Due on Monday, Jan 21.
Recall the following facts about complex numbers. Every \(z \in \mathbb{C}\) can be written as \(a+bi\) where \(a, b \in \mathbb{R}\). The values \(a\) and \(b\) are called the real and imaginary parts of \(z\), and are sometimes denoted \(\operatorname{Re} z\) and \(\operatorname{Im} z\) (Notice that the imaginary part of \(a+bi\) is the real number \(b\), it does not include the \(i\)). The absolute value of \(z\) is defined to be \(|z| = \sqrt{a^2 + b^2}\). The complex conjugate of \(z\) is \(\bar{z} = a - bi\).
For any complex number \(z = a + bi\) (where \(a, b \in \mathbb{R}\)), show that \(\bar{z} z = |z|^2\).
Show that if \(z \in \mathbb{C}\) and \(|z|=1\), then \(\bar{z} = \frac{1}{z}\).
Show that \(\operatorname{Re} z = \frac{1}{2}(z+\bar{z})\) and \(\operatorname{Im} z = \frac{1}{2i}(z-\bar{z})\).
Prove that for vectors in an inner product space, \[\|x\pm y\|^2 = \|x\|^2 + \|y\|^2 \pm \operatorname{Re} \langle x, y \rangle.\]
Prove the parallelogram identity for an inner product space. \[\|x+y\|^2 + \|x-y\|^2 = 2(\|x\|^2 + \|y\|^2).\]
Use the Law of Cosines to prove that \(\langle x, y \rangle = \|x\| \|y\| \cos \theta\), where \(\theta\) is the angle between the vectors \(x\) and \(y\) in the real subspace spanned by \(x\) and \(y\) (if you want, assume that \(x,y \in \mathbb{R}^2\)).