For following function \(w=f\relax (z) =u+iv\) find \(u\) and \(v\) as functions of \(x\) and \(y\). Sketch the graphs in the \((x,y)\) plane of the images of \(u=const\) and \(v=const\) for several values of \(u\) and several values of \(v\) where \(w=\frac {z+1}{2i}\)
Answer let \(z=x+iy\), hence \(w=\frac {z+1}{2i}=\frac {x+iy+1}{2i}=-i\frac {x+iy+1}{2}=\frac {-ix+y-i}{2}\ =\frac {y}{2}+i\left (\frac {-1-x}{2}\right ) \). So, since \(w=u+iv\) then \(\ u=\frac {y}{2}\ \) and \(\ v=\left (\frac {-1-x}{2}\right ) \ \) Then \(u=C\), where \(C\) is a constant, gives the equation \(\frac {y}{2}=C\). Which is the equation of a straight line \(y=C\).
\(v=\)constant, gives the equation \(\left (\frac {-1-x}{2}\right ) =C\), gives the equation of the straight line \(x=C-1.\)
These two equations are plotted for few points. The following shows the plots generated for the mapping from the z-plane to the w-plane, and then the image of u=const and the image of v=const back into the xy plane.
For following function \(w=f\relax (z) =u+iv\) find \(u\) and \(v\) as functions of \(x\) and \(y\). Sketch the graphs in the \((x,y)\) plane of the images of \(u=C\) and \(v=C\), where \(C\) is constant, for several values of \(u\) and several values of \(v\). \(w=\frac {1}{z}\)
Answer let \(z=x+iy\), hence \(w=\frac {1}{z}=\frac {\ 1}{x+iy}=\ \frac {1}{x+iy}\frac {x-iy}{x-iy}=\frac {x-iy}{x^{2}+y^{2}}\ =\ \frac {x\ }{x^{2}+y^{2}}+\frac {-iy}{x^{2}+y^{2}}\). Hence, since \(w=u+iv\) then \(\ u=\ \frac {x\ }{x^{2}+y^{2}}\ \) and \(\ v=\frac {-y}{x^{2}+y^{2}}.\) Then \(u=C\) gives the equation \(\frac {x\ }{x^{2}+y^{2}}=C\).
\(v=C\) gives the equation \(\frac {-y}{x^{2}+y^{2}}=C\).
These 2 equations were plotted for few points. The following shows the plots generated for the mapping from the z-plane to the \(w\)-plane, and then the image of \(u=C\) and the image of \(v=C\) back into the \(xy\) plane.
For the function \(w=f\relax (z) =u+iv\) shown below, find \(u\) and \(v\) as functions of \(x\) and \(y\). Sketch the graphs in the \((x,y)\) plane of the images of \(u=const\) and \(v=const\) for several values of \(u\) and several values of \(v\). \(w=e^{z}\)
Answer let \(z=x+iy\), hence \(w=e^{z}=e^{x+iy}=\ e^{x}e^{iy}=e^{x}\left (\cos y+i\sin y\right ) \ =\ e^{x}\cos \relax (y) +i\ e^{x}\sin \relax (y) \). Therefore, since \(w=u+iv\) then \(\ u=\ \ e^{x}\cos \relax (y) \ \) and \(\ v=e^{x}\sin \relax (y) .\) Then \(u=C\) gives the equation \(e^{x}\cos \relax (y) =C\) and \(v=C\) gives the equation \(e^{x}\sin \relax (y) =C\)
These 2 equations are plotted for few points. The following shows the plots generated for the mapping from the z-plane to the w-plane, and then the image of u=const and the image of v=const back into the xy plane.
For the function \(w=f\relax (z) =u+iv\) shown below, find \(u\) and \(v\) as functions of \(x\) and \(y\). Sketch the graphs in the \((x,y)\) plane of the images of \(u=C\) and \(v=C\) for several values of \(u\) and several values of \(v\). use \(w=\sin \relax (z) \)
Answer let \(z=x+iy\), hence \begin {align*} w & =\sin \relax (z) =\sin \left (x+iy\right ) \\ & =\sin \relax (x) \cos \left (iy\right ) +\cos \relax (x) \sin \left (iy\right ) \end {align*}
But \(\cos \left (iy\right ) =\cosh \relax (y) \) and \(\sin \left ( iy\right ) =i\sinh \relax (y) \), therefore \[ w=\sin \relax (x) \cosh \relax (y) +\cos \relax (x) i\sinh \relax (y) \]
Since \(w=u+iv\) then \(\ u=\sin \relax (x) \cosh \relax (y) \ \) and \(\ v=\cos \relax (x) \sinh \relax (y) \).\(\ \)Then \(u=C\) gives the equation \[ \sin \relax (x) \cosh \relax (y) =C \]
And \(v=C\) gives the equation \[ \cos \relax (x) \sinh \relax (y) =C \]
These 2 equations are plotted for few points. The following shows the plots generated for the mapping from the z-plane to the w-plane, and then the image of u=const and the image of v=const back into the xy plane.
