Use Cauchy-Riemann equations to determine if \(\left \vert z\right \vert \) analytic function of the complex variable \(z.\)
Solution\[ f\left ( z\right ) =\left \vert z\right \vert \] Let \(z=x+iy\), then\begin{align*} f\left ( z\right ) & =\left ( x^{2}+y^{2}\right ) ^{\frac{1}{2}}\\ & =u+iv \end{align*}
Hence \begin{align*} u & =\sqrt{x^{2}+y^{2}}\\ v & =0 \end{align*}
Cauchy-Riemann equations are\begin{align} \frac{\partial u}{\partial x} & =\frac{\partial v}{\partial y}\tag{1}\\ -\frac{\partial u}{\partial y} & =\frac{\partial v}{\partial x} \tag{2} \end{align}
First equation above gives \(\frac{\partial v}{\partial y}=0\) and \(\frac{\partial u}{\partial x}=\frac{1}{2}\frac{2x}{\sqrt{x^{2}+y^{2}}}\), which shows that \(\frac{\partial v}{\partial y}\neq \frac{\partial u}{\partial x}\). Therefore \(\left \vert z\right \vert \) is not analytic.
Use Cauchy-Riemann equations to determine if \(\operatorname{Re}\left ( z\right ) \) analytic function of the complex variable \(z.\)
Solution\[ f\left ( z\right ) =\operatorname{Re}\left ( z\right ) \] Let \(z=x+iy\), then\begin{align*} f\left ( z\right ) & =x\\ & =u+iv \end{align*}
Hence \begin{align*} u & =x\\ v & =0 \end{align*}
Cauchy-Riemann equations are\begin{align} \frac{\partial u}{\partial x} & =\frac{\partial v}{\partial y}\tag{1}\\ -\frac{\partial u}{\partial y} & =\frac{\partial v}{\partial x} \tag{2} \end{align}
First equation above gives \(\frac{\partial v}{\partial y}=0\) and \(\frac{\partial u}{\partial x}=1\), which shows that \(\frac{\partial v}{\partial y}\neq \frac{\partial u}{\partial x}\). Therefore \(\operatorname{Re}\left ( z\right ) \) is not analytic.
Use Cauchy-Riemann equations to determine if \(e^{\sin z}\) analytic function of the complex variable \(z.\)
Solution
\(f\left ( z\right ) =e^{\sin z}\) is analytic since we can show that \(\exp \left ( z\right ) \) is analytic by applying Cauchy-Riemann (C-R), and also show that \(\sin \left ( z\right ) \) is analytic using C-R. Theory of analytic functions it says that the composition of analytic functions is also an analytic function, which means \(e^{\sin z}\) is analytic.
But this problems seems to ask to use C-R equations directly to show this. Therefore we need to first determine the real and complex parts \(\left ( u,v\right ) \) of the function \(e^{\sin z}\). Since \[ \sin z=\frac{z-z^{-1}}{2i}\] Then \begin{align*} f\left ( z\right ) & =e^{\sin z}\\ & =\exp \left ( \frac{z-z^{-1}}{2i}\right ) \\ & =\exp \left ( \frac{z}{2i}\right ) \exp \left ( \frac{-1}{2iz}\right ) \end{align*}
But \(z=x+iy\) and the above expands to\begin{align*} \exp \left ( \sin z\right ) & =\exp \left ( \frac{1}{2i}\left ( x+iy\right ) \right ) \exp \left ( \frac{-1}{2i\left ( x+iy\right ) }\right ) \\ & =\exp \left ( \frac{-i}{2}x+\frac{1}{2}y\right ) \exp \left ( \frac{i}{2}\frac{1}{\left ( x+iy\right ) }\right ) \\ & =\exp \left ( \frac{-i}{2}x+\frac{1}{2}y\right ) \exp \left ( \frac{i}{2}\frac{x-iy}{\left ( x+iy\right ) \left ( x-iy\right ) }\right ) \\ & =\exp \left ( \frac{-i}{2}x+\frac{1}{2}y\right ) \exp \left ( \frac{i}{2}\frac{x-iy}{x^{2}+y^{2}}\right ) \\ & =\exp \left ( \frac{-i}{2}x+\frac{1}{2}y\right ) \exp \left ( \frac{i}{2}\left ( \frac{x}{x^{2}+y^{2}}-\frac{iy}{x^{2}+y^{2}}\right ) \right ) \\ & =\exp \left ( \frac{-i}{2}x+\frac{1}{2}y\right ) \exp \left ( \frac{i}{2}\frac{x}{x^{2}+y^{2}}+\frac{1}{2}\frac{y}{x^{2}+y^{2}}\right ) \\ & =\exp \left ( \frac{-i}{2}x\right ) \exp \left ( \frac{1}{2}y\right ) \exp \left ( \frac{i}{2}\frac{x}{x^{2}+y^{2}}\right ) \exp \left ( \frac{1}{2}\frac{y}{x^{2}+y^{2}}\right ) \end{align*}
Collecting terms gives\begin{align*} \exp \left ( \sin z\right ) & =\exp \left ( \frac{1}{2}y+\frac{1}{2}\frac{y}{x^{2}+y^{2}}\right ) \exp \left ( \frac{i}{2}\frac{x}{x^{2}+y^{2}}-\frac{i}{2}x\right ) \\ & =\exp \left ( \frac{1}{2}\frac{y\left ( 1+\left ( x^{2}+y^{2}\right ) \right ) }{x^{2}+y^{2}}\right ) \exp \left ( \frac{i}{2}\frac{x}{x^{2}+y^{2}}-\frac{i}{2\left ( x^{2}+y^{2}\right ) }x\left ( x^{2}+y^{2}\right ) \right ) \\ & =\exp \left ( \frac{1}{2}\frac{y\left ( 1+\left ( x^{2}+y^{2}\right ) \right ) }{x^{2}+y^{2}}\right ) \exp \left ( i\frac{1}{2}\frac{x\left ( 1-\left ( x^{2}+y^{2}\right ) \right ) }{x^{2}+y^{2}}\right ) \\ & =\exp \left ( \frac{1}{2}\frac{y\left ( 1+\left ( x^{2}+y^{2}\right ) \right ) }{x^{2}+y^{2}}\right ) \left [ \cos \left ( \frac{1}{2}\frac{x\left ( 1-\left ( x^{2}+y^{2}\right ) \right ) }{x^{2}+y^{2}}\right ) +i\sin \left ( \frac{1}{2}\frac{x\left ( 1-\left ( x^{2}+y^{2}\right ) \right ) }{x^{2}+y^{2}}\right ) \right ] \\ & =\exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \cos \left ( \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) +i\exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \sin \left ( \frac{1}{2}\frac{x-x\left ( x^{2}+y^{2}\right ) }{\left ( x^{2}+y^{2}\right ) }\right ) \end{align*}
