Question on a submanifold of $\mathbb{C}^2$ and tangent vectors Let $X\subset \mathbb{C}^2$ (with complex coordinates $z,w$) be defined by the equation $|z|^2=|w|^2$. 
If we see $\mathbb{C}^2$ as $\mathbb{R}^4$ with real coordinates $z=x+iy$, $w=u+iv$ then $X$ is a real submanifold of dimension 3 defined by $x^2+y^2-u^2-v^2=0$.  
Question 1: is $X$ also a complex submanifold?
Question 2: is $(i,0)$ a tangent vector to $X$ in $(1,1)\in X$?
I think the answer is no to both questions. In particular for the second one it should be true that $T_{(1,1)}X\simeq Ker(df_{(1,1)})$ with $f= x^2+y^2-u^2-v^2=0$ and $df_{(1,1)}=(2,2,-2,-2)$. But $(i,0)=(0,1,0,0)$ in real coordinates and $df_{(1,1)}(0,1,0,0)\neq 0$.
Am I right?
 A: Question 1
Since $X\subset \mathbb R^4$ is a real hypersuface it has real dimension $3$ and so cannot be a complex manifold nor even a complex space since complex spaces have even real dimension.
(Beware that, if  $O\in \mathbb C^2$ denotes  the origin,  $X\setminus \{O\}$ is indeed a real manifold , but $X$ itself is not a real manifold because  it has a singularity at $O$)  
Question 2
We have for $P=(a,b)=(p+iq,r+is)\neq O\in X\subset \mathbb C^2$ the relation $T_PX=\ker d_Pf$, where $$d_Pf:\mathbb C^2\to \mathbb R:(z=x+iy,w=u+iv)\mapsto 2(px+qy-ru-sv)$$ or in vector notation $$ d_Pf=(2p,2q,-2r,-2s)$$ If now $a=1,b=1$, we get $$d_Pf(z,w)=2(x-u) \quad [\operatorname {or in vector notation } d_Pf=(2,0-2,0)]$$ so that $T_Pf$ is the real hyperplane $$T_Pf=\{(z,w)\in \mathbb C^2\vert x=u\}$$ So, yes,  $(i,0)=(0+i,0+0i)\in T_Pf$ since for that vector $(i,0)$ one has $x=u=0$   
Supplementary remark
 One may notice that the only complex line $L$ included in the $3$-dimensional real vector space $T_{(1,1)}X\subset \mathbb C^2$ is $$L=\{(0,w)\in \mathbb C^2\vert w\in \mathbb C\}$$ 
