I am starting with a normed space $X$ over $\mathbb{R}$ with norm $||\cdot||_X$.
I have to prove that the following is a norm on $X\times X$ over $\mathbb{C}$: $$||(x,y)||=\sup_{0\leq\theta\leq 2\pi} || x \cos\theta + y\sin\theta ||_X$$ and with addition and scalar multiplication defined by: $$ (x,y)+(x_1, y_1)=(x+x_1,y+y_1) $$ $$ (a+ib)(x,y)=(ax-by, bx+ay) $$ I have already shown triangle inequality, and the fact that $||(x,y)||=0 \iff (x,y)=(0,0)$. But I am stuck on homogeneity, ie
Show that for any $(x,y)\in X\times X$ and $a+ib \in \mathbb{C}$, we have $$ || (a+ib)(x,y) || = |a+ib| \cdot ||(x,y)|| $$
My issue is that I can start off easily by following definitions, $$||(a+ib)(x,y)||=\sup_{0\leq\theta\leq 2\pi} ||(ax-by)\cos\theta + (bx+ay)\sin\theta ||_X $$ but I cannot for the life of me make this look like how I want, which is $$ \sqrt{a^2 + b^2} \,\, ||(x,y)|| $$ I'm not sure how to introduce the square root and squares..
I think it's a trap to go the direct route. Note that, clearly, if $r\geq 0,$ then $$ \| (rx,ry)\|=r\|(x,y)\| $$ Hence, let $\varphi\in \mathbb{R}$ and consider
\begin{align} \| e^{i\varphi}(x,y)\| &=\sup_{\theta} \|(\cos(\varphi)x-\sin(\varphi)y)\cos(\theta)+(\sin(\varphi)x+\cos(\varphi)y)\sin(\theta)\|_X \\ &= \sup_{\theta} \| x (\cos(\varphi)\cos(\theta)+\sin(\varphi)\sin(\theta))+y(\cos(\varphi)\sin(\theta)-\sin(\varphi)\cos(\theta))\|_X \\ &=\sup_{\theta} \| x(\cos(\varphi+\theta+\pi))-y(\sin(\varphi+\theta+\pi))\|_X \\ &= \sup_{\theta} \| x\cos(\theta)+y\sin(\theta)\|_X\\ &= \|(x,y)\| \end{align}
Hence, $$ \| re^{i\theta}(x,y)\|=r\| e^{i\theta}(x,y)\|=r\|(x,y)\|, $$ which is what we wanted.