Therefore, since \(\exp \left ( \sin z\right ) =u+iv\), then we see from above that \begin{align*} u & =\exp \left ( \frac{1}{2}\frac{y+y\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \cos \left ( \frac{1}{2}\frac{x-x\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \\ v & =\exp \left ( \frac{1}{2}\frac{y+y\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \sin \left ( \frac{1}{2}\frac{x-x\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \end{align*}
Now we need to check the Cauchy-Riemann equations on the above \(u,v\) functions we found.\begin{align} \frac{\partial u}{\partial x} & =\frac{\partial v}{\partial y}\tag{1}\\ -\frac{\partial u}{\partial y} & =\frac{\partial v}{\partial x} \tag{2} \end{align}
Evaluating each partial derivative gives\begin{align*} \frac{\partial u}{\partial x} & =\frac{d}{dx}\left ( \frac{1}{2}\frac{y+y\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \exp \left ( \frac{1}{2}\frac{y+y\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \cos \left ( \frac{1}{2}\frac{x-x\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \\ & +\exp \left ( \frac{1}{2}\frac{y+y\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \frac{d}{dx}\cos \left ( \frac{1}{2}\frac{x-x\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \\ & =\frac{1}{2}\frac{2yx\left ( x^{2}+y^{2}\right ) -\left ( y+y\left ( x^{2}+y^{2}\right ) \right ) 2x}{\left ( x^{2}+y^{2}\right ) ^{2}}\exp \left ( \frac{1}{2}\frac{y\left ( \left ( x^{2}+y^{2}\right ) +1\right ) }{x^{2}+y^{2}}\right ) \cos \left ( \frac{1}{2}\frac{x\left ( 1-\left ( x^{2}+y^{2}\right ) \right ) }{x^{2}+y^{2}}\right ) \\ & -\exp \left ( \frac{1}{2}\frac{y+y\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \sin \left ( \frac{1}{2}\frac{x-x\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \frac{d}{dx}\left ( \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \\ & =\frac{-xy}{\left ( x^{2}+y^{2}\right ) ^{2}}\exp \left ( \frac{1}{2}\frac{y\left ( \left ( x^{2}+y^{2}\right ) +1\right ) }{x^{2}+y^{2}}\right ) \cos \left ( \frac{x\left ( 1-\left ( x^{2}+y^{2}\right ) \right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \\ & -\exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \sin \left ( \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \left ( \frac{\left ( 1-3x^{2}-y^{2}\right ) \left ( x^{2}+y^{2}\right ) -\left ( x-x\left ( x^{2}+y^{2}\right ) \right ) 2x}{2\left ( x^{2}+y^{2}\right ) ^{2}}\right ) \\ & =\frac{-xy}{\left ( x^{2}+y^{2}\right ) ^{2}}\exp \left ( \frac{1}{2}\frac{y\left ( x^{2}+y^{2}+1\right ) }{x^{2}+y^{2}}\right ) \cos \left ( \frac{1}{2}\frac{x\left ( 1-\left ( x^{2}+y^{2}\right ) \right ) }{x^{2}+y^{2}}\right ) \\ & -\exp \left ( \frac{1}{2}\frac{y\left ( x^{2}+y^{2}+1\right ) }{x^{2}+y^{2}}\right ) \sin \left ( \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \left ( \frac{\left ( -x^{4}-2x^{2}y^{2}-x^{2}-y^{4}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) ^{2}}\right ) \end{align*}
The above can be simplified more to become \begin{multline} \frac{\partial u}{\partial x}=\frac{-1}{2\left ( x^{2}+y^{2}\right ) ^{2}}\exp \left ( \frac{y\left ( x^{2}+y^{2}+1\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \nonumber \\ \left [ 2xy\cos \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }+\left ( -x^{4}-2x^{2}y^{2}-x^{2}-y^{4}+y^{2}\right ) \sin \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ] \tag{3} \end{multline} Now we evaluate \(\frac{\partial v}{\partial y}\) to see if it the same as above. Since \(v=\exp \left ( \frac{1}{2}\frac{y+y\left ( x^{2}+y^{2}\right ) }{\left ( x^{2}+y^{2}\right ) }\right ) \sin \left ( \frac{1}{2}\frac{x-x\left ( x^{2}+y^{2}\right ) }{\left ( x^{2}+y^{2}\right ) }\right ) \) then\begin{align*} \frac{\partial v}{\partial y} & =\frac{d}{dy}\left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \sin \left ( \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \\ & +\exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \cos \left ( \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \frac{d}{dy}\left ( \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \\ & =\left ( \frac{1}{2}\frac{\left ( 1+x^{2}+3y^{2}\right ) \left ( x^{2}+y^{2}\right ) -\left ( y+y\left ( x^{2}+y^{2}\right ) \right ) 2y}{\left ( x^{2}+y^{2}\right ) ^{2}}\right ) \exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \sin \left ( \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \\ & +\exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \cos \left ( \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \left ( \frac{1}{2}\frac{\left ( -2xy\right ) \left ( x^{2}+y^{2}\right ) -\left ( x-x\left ( x^{2}+y^{2}\right ) \right ) \left ( 2y\right ) }{\left ( x^{2}+y^{2}\right ) ^{2}}\right ) \\ & =\left ( \frac{1}{2}\frac{x^{4}+2x^{2}y^{2}+x^{2}+y^{4}-y^{2}}{\left ( x^{2}+y^{2}\right ) ^{2}}\right ) \exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \sin \left ( \frac{1}{2}\frac{x-x\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \\ & +\exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \cos \left ( \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \left ( \frac{1}{2}\frac{-2xy}{\left ( x^{2}+y^{2}\right ) ^{2}}\right ) \\ & =\left ( \frac{1}{2}\frac{x^{4}+2x^{2}y^{2}+x^{2}+y^{4}-y^{2}}{\left ( x^{2}+y^{2}\right ) ^{2}}\right ) \exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \sin \left ( \frac{1}{2}\frac{x-x\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \\ & -\exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \cos \left ( \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \left ( \frac{xy}{\left ( x^{2}+y^{2}\right ) ^{2}}\right ) \end{align*}
Simplifying the above more gives \begin{multline} \frac{\partial v}{\partial y}=\frac{-1}{2\left ( x^{2}+y^{2}\right ) ^{2}}\exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \nonumber \\ \left [ 2xy\cos \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }+\left ( -x^{4}-2x^{2}y^{2}-x^{2}-y^{4}+y^{2}\right ) \sin \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ] \tag{4} \end{multline} Comparing (3) and (4) shows they are the same expressions. Therefore the first equation is verified. \[ \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\] Now we verify the second equation \(-\frac{\partial u}{\partial y}=\frac{\partial v}{\partial x}\). Since \(u=\exp \left ( \frac{1}{2}\frac{y+y\left ( x^{2}+y^{2}\right ) }{\left ( x^{2}+y^{2}\right ) }\right ) \cos \left ( \frac{1}{2}\frac{x-x\left ( x^{2}+y^{2}\right ) }{\left ( x^{2}+y^{2}\right ) }\right ) \) then\begin{align*} \frac{\partial u}{\partial y} & =\frac{d}{dy}\left ( \frac{1}{2}\frac{y+y\left ( x^{2}+y^{2}\right ) }{\left ( x^{2}+y^{2}\right ) }\right ) \exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \cos \left ( \frac{1}{2}\frac{x-x\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \\ & -\exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \sin \left ( \frac{1}{2}\frac{x-x\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \frac{d}{dy}\left ( \frac{1}{2}\frac{x-x\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \\ & =\frac{\left ( 1+x^{2}+3y^{2}\right ) \left ( x^{2}+y^{2}\right ) -\left ( y+y\left ( x^{2}+y^{2}\right ) \right ) 2y}{2\left ( x^{2}+y^{2}\right ) ^{2}}\exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \cos \left ( \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \\ & -\exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \sin \left ( \frac{1}{2}\frac{x-x\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \frac{\left ( -2y\right ) \left ( x^{2}+y^{2}\right ) -\left ( x-x\left ( x^{2}+y^{2}\right ) \right ) 2y}{2\left ( x^{2}+y^{2}\right ) ^{2}}\\ & =\frac{\left ( x^{4}+2x^{2}y^{2}+x^{2}+y^{4}-y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) ^{2}}\exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \cos \left ( \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \\ & -\exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \sin \left ( \frac{1}{2}\frac{x-x\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \frac{\left ( -2y\right ) \left ( x^{2}+y^{2}\right ) -2yx+2yx\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) ^{2}}\\ & =\frac{\left ( x^{4}+2x^{2}y^{2}+x^{2}+y^{4}-y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) ^{2}}\exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \cos \left ( \frac{1}{2}\frac{x-x\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \\ & +\exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \sin \left ( \frac{1}{2}\frac{x-x\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \frac{yx}{\left ( x^{2}+y^{2}\right ) ^{2}} \end{align*}
