^{2}+y^{2}\right ) ^{2}} \tag{2} \end{align}
Now that we found \(u_{xx}\) and \(u_{yy}\), we need to verify that \(u_{xx}+u_{yy}=0\). Adding (1,2) gives\begin{align*} u_{xx}+u_{yy} & =\frac{y^{2}-x^{2}}{\left ( x^{2}+y^{2}\right ) ^{2}}+\frac{x^{2}-y^{2}}{\left ( x^{2}+y^{2}\right ) ^{2}}\\ & =\frac{y^{2}-x^{2}+x^{2}-y^{2}}{\left ( x^{2}+y^{2}\right ) ^{2}}\\ & =0 \end{align*}
Hence \(u=\ln r\) is harmonic.
To find its conjugate. Let the conjugate be \(v\left ( x,y\right ) \). Let \(u\) be the real part of analytic function \[ f=u+iv \] Applying Cauchy Riemann equations to \(f\) results in\begin{align} \frac{\partial u}{\partial x} & =\frac{\partial v}{\partial y}\tag{3}\\ \frac{\partial u}{\partial y} & =-\frac{\partial v}{\partial x} \tag{4} \end{align}
From (3) and using the earlier result found for \(u_{x}\) gives\[ \frac{\partial v}{\partial y}=\frac{x}{x^{2}+y^{2}}\] Integrating the above w.r.t. \(y\) gives\begin{align} v & =\int \frac{x}{x^{2}+y^{2}}dy+\Phi \left ( x\right ) \nonumber \\ & =x\int \frac{1}{x^{2}+y^{2}}dy+\Phi \left ( x\right ) \nonumber \\ & =\frac{1}{x}\int \frac{1}{1+\left ( \frac{y}{x}\right ) ^{2}}dy+\Phi \left ( x\right ) \nonumber \end{align}
The above is integrated using substitution. Let \(u=\frac{y}{x}\), then \(\frac{du}{dy}=\frac{1}{x}\) and the integral becomes\begin{align*} v & =\frac{1}{x}\left ( \int \frac{1}{1+u^{2}}\left ( xdu\right ) \right ) +\Phi \left ( x\right ) \\ & =\int \frac{1}{1+u^{2}}du+\Phi \left ( x\right ) \end{align*}
But \(\int \frac{1}{1+u^{2}}du=\arctan \left ( u\right ) =\arctan \left ( \frac{y}{x}\right ) \), therefore the above becomes\begin{equation} v=\arctan \left ( \frac{y}{x}\right ) +\Phi \left ( x\right ) \tag{5} \end{equation} Taking derivative of (5) w.r.t. \(x\) gives an ODE to solve for \(\Phi \left ( x\right ) \) \begin{equation} \frac{\partial v}{\partial x}=\frac{d}{dx}\left ( \arctan \left ( \frac{y}{x}\right ) \right ) +\Phi ^{\prime }\left ( x\right ) \tag{5A} \end{equation} To find \(\frac{d}{dx}\arctan \left ( \frac{y}{x}\right ) \), let\[ w=\arctan \left ( \frac{y}{x}\right ) \] Now the goal is to find \(\frac{dw}{dx}\). The above is the same as \begin{equation} \tan \left ( w\right ) =\frac{y}{x} \tag{6} \end{equation} Taking derivative of both sides of the above w.r.t. \(x\) gives\[ \frac{d}{dx}\tan \left ( w\right ) =-\frac{y}{x^{2}}\] But \(\frac{d}{dx}\tan \left ( w\right ) =\sec ^{2}\left ( w\right ) \frac{dw}{dx}\), and the above can be written as \begin{align} \sec ^{2}\left ( w\right ) \frac{dw}{dx} & =-\frac{y}{x^{2}}\nonumber \\ \frac{dw}{dx} & =-\frac{y}{x^{2}}\frac{1}{\sec ^{2}\left ( w\right ) } \tag{7} \end{align}
