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\begin{center}
  {\bf University of Kentucky, Physics 306 \\
    Homework \#2, Rev. A, due Wednesday, 2024-01-24 }
\end{center}

\vspace{.3 in}

{\bf 1. Complex algebra}

$e^{i\phi}=\cos\phi+i\sin\phi$ leads to the beautiful identity $e^{i\pi}+1=0$ involving the three basic operators and five fundamental constants exactly once each.  The purpose of this exercise is to explore implications of this relationship between exponential and trigonometric functions and it's role in rotations.

{\bf \quad a)}  Use the power series of $e^x$, $\sin\phi$, $\cos\phi$ and the \mbox{property $i^{2}=-1$ to prove \href{https://en.wikipedia.org/wiki/Euler\%27s_formula}{\it Euler's formula}.}

{\bf \quad b)} Show if $z=x+iy$ where $(x,y)$ are the (real, imaginary) Cartesian coordinates of $z$ in the \href{https://en.wikipedia.org/wiki/Complex_plane}{\it complex plane}, then $z=\rho e^{i\phi}$, where $(\rho,\phi)$ are the (radius, azimuth) polar coordinates of $z$.
 
{\bf \quad c)} The \href{https://en.wikipedia.org/wiki/Complex_conjugate}{\it complex conjugate} $z^*$ is formed by replacing $i$ with $-i$ everywhere in $z$.  The \href{https://en.wikipedia.org/wiki/Absolute_value#Complex_numbers}{\it modulus} $|z| \equiv \sqrt{z^*z}$ is the complex analog of absolute value.  Use $z=x+iy=\rho e^{i\phi}$ to show the relations $|z|^{2} = z^*z = zz^* = x^{2}+y^{2} = \rho^{2}$.  Thus $|z|=\rho$ is the distance of $z$ from the origin (radius).

{\bf \quad d)} Show that $iz$ rotates $z$ by $90^{\circ}$ CCW about the origin.  Use the \href{https://en.wikipedia.org/wiki/Arc_length#Arcs_of_circles}{\it arc length formula} $ds=\rho d\phi$ to show that the operator $(1+i d\phi)$ multiplied by $z$ rotates it by an angle $d\phi$.  Formally integrate the equation $dz=izd\phi$ to show that the operator $e^{i\phi}$ rotates $z$ by the angle $\phi$.  Use this result to justify the identity {$\displaystyle e^x=\lim_{n\to\infty}\textstyle(1+\frac{x}{n})^n$}.

{\bf \quad e)} Expand $z^2$ in terms of $x$,$y$ and also $\rho$,$\phi$ to see why $|z|^2$ is generally more useful than $z^2$.

{\bf \quad f)} Show that the real and imaginary parts of $z_{1}^{*}z_{2} = z_{2}z_{1}^{*} = (x_{1}+iy_{1})^{*}(x_{2}+iy_{2})$ match the dot and cross ($z$-component) products of the vectors $(x_{1},y_{1})^{T}$ and $(x_{2},y_{2})^{T}$.

{\bf \quad g)} Multiply $e^{i\phi}$ by its complex conjugate and expand using Euler's formula to prove the relation $\sin^2\phi + \cos^2\phi = 1$.  This shows that $e^{i\phi}$ traces out the unit circle in the complex plane.

{\bf \quad h)} Use Euler's formula on $e^{\pm i\phi}$ to express $\cos\phi$, $\sin\phi$ and $\tan\phi$ in terms of $e^{i\phi}$ and $e^{-i\phi}$.

{\bf \quad i)} Using the similar definition $\cosh\alpha \equiv \frac12(e^\alpha+e^{-\alpha})$ and
$\sinh\alpha \equiv \frac12(e^\alpha-e^{-\alpha})$, derive the analog of Euler's formula for the \href{https://en.wikipedia.org/wiki/Hyperbolic_functions}{\it hyperbolic functions}.  {\it Hint: $i$ becomes $\pm$.}

{\bf \quad j)} Multiply and expand $e^\alpha$ and $e^{-\alpha}$ to derive a similar formula as in part f) for $\cosh\alpha$ and $\sinh\alpha$.  This shows that $(\cosh\alpha,\sinh\alpha)$ traces out a hyperbola in the plane.

{\bf \quad k)} Derive addition formulas for $\cos(\alpha\pm\beta)$ and $\sin(\alpha\pm\beta)$ by multiplying and expanding $e^{i\alpha} \cdot e^{\pm i \beta}$ and then separating the real and imaginary parts.  Do the same for the hyperbolic functions.

{\bf \quad l)} Use $e^{i m\phi}$ to obtain \href{https://en.wikipedia.org/wiki/De_Moivre\%27s_formula}{\it de Moivre's formula} for $\cos(m\phi)+i\sin(m\phi)$ and $\cosh(m\alpha)\pm\sinh(m\alpha)$.  Use this formula to expand $\cos 2\phi$ and $\sin 2\phi$.

{\bf \quad m)} Obtain the derivatives of $\sin \phi$, $\cos \phi$, and $\sinh \alpha$, $\cosh \alpha$ from the derivative of $e^{x}$.
   

{\bf 2. Vectors}


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