The above can simplified more to give \begin{multline*} \frac{\partial u}{\partial y}=\frac{1}{2\left ( x^{2}+y^{2}\right ) ^{2}}\exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \\ \left [ \left ( x^{4}+2x^{2}y^{2}+x^{2}+y^{4}-y^{2}\right ) \cos \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }+2xy\sin \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ] \end{multline*} Hence\begin{multline} \frac{-\partial u}{\partial y}=\frac{1}{2\left ( x^{2}+y^{2}\right ) ^{2}}\exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \nonumber \\ \left [ -\left ( x^{4}+2x^{2}y^{2}+x^{2}+y^{4}-y^{2}\right ) \cos \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }-2xy\sin \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ] \tag{5} \end{multline} And since \(v=\exp \left ( \frac{1}{2}\frac{y+y\left ( x^{2}+y^{2}\right ) }{\left ( x^{2}+y^{2}\right ) }\right ) \sin \left ( \frac{1}{2}\frac{x-x\left ( x^{2}+y^{2}\right ) }{\left ( x^{2}+y^{2}\right ) }\right ) \) then\begin{align*} \frac{\partial v}{\partial x} & =\frac{d}{dx}\left ( \frac{1}{2}\frac{y+y\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \sin \left ( \frac{1}{2}\frac{x-x\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \\ & +\exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \cos \left ( \frac{1}{2}\frac{x-x\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \frac{d}{dx}\left ( \frac{1}{2}\frac{x-x\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \\ & =\frac{1}{2}\left ( \frac{2xy\left ( x^{2}+y^{2}\right ) -\left ( y+y\left ( x^{2}+y^{2}\right ) \right ) 2x}{\left ( x^{2}+y^{2}\right ) ^{2}}\right ) \exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \sin \left ( \frac{1}{2}\frac{x-x\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \\ & +\exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \cos \left ( \frac{1}{2}\frac{x-x\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \left ( \frac{\left ( 1-3x^{2}-y^{2}\right ) \left ( x^{2}+y^{2}\right ) -\left ( x-x\left ( x^{2}+y^{2}\right ) \right ) 2x}{2\left ( x^{2}+y^{2}\right ) ^{2}}\right ) \\ & =\frac{-xy}{\left ( x^{2}+y^{2}\right ) ^{2}}\exp \left ( \frac{1}{2}\frac{y+y\left ( x^{2}+y^{2}\right ) }{x^{2}+y^{2}}\right ) \sin \left ( \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \\ & +\exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \cos \left ( \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \left ( \frac{-x^{4}-2x^{2}y^{2}-x^{2}-y^{4}+y^{2}}{2\left ( x^{2}+y^{2}\right ) ^{2}}\right ) \end{align*}
The above can simplified more to give \begin{multline} \frac{\partial v}{\partial x}=\frac{1}{2\left ( x^{2}+y^{2}\right ) ^{2}}\exp \left ( \frac{y+y\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ) \nonumber \\ \left [ -\left ( x^{4}+2x^{2}y^{2}+x^{2}+y^{4}-y^{2}\right ) \cos \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }-2xy\sin \frac{x-x\left ( x^{2}+y^{2}\right ) }{2\left ( x^{2}+y^{2}\right ) }\right ] \tag{6} \end{multline} Comparing (5,6) shows they are the same, i.e. \[ \frac{-\partial u}{\partial y}=\frac{\partial v}{\partial x}\] C-R equations are satisfied, and because it is clear that all partial derivatives \(\frac{\partial v}{\partial x},\frac{\partial v}{\partial y},\frac{\partial u}{\partial x},\frac{\partial u}{\partial y}\) are continuous functions in \(x,y\) as they are made up of exponential and trigonometric functions which are continuous, then we conclude that \(f\left ( z\right ) =e^{\sin z}\) is analytic function everywhere.
Represent \(\frac{z+3}{z-3}\) by its Maclaurin series and give the region of validity for the representation. Next expand this in powers of \(\frac{1}{z}\) to find a Laurent series. What is the range of validity of the Laurent series?
Solution
Maclaurin series is expansion of \(f\left ( z\right ) \) around \(z=0\). Since \(f\left ( z\right ) \) has simple pole at \(z=3\), then the region of validity will be a disk centered at \(z=0\) up to the nearest pole, which is at \(z=3.\) Hence \(\left \vert z\right \vert <3\) is the region. \begin{align*} f\left ( z\right ) & =\frac{z+3}{z-3}\\ & =\frac{z+3}{-3\left ( 1-\frac{z}{3}\right ) }\\ & =\frac{z+3}{-3}\left ( \frac{1}{1-\frac{z}{3}}\right ) \end{align*}
Now we can expand using Binomial to obtain\begin{align*} f\left ( z\right ) & =\frac{3+z}{-3}\left ( 1+\frac{z}{3}+\left ( \frac{z}{3}\right ) ^{2}+\left ( \frac{z}{3}\right ) ^{3}+\cdots \right ) \\ & =\left ( -1-\frac{z}{3}\right ) \left ( 1+\frac{z}{3}+\left ( \frac{z}{3}\right ) ^{2}+\left ( \frac{z}{3}\right ) ^{3}+\cdots \right ) \\ & =\left ( -1-\frac{z}{3}\right ) +\left ( -1-\frac{z}{3}\right ) \left ( \frac{z}{3}\right ) +\left ( -1-\frac{z}{3}\right ) \left ( \frac{z}{3}\right ) ^{2}+\left ( -1-\frac{z}{3}\right ) \left ( \frac{z}{3}\right ) ^{3}+\cdots \\ & =-1-\frac{z}{3}-\frac{z}{3}-\left ( \frac{z}{3}\right ) ^{2}-\left ( \frac{z}{3}\right ) ^{2}-\left ( \frac{z}{3}\right ) ^{3}-\left ( \frac{z}{3}\right ) ^{3}-\left ( \frac{z}{3}\right ) ^{4}+\cdots \\ & =-1-\frac{2}{3}z-\frac{2}{9}z^{2}-\frac{2}{27}z^{3}-\frac{2}{81}z^{4}-\cdots \end{align*}
