Welcome to the class!
We would like to get to know you a bit and learn about your expectations for this course. Please take a few minutes to share your thoughts on the following questions.
\begin{align} z_1-z_2&=(3-4i)-(7+2i)\\ &=(3-7)+(-4-2)i\\ &=-4-6i \end{align}
\begin{align} z_1 \, z_2&=(3-4i)(7+2i)\\ &=21+6i-28i+8\\ &=29-22i \end{align}
\begin{align} \frac{z_1}{z_2}&=\frac{3-4i}{7+2i}\\ &=\frac{3-4i}{7+2i}\,\frac{7-2i}{7-2i}\\ &=\frac{21-6i-28i-8}{49-14i+14i+4}\\ &= \frac{13-34i}{53}\\ &=\frac{13}{53}-\frac{34i}{53} \end{align}
\[AB\doteq \begin{pmatrix} 1&0&0\\ 0&0&1\\ 0&-1&0\\ \end{pmatrix} \begin{pmatrix} a&b&c\\ d&e&f\\ g&h&j\\ \end{pmatrix} = \begin{pmatrix} a&b&c\\ g&h&j\\ -d&-e&-f\\ \end{pmatrix} \]
\[{\rm tr} (B)\doteq{\rm tr} \begin{pmatrix} a&b&c\\ d&e&f\\ g&h&j\\ \end{pmatrix} =a+e+j\]
\[A\vert D\rangle\doteq \begin{pmatrix} 1&0&0\\ 0&0&1\\ 0&-1&0\\ \end{pmatrix} \begin{pmatrix} 1\\ i\\ -1\\ \end{pmatrix} = \begin{pmatrix} 1\\ -1\\ -i\\ \end{pmatrix} \]
\[\det(\lambda{\cal I}-A) \doteq\left\vert \begin{matrix} \lambda-1&0&0\\ 0&\lambda&-1\\0&1&\lambda \end{matrix} \right\vert =\lambda^2(\lambda-1)+(\lambda-1)=(\lambda-1)(\lambda-i)(\lambda+i)\]
\(C\) is the matrix that rotates two-dimensional vectors counterclockwise through the angle \(\theta\). Notice that \(C^{-1}\) is just the matrix that rotates these vectors back the other way (clockwise) by the angle \(\theta\). This inverse matrix “undoes” the original transformation.
\begin{align*} C^{-1} &=\begin{pmatrix} \cos(-\theta) & -\sin(-\theta) \\[12pt] \sin(-\theta) & \cos(-\theta) \end{pmatrix} &= \begin{pmatrix} \cos\theta & \sin\theta \\[12pt] -\sin\theta & \cos\theta \end{pmatrix} \end{align*}
In general, Inverses can be found by creating a matrix \([I][C]\) and then performing row reduction operations until we get the matrix \([C^{-1}][I]\)
\begin{align*} \begin{pmatrix} 1 & 0 & \cos\theta & -\sin\theta \\[12pt] 0 & 1 & \sin\theta & \cos\theta \end{pmatrix} &\rightarrow \begin{pmatrix} \cos\theta & 0 & \cos^2\theta & -\sin\theta\cos\theta \\ 0 & \sin\theta & \sin^2\theta & \sin\theta\cos\theta \end{pmatrix} & \text{Multiply top row by \(\cos\theta\) and bottom row by \(\sin\theta\)} \\[12pt] &\rightarrow \begin{pmatrix} \cos\theta & \sin\theta & \cancelto{1}{\cos^2\theta + \sin^2\theta} & 0 \\ 0 & 1 & \sin\theta & \cos\theta \end{pmatrix} &\text{Add bottom row to top, divide bottom row by \(\sin\theta\)}\\[12pt] &\rightarrow \begin{pmatrix} -sin\theta\cos\theta & -\sin^2\theta & -\sin\theta & 0 \\ 0 & 1 & \sin\theta & \cos\theta \end{pmatrix}&\text{Multiply top row by \(-\sin\theta\)}\\[12pt] &\rightarrow \begin{pmatrix} \cos\theta & \sin\theta & 1 & 0 \\ -\sin\theta\cos\theta & \cancelto{\cos^2\theta}{1-\sin^\theta} & 0 & \cos\theta \end{pmatrix}&\text{Add top row to bottom. Divide top row by \(-\sin\theta\)} \\[12pt] &\rightarrow \begin{pmatrix} \cos\theta & \sin\theta & 1 & 0 \\ -\sin\theta & \cos\theta &0 & 1 \end{pmatrix} &\text{Divide bottom row by \(\cos\theta\)} \end{align*}
\[C^{-1} = \begin{pmatrix} \cos\theta &\sin\theta\\ -\sin\theta&\cos\theta\\ \end{pmatrix} \]
\(C\) is the matrix that rotates two-dimensional vectors counterclockwise through the angle \(\theta\). Notice that \(C^{-1}\) is just the matrix that rotates these vectors back the other way (clockwise) by the angle \(\theta\). This inverse matrix “undoes” the original transformation.
Use Euler's formula \(e^{i\phi}=\cos\phi+i\sin\phi\) and its complex conjugate to find formulas for \(\sin\phi\) and \(\cos\phi\). In your physics career, you will often need to read these formula “backwards,” (i.e. notice one of these combinations of exponentials in a sea of other symbols and say, Ah ha! that is \(\cos\phi\)). So, pay attention to the result of the homework problem!
\[e^{i\phi}=\cos\phi+i\sin\phi\] \[e^{-i\phi}=\cos\phi-i\sin\phi\]
When these are added: \[e^{i\phi}+e^{-i\phi}=2\cos\phi\] \[\Longrightarrow \cos\phi=\frac{e^{i\phi}+e^{-i\phi}}{2}\] When they are subtracted: \[e^{i\phi}=\cos\phi+i\sin\phi\] \[-e^{-i\phi}=-\cos\phi+i\sin\phi\]
\[\Longrightarrow e^{i\phi}-e^{-i\phi}=2i\sin\phi\]
\[\Longrightarrow \sin\phi=\frac{e^{i\phi}-e^{-i\phi}}{2i}\]
Show that Euler's formula:
\[e^{i\phi} = \cos\phi +i \sin\phi\]
is true, by comparing the power series for the various terms.
\begin{align*} e^{i\phi} =& 1+i\phi+\frac{1}{2!}(i\phi)^2+\frac{1}{3!}(i\phi)^3+\frac{1}{4!}(i\phi)^4+\frac{1}{5!}(i\phi)^5+...\\ =& \left(1-\frac{1}{2!}\, \phi^2+\frac{1}{4!}\,\phi^4-...\right)+i\left(\phi-\frac{1}{3!}\,\phi^3+\frac{1}{5!}\,\phi^5+...\right)\\ =& \cos\phi+i\sin\phi \end{align*}