But \(\sec ^{2}\left ( w\right ) =\frac{1}{\cos ^{2}w}\) and \(\cos ^{2}w+\sin ^{2}w=1\). Therefore dividing by \(\cos ^{2}w\) gives \(1+\frac{\sin ^{2}w}{\cos ^{2}w}=\sec ^{2}\left ( w\right ) \) or \(1+\tan ^{2}w=\sec ^{2}\left ( w\right ) \). But from (6) we know that \(\tan \left ( w\right ) =\frac{y}{x}\), therefore \(1+\left ( \frac{y}{x}\right ) ^{2}=\sec ^{2}\left ( w\right ) \). Replacing this expression for \(\sec ^{2}\left ( w\right ) \) in (7) gives\begin{align*} \frac{dw}{dx} & =-\frac{y}{x^{2}}\frac{1}{1+\left ( \frac{y}{x}\right ) ^{2}}\\ & =-\frac{y}{x^{2}}\frac{x^{2}}{x^{2}+y^{2}}\\ & =\frac{-y}{x^{2}+y^{2}} \end{align*}
Now that we found \(\frac{dw}{dx}\) which is \(\frac{d}{dx}\arctan \left ( \frac{y}{x}\right ) \), then 5A becomes\[ \frac{\partial v}{\partial x}=\frac{-y}{x^{2}+y^{2}}+\Phi ^{\prime }\left ( x\right ) \] But from Cauchy Riemann equation (4) above, we know that \(\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}\), therefore the above is the same as\[ \frac{\partial u}{\partial y}=-\left ( \frac{-y}{x^{2}+y^{2}}+\Phi ^{\prime }\left ( x\right ) \right ) \] We know what \(\frac{\partial u}{\partial y}\) is. We found this earlier which is \(\frac{\partial u}{\partial y}=\frac{y}{x^{2}+y^{2}}\). Hence the above equation becomes\begin{align*} \frac{y}{x^{2}+y^{2}} & =\frac{y}{x^{2}+y^{2}}-\Phi ^{\prime }\left ( x\right ) \\ \Phi ^{\prime }\left ( x\right ) & =0 \end{align*}
Therefore \(\Phi \) is constant, say \(C_{1}\). Equation (5) becomes\begin{equation} \fbox{$v\left ( x,y\right ) =\arctan \left ( \frac{y}{x}\right ) +C_1$} \tag{8} \end{equation} Which is the conjugate of \(u=\frac{1}{2}\ln \left ( x^{2}+y^{2}\right ) \). To verify the result in (8), we now check that \(v\left ( x,y\right ) \) is indeed harmonic by checking that it satisfies the Laplace PDE.\begin{align*} v_{x} & =\frac{-y}{x^{2}+y^{2}}\\ v_{xx} & =\frac{y\left ( 2x\right ) }{\left ( x^{2}+y^{2}\right ) ^{2}} \end{align*}
And\begin{align*} v_{y} & =\frac{x}{x^{2}+y^{2}}\\ v_{yy} & =\frac{-x\left ( 2y\right ) }{\left ( x^{2}+y^{2}\right ) ^{2}} \end{align*}
Using the above we see that \begin{align*} v_{xx}+v_{yy} & =\frac{y\left ( 2x\right ) }{\left ( x^{2}+y^{2}\right ) ^{2}}-\frac{x\left ( 2y\right ) }{\left ( x^{2}+y^{2}\right ) ^{2}}\\ & =0 \end{align*}
This shows that \(v\left ( x,y\right ) \) obtained above is harmonic. It is the conjugate of \(u\left ( x,y\right ) \,\).
\(v\left ( x,y\right ) \) is not a unique conjugate of \(u\left ( x,y\right ) \), since the constant \(C_{1}\) is arbitrary.