Or \[ f\left ( z\right ) =-1-\sum _{n=1}^{\infty }\frac{2}{3^{n}}z^{n}\] To expand in negative powers of \(z\), or in \(\frac{1}{z}\), then\begin{align*} f\left ( z\right ) & =\frac{z+3}{z\left ( 1-\frac{3}{z}\right ) }\\ & =\frac{z+3}{z}\left ( \frac{1}{1-\frac{3}{z}}\right ) \end{align*}
For \(\left \vert \frac{3}{z}\right \vert <1\) or \(\left \vert z\right \vert <3\) the above becomes\begin{align*} f\left ( z\right ) & =\frac{z+3}{z}\left ( 1+\frac{3}{z}+\left ( \frac{3}{z}\right ) ^{2}+\left ( \frac{3}{z}\right ) ^{3}+\cdots \right ) \\ & =\left ( 1+\frac{3}{z}\right ) \left ( 1+\frac{3}{z}+\left ( \frac{3}{z}\right ) ^{2}+\left ( \frac{3}{z}\right ) ^{3}+\cdots \right ) \\ & =\left ( 1+\frac{3}{z}\right ) +\left ( 1+\frac{3}{z}\right ) \frac{3}{z}+\left ( 1+\frac{3}{z}\right ) \left ( \frac{3}{z}\right ) ^{2}+\left ( 1+\frac{3}{z}\right ) \left ( \frac{3}{z}\right ) ^{3}+\cdots \\ & =1+\frac{3}{z}+\frac{3}{z}+\left ( \frac{3}{z}\right ) ^{2}+\left ( \frac{3}{z}\right ) ^{2}+\left ( \frac{3}{z}\right ) ^{3}+\left ( \frac{3}{z}\right ) ^{3}+\left ( \frac{3}{z}\right ) ^{4}+\cdots \\ & =1+\frac{6}{z}+\frac{18}{z^{2}}+\frac{54}{z^{3}}+\cdots \end{align*}
This is valid for \(\left \vert z\right \vert >3\). The residue is \(6\), which can be confirmed using\begin{align*} \operatorname{Residue}\left ( 3\right ) & =\lim _{z\rightarrow 3}\left ( z-3\right ) f\left ( z\right ) \\ & =\lim _{z\rightarrow 3}\left ( z-3\right ) \frac{z+3}{z-3}\\ & =\lim _{z\rightarrow 3}\left ( z+3\right ) \\ & =6 \end{align*}
Summary
\[ \fbox{$f\left ( z\right ) =\frac{z+3}{z-3}=\left \{ \begin{array} [c]{lll}-1-\frac{2}{3}z-\frac{2}{9}z^{2}-\frac{2}{27}z^{3}-\frac{2}{81}z^{4}-\cdots & & \left \vert z\right \vert <3\\ 1+\frac{6}{z}+\frac{18}{z^{2}}+\frac{54}{z^{3}}+\cdots & & \left \vert z\right \vert >3 \end{array} \right . $}\]
Find Laurent series for \(\frac{z}{\left ( z+1\right ) \left ( z-3\right ) }\) in each of the following domains (i) \(\left \vert z\right \vert <1\) (ii) \(1<\left \vert z\right \vert <3\) (iii) \(\left \vert z\right \vert >3\)
Solution
The possible region are shown below. Since there is a pole at \(z=-1\) and pole at \(z=3\), then there are three different regions. They are named\(\ A,B,C\) in the following diagram
First the expression \(\frac{z}{\left ( z+1\right ) \left ( z-3\right ) }\) is expanded using partial fractions \begin{equation} \frac{z}{\left ( z+1\right ) \left ( z-3\right ) }=\frac{A}{\left ( z+1\right ) }+\frac{B}{\left ( z-3\right ) } \tag{1} \end{equation} Hence\begin{align*} z & =A\left ( z-3\right ) +B\left ( z+1\right ) \\ & =z\left ( A+B\right ) -3A+B \end{align*}
The above gives two equations \begin{align*} A+B & =1\\ 0 & =-3A+B \end{align*}
First equation gives \(A=1-B\). Substituting in the second equation gives \(0=-3\left ( 1-B\right ) +B\) or \(0=-3+4B\), hence \(B=\frac{3}{4}\), which implies \(A=1-\frac{3}{4}=\frac{1}{4}\), therefore (1) becomes\[ \frac{z}{\left ( z+1\right ) \left ( z-3\right ) }=\frac{1}{4}\frac{1}{\left ( z+1\right ) }+\frac{3}{4}\frac{1}{\left ( z-3\right ) }\] Considering each term in turn. For \(\frac{1}{4}\frac{1}{\left ( z+1\right ) }\), we can expand this as\begin{align} \frac{1}{4}\frac{1}{\left ( z+1\right ) } & =\frac{1}{4}\left ( 1-z+z^{2}-z^{3}+z^{4}+\cdots \right ) \qquad \left \vert z\right \vert <1\tag{2a}\\ \frac{1}{4}\frac{1}{\left ( z+1\right ) } & =\frac{1}{4z}\frac{1}{\left ( 1+\frac{1}{z}\right ) }=\frac{1}{4z}\left ( 1-\left ( \frac{1}{z}\right ) +\left ( \frac{1}{z}\right ) ^{2}-\left ( \frac{1}{z}\right ) ^{3}+\left ( \frac{1}{z}\right ) ^{4}-\cdots \right ) \qquad \left \vert z\right \vert >1 \tag{2b} \end{align}
And for the term \(\frac{3}{4}\frac{1}{\left ( z-3\right ) }\), we can expand this as\begin{align} \frac{3}{4}\frac{1}{\left ( z-3\right ) } & =-\frac{1}{4}\frac{1}{\left ( 1-\frac{z}{3}\right ) }=-\frac{1}{4}\left ( 1+\frac{z}{3}+\left ( \frac{z}{3}\right ) ^{2}+\left ( \frac{z}{3}\right ) ^{3}+\left ( \frac{z}{3}\right ) ^{4}+\cdots \right ) \qquad \left \vert z\right \vert <3\tag{3a}\\ \frac{3}{4}\frac{1}{\left ( z-3\right ) } & =\frac{3}{4z}\frac{1}{\left ( 1-\frac{3}{z}\right ) }=\frac{3}{4z}\left ( 1+\left ( \frac{3}{z}\right ) +\left ( \frac{3}{z}\right ) ^{2}+\left ( \frac{3}{z}\right ) ^{3}+\cdots \right ) \qquad \left \vert z\right \vert >3 \tag{3b} \end{align}
Now that we expanded all the terms in the two possible ways for each each, we now consider each region of interest, and look at the above 4 expansions, and simply pick for each region the expansion which is valid in for that region of interest.
