#### Question 2 Find the complex number $z_{0}$ in polar form which has the property that multiplication by $z_0$ gives a function which rotates all complex numbers by the angle $\pi/6$ counterclockwise and also scales them by $6$. Call this function $f(z_{0})$ , so that for any complex number $z$ the function gives $f_{z_0} (z) = z_0z$. Find the cartesian form of $z_0$. Find the 2D matrix which performs the same operation on points of the plane $\R^{2}$ as $f(z_0)$ performs on the complex plane. $\huge \begin{align} f_{z_{0}}(z)&=6e^{\frac{\pi}{6}i}z \\ \\ \end{align} $ $\huge f: \mat{x\\y} \mapsto \mat{ 6\cos \frac{\pi}{6} & -6\sin \frac{\pi}{6} \\ 6\sin \frac{\pi}{6} & 6\cos \frac{\pi}{6} }\mat{x\\y} $ #### Question 4 Show that complex numbers $w$ and $z$ are linearly dependent (as real vectors) if and only if $\bar{w}z\in\R$. (Note: the linear dependence statement uses only the vector space properties of $\C$, but the criterion in this case uses the multiplication of $\mathbb{C}$. To show the if and only if statement is true, you need to prove two implications. For example: “A if and only if B” is true if A implies B and B implies A. So, first assume A and show that B is true, then second assume B and show that A is true.) Let $\mathcal L_{I}$ be the [[Set|set]] of all [[Linear Independence|Linearly Independent]] [[Set|sets]]. $\huge \begin{align} \let a,b,c,d &\in \R \\ w &= a+bi \\ z &= c+bi\\ \vec w &= \mat{a\\b} \\ \\ \vec z &= \mat{c\\d} \end{align} $ If the vectors $\vec w$ and $\vec z$ are linearly dependent, then a matrix composing them as column vectors $A$ must be singular, meaning $\det(A)=0$. $\huge \begin{align} A &= \mat{a&c\\b&d} \\ \det(A) &= 0 \\ ad-bc &= 0 \\ \\ \end{align} $ $\huge ad-bc =0 \iff \left\{ \mat{a\\ b},\mat{c\\d} \right\} \notin \mathcal L_{I} $ $\huge \begin{align} \bar{w} z &=(a-bi)(c+di) \\ \\ &=ac+adi-bci+bd \\ &= \underbrace{(ac+bd)}_{{\R}}+ \underbrace{(ad-bc)}_{\R}i \\ \end{align} $ $\huge \begin{align} ac+bd &\in \R \\ ad-bc &\in \R \\ \therefore \bar{w}z \in \R &\iff ad-bc =0 \end{align} $ $\huge\boxed{ \bar{w}z\in \R \iff ad-bc=0 \iff \set{\vec w,\vec z } \notin \mathcal{L}_{I}} $