i'd make it a bit easier upon yourself then and look at a special case where $a = f(t)$ and $b = g(t)$, so you reduce the "two degrees of freedom" to one
there is a particular choice of $f$ and $g$ that will make a lot of sense (it's should hopefully not be very hard to figure out)
Now if you look at $T/\sqrt{a^2+b^2}$, you'll find you can rewrite it as $[[sin{\theta}, \cos{\theta}], [-cos{\theta}, \sin{\theta}]]$ for some $\theta$
Which axis of the complex plane should the locus $\mathbf{Im}(z)>1$ be plotted on? $\mathbf{Im}(z)$ itself is just going to be a real number, that's where my confusion lies
I have tried to draw the graph of the $f(\theta)$ function.
Firstly the domain of the function is,
$$D(\sin\theta): -\infty<\sin\theta<\infty$$
$$D(\cos\theta+1): -\infty<\cos\theta<\infty \Rightarrow -\infty<\cos\theta + 1<\infty$$
The intersection of these domains is $-\infty<f(\theta)<\inf...
The general statement here is that if $(X,\Sigma,\mu)$ is a finite measure space, then for $1\leq p<q\leq\infty$ we have $L^q(X)\subseteq L^p(X)$ and $\|f\|_p\leq \mu(X)^{1/r}\|f\|_q$ where $r>0$ is such that $1/p=1/q+1/r$
That is a difficult question. For the beginning math student, understanding the difference among finite, countably infinite, and uncountable is more than enough. But then there's a whole thing in the field of set theory, where you learn about (infinite) cardinal numbers and ordinal numbers.
I would recommend you not go there until you're much older.
For $E[XY]$, if I understood correctly, there is a mistake. You are first choosing $X$ and than $Y=h(X)$, with $h$ deterministic. So you have to evaluate:
$E[XY]=\int f_X(x) xh(x)dx$
Instead it looks that you integrated over $y$. But consider when $h$ is invertible:
$\int f_X(x) xh(x)dx \ne \i...
I have tried to draw the graph of the $f(\theta)$ function.
Firstly the domain of the function is,
$$D(\sin\theta): -\infty<\sin\theta<\infty$$
$$D(\cos\theta+1): -\infty<\cos\theta<\infty \Rightarrow -\infty<\cos\theta + 1<\infty$$
The intersection of these domains is $-\infty<f(\theta)<\inf...
For $E[XY]$, if I understood correctly, there is a mistake. You are first choosing $X$ and than $Y=h(X)$, with $h$ deterministic. So you have to evaluate:
$E[XY]=\int f_X(x) xh(x)dx$
Instead it looks that you integrated over $y$. But consider when $h$ is invertible:
$\int f_X(x) xh(x)dx \ne \i...
