3.10 HW 9

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3.10.1 chapter 14, problem 9.2

For following function $w=f(z)=u+iv$ 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.

3.10.2 chapter 14, problem 9.3

For following function $w=f(z)=u+iv$ 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.

3.10.3 chapter 14, problem 9.4

For the function $w=f(z)=u+iv$ 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 $w=e^z=e^{x+iy}=e^x{e}^{iy}=e^x(\cos y+i\sin y)=e^x\cos (y)+i{e}^x\sin (y)$. Therefore, since $w=u+iv$ then $u=e^x\cos (y)$ and $v=e^x\sin (y).$ Then $u=C$ gives the equation $e^x\cos (y)=C$ and $v=C$ gives the equation $e^x\sin (y)=C$

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.

3.10.4 chapter 14, problem 9.7

For the function $w=f(z)=u+iv$ 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 $w=\sin (z)$

Answer let $z=x+iy$, hence $$\begin{array}{rl}w& =\sin (z)=\sin (x+iy)\\ & =\sin (x)\cos \left(iy\right)+\cos (x)\sin \left(iy\right)\end{array}$$

But $\cos \left(iy\right)=\cosh (y)$ and $\sin \left(iy\right)=i\sinh (y)$, therefore $$w=\sin (x)\cosh (y)+\cos (x)i\sinh (y)$$

Since $w=u+iv$ then $u=\sin (x)\cosh (y)$ and $v=\cos (x)\sinh (y)$.$$Then $u=C$ gives the equation $$\sin (x)\cosh (y)=C$$

And $v=C$ gives the equation $$\cos (x)\sinh (y)=C$$

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.

3.10.5 chapter 7, problem 3.4

Draw a graph over a whole period for $f(t)=\cos \left(2\pi t\right)+\cos \left(4\pi t\right)+\frac{1}{2}\cos \left(6\pi t\right)$

Answer First, find the period of the above function. A function is periodic with period $p$ if $f(t+p)=f(t)$ for all $t.$ We know that $\cos \left(nt\right)$ has the same period as $\cos (nt+2n\pi )$ for $n$ integer, since the function $\cos (x)$ has a period of $2\pi$. So a period of $f\left(t\right)$ is $2\pi$ since it is the sum of $\cos (x)$ 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

3.10.6 chapter 7, problem 3.6

Draw a graph of $f(x)=\sin \left(2x\right)+\sin 2(x+\frac{\pi }{3}).$ what are the period and amplitude? Write as a single harmonic.

Answer

Since $\sin (x+y)=\sin (x)\cos (y)+\cos (x)\sin (y)$ then

$$\sin 2(x+\frac{\pi }{3})=\sin (2x+\frac{2}{3}\pi )=\sin \left(2x\right)\cos \left(\frac{2}{3}\pi \right)+\cos \left(2x\right)\sin \left(\frac{2}{3}\pi \right)$$

Hence

$$f(x)=\sin \left(2x\right)+\sin 2(x+\frac{\pi }{3})=\sin \left(2x\right)+\sin \left(2x\right)\cos \left(\frac{2}{3}\pi \right)+\cos \left(2x\right)\sin \left(\frac{2}{3}\pi \right)$$

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

$$\begin{array}{rl}f(x)& =\sin \left(2x\right)-\frac{1}{2}\sin \left(2x\right)+\frac{\sqrt{3}}{2}\cos \left(2x\right)\\ f(x)& =\frac{1}{2}\sin \left(2x\right)+\frac{\sqrt{3}}{2}\cos \left(2x\right)\end{array}$$

But $\sin (2x)=\cos (\frac{\pi }{2}-2x)$

$$f(x)=\frac{1}{2}\cos (\frac{\pi }{2}-2x)+\frac{\sqrt{3}}{2}\cos \left(2x\right)$$

Now this is in term of a single harmonic function. Hence, we see that $f(x)$ is the sum of harmonics of the same periods (the $\cos$ function have period of $2\pi ).$hence the period of $f(x)$ is $2\pi$. To find Max amplitude, this is a problem of finding a maximum of a function.

$$\begin{array}{rl}\frac{d}{dx}f(x)& =-\frac{1}{2}\sin (\frac{\pi }{2}-2x)(-2)-\frac{\sqrt{3}}{2}\sin \left(2x\right)\left(2\right)\\ & =\sin (\frac{\pi }{2}-2x)-\sqrt{3}\sin \left(2x\right)\\ & =\cos (2x)-\sqrt{3}\sin \left(2x\right)\end{array}$$

$$

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 $f(x)$ 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

3.10.7 First part of HW9 was scanned

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