The length of the vector $y$ is $||y||=4$, inner product $x\cdot y=-5+2i$ and $||x-y||=5$. How do I find $||x||$?
It's probably a very easy thing to do but I just don't understand how to get to the solution. I tried:
$$||x-y||^2 = ||x||^2+||y||^2-2x\cdot y=5^2$$
and now solving for $||x||$ gives me a square root of a complex number...
Ok, I see now that there are some problems. First, given two complex numbers $a,b\in \mathbb C$ the product $a\cdot b$ is not an inner product. That is simply verified by considering that an inner product must be positive definite: $$<\cdot,\cdot>:V \times V \longrightarrow \mathbb C$$ $$<v,v> \geq 0\quad \forall v\in V$$ and just taking $v=i$ you get $i\cdot i = i^2 = -1\ngeq 0$
In the complex field, i.e. $V = \mathbb C^n$, the inner product is usually defined as follows: $$<a,b> = a\cdot \bar b$$ where $\bar b$ is the complex conjugate of $b\in\mathbb C$. This way you can define properly the norm of the vectors of your space, which in this case [$\star$] ($V = \mathbb C$) coincides with the absolute value of the complex number: $$||x||^2 = <x,x> \stackrel{\star}{=} x\cdot \bar x = |x|^2$$ This leads to the following identity: \begin{align} ||x-y||^2 &= <x-y,x-y> = <x,x> - <y,x> - <x,y> + <y,y>\\ &= ||x||^2 -\overline{<x,y>} -<x,y> + ||y||^2\\ &= ||x||^2 -2Re[<x,y>] + ||y||^2\\ &\stackrel{\star}{=} |x|^2 + |y|^2 -2|x\cdot\bar y|\cos(\theta) \end{align} where $\theta = \angle x -\angle y$ is the phase difference between $x$ and $y$.
I hope this will help you clarify the problem.