Draw a graph over a whole period for \(f\relax (t) =\cos \left (2\pi t\right ) +\cos \left (4\pi t\right ) +\frac {1}{2}\cos \left (6\pi t\right ) \)
Answer First, find the period of the above function. A function is periodic with period \(p\) if \(f\left (t+p\right ) =f\relax (t) \) for all \(t.\) We know that \(\cos \left (nt\right ) \) has the same period as \(\cos \left (nt+2n\pi \right ) \) for \(n\) integer, since the function \(\cos \relax (x) \) has a period of \(2\pi \). So a period of \(f\left ( t\right ) \) is \(2\pi \) since it is the sum of \(\cos \relax (x) \) functions. To plot this function, we plot each of its components over the same period of \(2\pi \) and then sum them together.
This plot below shows the result
Draw a graph of \(f\relax (x) =\sin \left (2x\right ) +\sin 2\left ( x+\frac {\pi }{3}\right ) .\) what are the period and amplitude? Write as a single harmonic.
Answer
Since \(\sin \left (x+y\right ) =\sin \relax (x) \cos \relax (y) +\cos \relax (x) \sin \relax (y) \) then
\[ \sin 2\left (x+\frac {\pi }{3}\right ) =\sin \left (2x+\frac {2}{3}\pi \right ) =\sin \left (2x\right ) \cos \left (\frac {2}{3}\pi \right ) +\cos \left ( 2x\right ) \sin \left (\frac {2}{3}\pi \right ) \]
Hence
\[ f\relax (x) =\sin \left (2x\right ) +\sin 2\left (x+\frac {\pi }{3}\right ) =\ \ \sin \left (2x\right ) +\sin \left (2x\right ) \cos \left ( \frac {2}{3}\pi \right ) +\cos \left (2x\right ) \sin \left (\frac {2}{3}\pi \right ) \]
Now \(\cos \left (\frac {2}{3}\pi \right ) =-\frac {1}{2}\ \)and \(\sin \left ( \frac {2}{3}\pi \right ) =\frac {\sqrt {3}}{2}\) so above can be written as
\begin {align*} f\relax (x) & =\ \sin \left (2x\right ) -\frac {1}{2}\sin \left ( 2x\right ) \ +\frac {\sqrt {3}}{2}\cos \left (2x\right ) \ \\ f\relax (x) & =\ \frac {1}{2}\sin \left (2x\right ) \ +\frac {\sqrt {3}}{2}\cos \left (2x\right ) \end {align*}
But \(\sin (2x)=\cos (\frac {\pi }{2}-2x)\)
\[ f\relax (x) =\ \frac {1}{2}\cos (\frac {\pi }{2}-2x)\ +\frac {\sqrt {3}}{2}\cos \left (2x\right ) \]
Now this is in term of a single harmonic function. Hence, we see that \(f\relax (x) \) is the sum of harmonics of the same periods (the \(\cos \) function have period of \(2\pi ).\)hence the period of \(f\relax (x) \) is \(2\pi \). To find Max amplitude, this is a problem of finding a maximum of a function.
\begin {align*} \frac {d}{dx}f\relax (x) & =-\frac {1}{2}\sin (\frac {\pi }{2}-2x)\ \left (-2\right ) -\frac {\sqrt {3}}{2}\sin \left (2x\right ) \left ( 2\right ) \\ & =\ \sin (\frac {\pi }{2}-2x)\ \ -\sqrt {3}\sin \left (2x\right ) \ \\ & =\cos (2x)\ \ -\sqrt {3}\sin \left (2x\right ) \ \end {align*}
\(\ \)
Hence for a maximum, \(\cos (2x)\ \ -\sqrt {3}\sin \left (2x\right ) \ \ =0\). A root for this equation is found at \(x=0.261799\) so I use this value in \(f\relax (x) \) to find the amplitude.
\(f\left (0.261799\right ) =1\). This is the maximum value, or the amplitude\(.\)The following is a plot of this function