For (i), region \(A\): In this region, we want \(\left \vert z\right \vert <1\). From (2,3) we see that (2a) and (3a) are valid expansions in \(\left \vert z\right \vert <1\). Hence\begin{align*} \frac{z}{\left ( z+1\right ) \left ( z-3\right ) } & =\frac{1}{4}\left ( 1-z+z^{2}-z^{3}+z^{4}+\cdots \right ) -\frac{1}{4}\left ( 1+\frac{z}{3}+\left ( \frac{z}{3}\right ) ^{2}+\left ( \frac{z}{3}\right ) ^{3}+\left ( \frac{z}{3}\right ) ^{4}+\cdots \right ) \\ & =\frac{1}{4}\left ( 1-z+z^{2}-z^{3}+z^{4}-\cdots \right ) -\frac{1}{4}\left ( 1+\frac{z}{3}+\frac{z^{2}}{9}+\frac{z^{3}}{27}+\frac{z^{4}}{81}+\cdots \right ) \\ & =\left ( \frac{1}{4}-\frac{1}{4}z+\frac{1}{4}z^{2}-\frac{1}{4}z^{3}+\frac{1}{4}z^{4}-\cdots \right ) -\left ( \frac{1}{4}+\frac{z}{12}+\frac{z^{2}}{36}+\frac{z^{3}}{108}+\frac{z^{4}}{324}+\cdots \right ) \\ & =-\frac{1}{4}z-\frac{z}{12}+\frac{1}{4}z^{2}-\frac{z^{2}}{36}-\frac{1}{4}z^{3}-\frac{z^{3}}{108}+\frac{1}{4}z^{4}-\frac{z^{4}}{324}-\cdots \\ & -\frac{1}{3}z+\frac{2}{9}z^{2}-\frac{7}{27}z^{3}+\frac{20}{81}z^{4}-\cdots \end{align*}
For (ii), region \(B\): This is for \(1<\left \vert z\right \vert <3\). From equations (2,3) we see that (2b) and (3a) are valid in this region. Hence
\begin{align*} \frac{z}{\left ( z+1\right ) \left ( z-3\right ) } & =\frac{1}{4z}\left ( 1-\left ( \frac{1}{z}\right ) +\left ( \frac{1}{z}\right ) ^{2}-\left ( \frac{1}{z}\right ) ^{3}+\left ( \frac{1}{z}\right ) ^{4}-\cdots \right ) -\frac{1}{4}\left ( 1+\frac{z}{3}+\left ( \frac{z}{3}\right ) ^{2}+\left ( \frac{z}{3}\right ) ^{3}+\left ( \frac{z}{3}\right ) ^{4}+\cdots \right ) \\ & =\frac{1}{4z}\left ( 1-\frac{1}{z}+\frac{1}{z^{2}}-\frac{1}{z^{3}}+\frac{1}{z^{4}}-\cdots \right ) -\frac{1}{4}\left ( 1+\frac{z}{3}+\frac{z^{2}}{9}+\frac{z^{3}}{27}+\frac{z^{4}}{81}+\cdots \right ) \\ & =\left ( \frac{1}{4z}-\frac{1}{4z^{2}}+\frac{1}{4z^{3}}-\frac{1}{4z^{4}}+\frac{1}{4z^{5}}-\cdots \right ) -\left ( \frac{1}{4}+\frac{z}{12}+\frac{z^{2}}{36}+\frac{z^{3}}{108}+\frac{z^{4}}{324}+\cdots \right ) \\ & =\overset{\text{principal part}}{\overbrace{\cdots +\frac{1}{4z^{5}}-\frac{1}{4z^{4}}+\frac{1}{4z^{3}}-\frac{1}{4z^{2}}+\frac{1}{4z}}}-\overset{\text{analytical part}}{\overbrace{\frac{1}{4}-\frac{z}{12}-\frac{z^{2}}{36}-\frac{z^{3}}{108}-\frac{z^{4}}{324}-\cdots }} \end{align*}
The residue is \(\frac{1}{4}\) by looking at the above. The value for the residue can be verified as follows. Using\[ b_{n}=\frac{1}{2\pi i}{\displaystyle \oint \limits _{C}} \frac{f\left ( z\right ) }{\left ( z-z_{0}\right ) ^{-n+1}}dz \] Where in the above \(z_{0}\) is the location of the pole and \(n\) is the coefficient of the \(\frac{1}{z^{n}}\) is the principal part. Since we want the residue, then \(n=1\) and the above becomes\[ b_{1}=\frac{1}{2\pi i}{\displaystyle \oint \limits _{C}} f\left ( z\right ) dz \] In the above, the contour \(C\) is circle somewhere inside the annulus \(1<\left \vert z\right \vert <3\). It does not matter that the radius is, as long as it is located in this range. For example, choosing radius \(2\) will work. The above then becomes\begin{equation} b_{1}=\frac{1}{2\pi i}{\displaystyle \oint \limits _{C}} \frac{z}{\left ( z+1\right ) \left ( z-3\right ) }dz \tag{5} \end{equation} However, since \(f\left ( z\right ) \) is analytic in this region, then \({\displaystyle \oint \limits _{C}} f\left ( z\right ) dz=2\pi i\sum \) \(\left ( \text{residues inside}\right ) \). There is only one pole now inside \(C\), which is at \(z=-1\). So all what we have to do is find the residue at \(z=-1\).\begin{align*} \operatorname{Residue}\left ( -1\right ) & =\lim _{z\rightarrow -1}\left ( z+1\right ) f\left ( z\right ) \\ & =\lim _{z\rightarrow -1}\left ( z+1\right ) \frac{z}{\left ( z+1\right ) \left ( z-3\right ) }\\ & =\lim _{z\rightarrow -1}\frac{z}{\left ( z-3\right ) }\\ & =\frac{-1}{\left ( -1-3\right ) }\\ & =\frac{1}{4} \end{align*}
Using this in (5) gives\begin{align*} b_{1} & =\frac{1}{2\pi i}\left ( 2\pi i\frac{1}{4}\right ) \\ & =\frac{1}{4} \end{align*}
Which agrees with what we found in (4) above.