Here \(z=re^{i\theta }\) and we are told that \(u\left ( r,\theta \right ) =\ln r\). To show this is harmonic in polar coordinates, we need to show it satisfies Laplacian in polar coordinates, which is\[ u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\theta \theta }=0 \] But \(u_{r}=\frac{d}{dr}\ln r=\frac{1}{r}\) and \(u_{rr}=-\frac{1}{r^{2}}\) and \(u_{\theta \theta }=0\). Substituting these into the above gives\begin{align*} -\frac{1}{r^{2}}+\frac{1}{r}\frac{1}{r} & =0\\ 0 & =0 \end{align*}
Therefore \(u=\ln r\) is harmonic since it satisfies the Laplacian in polar coordinates. To find its conjugate, we use C-R in polar coordinates, and these are given by\begin{align} \frac{\partial u}{\partial r} & =\frac{1}{r}\frac{\partial v}{\partial \theta }\tag{1}\\ \frac{\partial u}{\partial \theta } & =-r\frac{\partial v}{\partial r} \tag{2} \end{align}
From (1), and since we know that \(\frac{\partial u}{\partial r}=\frac{1}{r}\), then this gives\begin{align*} \frac{1}{r} & =\frac{1}{r}\frac{\partial v}{\partial \theta }\\ \frac{\partial v}{\partial \theta } & =1 \end{align*}
Or by integration w.r.t. \(\theta \)\[ v=\theta +\Phi \left ( r\right ) \] Where \(\Phi \left ( r\right ) \) is the constant of integration (a function). Taking derivative of the above w.r.t. \(r\) gives\[ \frac{\partial v}{\partial r}=\Phi ^{\prime }\left ( r\right ) \] But from (2) \(\frac{\partial v}{\partial r}=-\frac{1}{r}\frac{\partial u}{\partial \theta }=0\). (Because \(u\) does not depend on \(\theta \)). Hence the above results in \(\Phi ^{\prime }\left ( r\right ) =0\) or \(\Phi =C_{1}\) a constant. Therefore the conjugate harmonic function is\[ \fbox{$v\left ( r,\theta \right ) =\theta +C_1$}\] Now we verify this satisfies Laplacian in Polar. From \[ v_{rr}+\frac{1}{r}v_{r}+\frac{1}{r^{2}}v_{\theta \theta }=0 \] We see since \(v_{r}=0\) and \(v_{rr}=0\) and \(v_{\theta }=1\) and \(v_{\theta \theta }=0\), therefore we obtain \(0=0\) also. Hence \(v=\theta +C_{1}\) satisfies the Laplacian.
Find the value of \(\int _{C}f\left ( z\right ) dz\) where \(f\left ( z\right ) =e^{z}\) for two different contours. \(C_{1}\) is straight line from the origin to the point \(\left ( 2,1\right ) \). \(C_{2}\) is a straight line from the origin to the point \(\left ( 2,0\right ) \) followed by another straight line from \(\left ( 2,0\right ) \) to \(\left ( 2,1\right ) \)
Solution
Using contour \(C_{1}\). The line starts from \(\left ( x_{0},y_{0}\right ) =\left ( 0,0\right ) \) and ends at \(\left ( x_{1},y_{1}\right ) =\left ( 2,1\right ) \). Hence the parametrization for this line is given by\begin{align*} x\left ( t\right ) & =\left ( 1-t\right ) x_{0}+tx_{1}\\ & =2t \end{align*}
And\begin{align*} y\left ( t\right ) & =\left ( 1-t\right ) y_{0}+ty_{1}\\ & =t \end{align*}
Now \(f\left ( z\right ) =e^{z}=e^{x+iy}\), Therefore in terms of \(t\) this becomes \begin{align*} f\left ( t\right ) & =e^{2t+it}\\ & =e^{t\left ( 2+i\right ) } \end{align*}
Hence \begin{align*} \int _{C_{1}}f\left ( z\right ) dz & =\int _{t=0}^{t=1}f\left ( t\right ) z^{\prime }\left ( t\right ) dt\\ & =\int _{0}^{1}e^{t\left ( 2+i\right ) }z^{\prime }\left ( t\right ) dt \end{align*}
But \(z\left ( t\right ) =x\left ( t\right ) +iy\left ( t\right ) =2t+it\), hence \(z^{\prime }\left ( t\right ) =2+i\) and the above becomes \begin{align*} \int _{C_{1}}f\left ( z\right ) dz & =\int _{0}^{1}e^{t\left ( 2+i\right ) }\left ( 2+i\right ) dt\\ & =\left ( 2+i\right ) \int _{0}^{1}e^{t\left ( 2+i\right ) }dt\\ & =\left ( 2+i\right ) \left ( \frac{e^{t\left ( 2+i\right ) }}{\left ( 2+i\right ) }\right ) _{0}^{1}\\ & =\left ( e^{t\left ( 2+i\right ) }\right ) _{0}^{1} \end{align*}