For (iii), region \(C\): This is for \(\left \vert z\right \vert >3\). From (2,3) we see that (2b) and (3b) are valid expansions in \(z>3\), Hence\begin{align*} \frac{z}{\left ( z+1\right ) \left ( z-3\right ) } & =\frac{1}{4z}\left ( 1-\left ( \frac{1}{z}\right ) +\left ( \frac{1}{z}\right ) ^{2}-\left ( \frac{1}{z}\right ) ^{3}+\left ( \frac{1}{z}\right ) ^{4}-\cdots \right ) +\frac{3}{4z}\left ( 1+\left ( \frac{3}{z}\right ) +\left ( \frac{3}{z}\right ) ^{2}+\left ( \frac{3}{z}\right ) ^{3}+\cdots \right ) \\ & =\left ( \frac{1}{4z}-\frac{1}{4z^{2}}+\frac{1}{4z^{3}}-\frac{1}{4z^{4}}+\frac{1}{4z^{5}}-\cdots \right ) +\frac{3}{4z}\left ( 1+\frac{3}{z}+\frac{9}{z^{2}}+\frac{27}{z^{3}}+\cdots \right ) \\ & =\left ( \frac{1}{4z}-\frac{1}{4z^{2}}+\frac{1}{4z^{3}}-\frac{1}{4z^{4}}+\frac{1}{4z^{5}}-\cdots \right ) +\left ( \frac{3}{4z}+\frac{9}{4z^{2}}+\frac{27}{4z^{3}}+\frac{81}{4z^{4}}+\cdots \right ) \\ & =\cdots +\frac{20}{z^{4}}+\frac{7}{z^{3}}+\frac{2}{z^{2}}+\frac{1}{z} \end{align*}
This is as expected contains only a principal part and no analytical part. The residue is \(1\). This above value for the residue can be verified as follows. Using\[ b_{n}=\frac{1}{2\pi i}{\displaystyle \oint \limits _{C}} \frac{f\left ( z\right ) }{\left ( z-z_{0}\right ) ^{-n+1}}dz \] Where in the above \(z_{0}\) is the location of the pole and \(n\) is the coefficient of the \(\frac{1}{z^{n}}\) is the principal part. Since we want the residue, then \(n=1\) and the above becomes\[ b_{1}=\frac{1}{2\pi i}{\displaystyle \oint \limits _{C}} f\left ( z\right ) dz \] In the above, the contour \(C\) is circle somewhere in \(\left \vert z\right \vert >3\). It does not matter that the radius is. The above integral then becomes\begin{equation} b_{1}=\frac{1}{2\pi i}{\displaystyle \oint \limits _{C}} \frac{z}{\left ( z+1\right ) \left ( z-3\right ) }dz \tag{7} \end{equation} However, since \(f\left ( z\right ) \) is analytic in \(\left \vert z\right \vert >3\), then \({\displaystyle \oint \limits _{C}} f\left ( z\right ) dz=2\pi i\sum \) \(\left ( \text{residues inside}\right ) \). There are now two poles inside \(C\), one at \(z=-1\) and one at \(z=3\). So all what we have to do is find the residues at each. We found earlier that \(\operatorname{Residue}\left ( -1\right ) =\frac{1}{4}\). Now\begin{align*} \operatorname{Residue}\left ( 3\right ) & =\lim _{z\rightarrow 3}\left ( z-3\right ) f\left ( z\right ) \\ & =\lim _{z\rightarrow 3}\left ( z-3\right ) \frac{z}{\left ( z+1\right ) \left ( z-3\right ) }\\ & =\lim _{z\rightarrow 3}\frac{z}{\left ( z+1\right ) }\\ & =\frac{3}{4} \end{align*}
Therefore the sum of residues is \(1\). Using this result in (7) gives\begin{align*} b_{1} & =\frac{1}{2\pi i}\left ( 2\pi i\left ( \frac{1}{4}+\frac{3}{4}\right ) \right ) \\ & =1 \end{align*}
Which agrees with what result from (6) above.