Hence the final result is\[ \int _{C_{1}}f\left ( z\right ) dz=e^{2+i}-1 \]
Using \(C_{2}\). The first line starts from \(\left ( x_{0},y_{0}\right ) =\left ( 0,0\right ) \) and ends at \(\left ( x_{1},y_{1}\right ) =\left ( 2,0\right ) \). Hence the parametrization for this line is given by \begin{align*} x\left ( t\right ) & =\left ( 1-t\right ) x_{0}+tx_{1}\\ & =2t \end{align*}
And\begin{align*} y\left ( t\right ) & =\left ( 1-t\right ) y_{0}+ty_{1}\\ & =0 \end{align*}
Now \(f\left ( z\right ) =e^{z}=e^{x+iy}\), Therefore in terms of \(t\) the function \(f\left ( z\right ) \) becomes \[ f\left ( t\right ) =e^{2t}\] Hence, for the line from \(\left ( 0,0\right ) \,\) to \(\left ( 2,0\right ) \) we have\begin{align*} \int _{C_{2_{1}}}f\left ( z\right ) dz & =\int _{t=0}^{t=1}f\left ( t\right ) z^{\prime }\left ( t\right ) dt\\ & =\int _{0}^{1}e^{2t}z^{\prime }\left ( t\right ) dt \end{align*}
But \(z=x+iy=2t\) since \(y\left ( t\right ) =0\). hence \(z^{\prime }\left ( t\right ) =2\) and the above becomes \begin{align} \int _{C_{2_{1}}}f\left ( z\right ) dz & =2\int _{0}^{1}e^{2t}dt\nonumber \\ & =2\left ( \frac{e^{2t}}{2}\right ) _{0}^{1}\nonumber \\ & =e^{2}-1 \tag{1} \end{align}
The second line starts from \(\left ( x_{0},y_{0}\right ) =\left ( 2,0\right ) \) and ends at \(\left ( x_{1},y_{1}\right ) =\left ( 2,1\right ) \). Hence the parametrization for this line is given by \begin{align*} x\left ( t\right ) & =\left ( 1-t\right ) x_{0}+tx_{1}\\ & =\left ( 1-t\right ) 2+2t\\ & =2 \end{align*}
And\begin{align*} y\left ( t\right ) & =\left ( 1-t\right ) y_{0}+ty_{1}\\ & =t \end{align*}
Now \(f\left ( z\right ) =e^{z}=e^{x+iy}\), Therefore in terms of \(t\) this becomes \[ f\left ( t\right ) =e^{2+it}\] Hence, for the line from \(\left ( 2,0\right ) \,\) to \(\left ( 2,1\right ) \) we have\begin{align*} \int _{C_{2_{2}}}f\left ( z\right ) dz & =\int _{t=0}^{t=1}f\left ( t\right ) z^{\prime }\left ( t\right ) dt\\ & =\int _{0}^{1}e^{2+it}z^{\prime }\left ( t\right ) dt \end{align*}
But \(z=x+iy=2+it\). hence \(z^{\prime }\left ( t\right ) =i\) and the above becomes \begin{align} \int _{C_{2_{2}}}f\left ( z\right ) dz & =\int _{0}^{1}ie^{2+it}dt\nonumber \\ & =i\left ( \frac{e^{2+it}}{i}\right ) _{0}^{1}\nonumber \\ & =\left ( e^{2+it}\right ) _{0}^{1}\nonumber \\ & =e^{2+i}-e^{2} \tag{2} \end{align}
Therefore the total is the sum of (1) and (2)\begin{equation} \int _{C_{2}}f\left ( z\right ) dz=e^{2}-1+e^{2+i}-e^{2}\nonumber \end{equation} Hence the final result is \begin{equation} \int _{C_{2}}f\left ( z\right ) dz=e^{2+i}-1 \tag{3} \end{equation} To verify this, since \(e^{z}\) is analytic then \(\int _{C_{2}}f\left ( z\right ) dz-\int _{C_{1}}f\left ( z\right ) dz\) should come out to be zero (By Cauchy theorem). This is because \({\displaystyle \oint } f\left ( z\right ) dz=0\) around the closed contour, going clockwise. Let us see if this is true:\begin{align*} \int _{C_{2}}f\left ( z\right ) dz-\int _{C_{1}}f\left ( z\right ) dz & =\left [ e^{2+i}-1\right ] -\left [ e^{2+i}-1\right ] \\ & =0\\ & ={\displaystyle \oint } f\left ( z\right ) dz \end{align*}
Verified. A small note: \({\displaystyle \oint _{C}} f\left ( z\right ) dz=0\) does not necessarily mean that \(f\left ( z\right ) \) is analytic on and inside \(C\) as some non analytic function can give zero, depending on \(C\). But if \(f\left ( z\right ) \) happened to be analytic, then \({\displaystyle \oint _{C}} f\left ( z\right ) dz\) is always zero. But here we now that \(e^{az}\) is analytic.