Summary of results\[ \fbox{$f\left ( z\right ) =\frac{z}{\left ( z+1\right ) \left ( z-3\right ) }=\left \{ \begin{array} [c]{lll}-\frac{1}{3}z+\frac{2}{9}z^{2}-\frac{7}{27}z^{3}+\frac{20}{81}z^{4}-\cdots & & \left \vert z\right \vert <1\\ \cdots +\frac{1}{4z^{5}}-\frac{1}{4z^{4}}+\frac{1}{4z^{3}}-\frac{1}{4z^{2}}+\frac{1}{4z}-\frac{1}{4}-\frac{z}{12}-\frac{z^{2}}{36}-\frac{z^{3}}{108}-\frac{z^{4}}{324}-\cdots & & \,1<\left \vert z\right \vert <3\\ \cdots +\frac{20}{z^{4}}+\frac{7}{z^{3}}+\frac{2}{z^{2}}+\frac{1}{z} & & \left \vert z\right \vert >3 \end{array} \right . $}\]
Use residue theorem to evaluate \({\displaystyle \oint \limits _{C}} \frac{e^{-2z}}{z^{2}}dz\) on contour \(C\) which is circle \(\left \vert z\right \vert =1\) in positive sense.
Solution
For \(f\left ( z\right ) \) which is analytic on and inside \(C\), the Cauchy integral formula says\begin{equation}{\displaystyle \oint \limits _{C}} f\left ( z\right ) dz=2\pi i\sum _{j}\operatorname{Residue}\left ( z=z_{j}\right ) \tag{1} \end{equation} Where the sum is over all residues located inside \(C\). for \(f\left ( z\right ) =\frac{e^{-2z}}{z^{2}}\) there is a simple pole at \(z=0\) of order \(2\). To find the residue, we use the formula for pole or order \(m\) given by\[ \operatorname{Residue}\left ( z_{0}\right ) =\lim _{z\rightarrow z_{0}}\frac{d^{m-1}}{dz^{m-1}}\frac{\left ( z-z_{0}\right ) ^{m}}{\left ( m-1\right ) !}f\left ( z\right ) \] Hence for \(m=2\) and \(z_{0}=0\) the above becomes\begin{align*} \operatorname{Residue}\left ( 0\right ) & =\lim _{z\rightarrow 0}\frac{d}{dz}z^{2}f\left ( z\right ) \\ & =\lim _{z\rightarrow 0}\frac{d}{dz}z^{2}\frac{e^{-2z}}{z^{2}}\\ & =\lim _{z\rightarrow 0}\frac{d}{dz}e^{-2z}\\ & =\lim _{z\rightarrow 0}\left ( -2e^{-2z}\right ) \\ & =-2 \end{align*}
Therefore (1) becomes\begin{align*}{\displaystyle \oint \limits _{C}} \frac{e^{-2z}}{z^{2}}dz & =2\pi i\left ( -2\right ) \\ & =-4\pi i \end{align*}
Use residue theorem to evaluate \({\displaystyle \oint \limits _{C}} ze^{\frac{1}{z}}dz\) on contour \(C\) which is circle \(\left \vert z\right \vert =1\) in positive sense.
Solution
The singularity is at \(z=0\), but we can not use the simple pole residue finding method here, since this is an essential singularity now due to the \(e^{\frac{1}{z}}\) term. To find the residue, we expand \(f\left ( z\right ) \) around \(z=0\) in Laurent series and look for the coefficient of \(\frac{1}{z}\) term.\begin{align*} f\left ( z\right ) & =ze^{\frac{1}{z}}\\ & =z\left ( 1+\frac{1}{z}+\frac{1}{2}\frac{1}{z^{2}}+\frac{1}{3!}\frac{1}{z^{3}}+\cdots \right ) \\ & =z+1+\frac{1}{2}\frac{1}{z}+\frac{1}{3!}\frac{1}{z^{2}}+\cdots \end{align*}
Hence residue is \(\frac{1}{2}\). Therefore\begin{align*}{\displaystyle \oint \limits _{C}} ze^{\frac{1}{z}}dz & =2\pi i\left ( \frac{1}{2}\right ) \\ & =\pi i \end{align*}
Use residue theorem to evaluate \({\displaystyle \oint \limits _{C}} \frac{z+2}{z^{2}-\frac{z}{2}}dz\) on contour \(C\) which is circle \(\left \vert z\right \vert =1\) in positive sense.
Solution
\begin{align*} f\left ( z\right ) & =\frac{z+2}{z^{2}-\frac{z}{2}}\\ & =\frac{z+2}{z\left ( z-\frac{1}{2}\right ) } \end{align*}
Hence there is a simple pole at \(z=0\) and simple pole at \(z=\frac{1}{2}\)\begin{align*} \operatorname{Residue}\left ( 0\right ) & =\lim _{z\rightarrow 0}\left ( z\right ) f\left ( z\right ) \\ & =\lim _{z\rightarrow 0}z\frac{z+2}{z\left ( z-\frac{1}{2}\right ) }\\ & =\lim _{z\rightarrow 0}\frac{z+2}{\left ( z-\frac{1}{2}\right ) }\\ & =\frac{2}{-\frac{1}{2}}\\ & =-4 \end{align*}
And\begin{align*} \operatorname{Residue}\left ( \frac{1}{2}\right ) & =\lim _{z\rightarrow \frac{1}{2}}\left ( z-\frac{1}{2}\right ) f\left ( z\right ) \\ & =\lim _{z\rightarrow \frac{1}{2}}\left ( z-\frac{1}{2}\right ) \frac{z+2}{z\left ( z-\frac{1}{2}\right ) }\\ & =\lim _{z\rightarrow \frac{1}{2}}\frac{z+2}{z}\\ & =\frac{\frac{1}{2}+2}{\frac{1}{2}}\\ & =5 \end{align*}
Therefore\begin{align*}{\displaystyle \oint \limits _{C}} \frac{z+2}{z^{2}-\frac{z}{2}}dz & =2\pi i\left ( 5-4\right ) \\ & =2\pi i \end{align*}