Series of short questions intended to clarify $p$-norm of vectors with complex elements Note: 
My central question is Q.2.B. I'd really like an answer to that question.
Context:
From [1], we find that
``Let $ p \geq 1$ be a real number.  The $p$-norm  of vectors $\mathbf{x} = (x_1, \ldots, x_n)$  is
$$ \left\| \mathbf{x} \right\| _p := \bigg( \sum_{i=1}^n \left| x_i \right| ^p \bigg) ^{1/p}.$$
Questions:
[Q.1.A]
Is the $1$-norm  of vector $\mathbf{x} = ([1 + i], [3 + 2\,i], [6-20\,i])$ given as follows:
\begin{align}
 \left\| \mathbf{x} \right\| _1 
= 
&
\bigg( \sum_{i=1}^3 \left| x_i \right| ^1 \bigg) ^{1/1}
\\
=
&
\bigg(   \left|  1 + i  \right| + \left|  3 + 2\,i   \right|+ \left|    6-20\,i  \right|\bigg) 
\\
=
&
\bigg(   \left|  1 \right|+ \left|i  \right| + \left|  3 \right|+ \left| 2\,i   \right|+ \left|    6 \right| + \left|- 20\,i  \right|\bigg) 
\\
=
&
\bigg(     1 + 1    +   3  +   2    +     6   +   20 \bigg) 
\\
=
& 
\, 33? 
\end{align}
[Q.1.B]
Or, is the $1$-norm  of vector $\mathbf{x} = ([1 + i], [3 + 2\,i], [6-20\,i])$ given as follows:
\begin{align}
 \left\| \mathbf{x} \right\| _1 
= 
&
\bigg( \sum_{i=1}^3 \left| x_i \right| ^1 \bigg) ^{1/1}
\\
=
&
\bigg(   \left|  1 + i  \right| + \left|  3 + 2\,i   \right|+ \left|    6-20\,i  \right|\bigg) 
\\
=
&
\bigg(   \sqrt{1^2 + 1^2}  +  \sqrt{3^2 + 2^2} +  \sqrt{6^2 + 20^2} \bigg) ?
\end{align}
[Q.2.A]
Is there a named norm such that the named norm  of vector $\mathbf{x} = ([1 + i], [3 + 2\,i], [6-20\,i])$ is equal to 33 (see Q.1.A)? 
[Q.2.b]
If there is a such a named norm, what is the name of the norm?
[Q.3]
Is the $2$-norm  of vector $\mathbf{x} = ([1 + i], [3 + 2\,i], [6-20\,i])$ given as follows:
\begin{align}
 \left\| \mathbf{x} \right\| _2 
= 
&
\bigg( \sum_{i=1}^3 \left| x_i \right| ^2 \bigg) ^{1/2}
\\
=
&
\sqrt{   \left|  1 + i  \right|^2 + \left|  3 + 2\,i   \right|^2+ \left|    6-20\,i  \right|^2} 
\\
=
&
\sqrt{   \left( \sqrt{ 1^2 + 1^2 }\right)^2 + \left( \sqrt{ 3^2 + 2^2 }\right)^2 + \left( \sqrt{ 6^2 + (-20)^2 }\right)^2}
\\
=
&
\sqrt{      1^2 + 1^2    +   3^2 + 2^2   +   6^2 + (-20)^2  }
\\
=
& 
\sqrt{      1 + 1    +   9 + 4   +   36  + 400  }
\\
=
& 
\sqrt{           451  }?
\end{align}
Bibliography
[1]
https://en.wikipedia.org/wiki/Norm_(mathematics)#Generalizations
 A: Here are my answers:
Answer to Q.1.A:
No, the 1-norm of $x=([1+i],[3+2i],[6−20i])$ is not equal to 33.
Answer to Q.1.B:
Yes, the 1-norm of $x=([1+i],[3+2i],[6−20i])$ is as given.
Answer to Q.2:
Pursuant to [2], "given a vector space $V$ over a subfield $F$ of the complex numbers, a norm on $V$ is a function $p$: $V → R$ with the following properties:
For all $a \in F$ and all $\mathbf{u}, \mathbf{v} \in V$,
$p(a\,\mathbf{v}) = |a| \, p(\mathbf{v})$ (being absolutely homogeneous or absolutely scalable).
$p(\mathbf{u} + \mathbf{v}) \leq p(\mathbf{u}) + p(\mathbf{v})$ (being subadditive or satisfying the triangle inequality).
$p(\mathbf{v}) \geq 0$ (being positive or more precisely non-negative).
If $p(\mathbf{v}) = 0$ then $\mathbf{v}=0$ is the zero vector (being definite or being point-separating)."
Here I define a function $p_o(\mathbf{v})$ as
$$p_o(\mathbf{v}) :=    \left\| \textrm{Re}(\mathbf{v}) \right\|_1    + \left\| \textrm{Im}(\mathbf{v}) \right\|_1;$$
where
$$\left\| \mathbf{x} \right\| _p := \bigg( \sum_{i=1}^n \left| x_i \right| ^p \bigg) ^{1/p}$$
and $\textrm{Re}(\mathbf{v})$ and $ \textrm{Im}(\mathbf{v})$ are the real parts of $\mathbf{v} $ and imaginary parts of $\mathbf{v}$ , respectively.
Now, I have to check that the function $p_o$ has the properties of a norm. If so, the function is a norm. If not, the function is not a norm.
Is function $p_o$ absolutely homogeneous or absolutely scalable?
\begin{align}
p_o(a\,\mathbf{v}) =
&
    \left\| \textrm{Re}(a\,\mathbf{v}) \right\|_1    + \left\| \textrm{Im}(a\,\mathbf{v}) \right\|_1
\\
&
    \left\| \textrm{Re}(a)\, \textrm{Re}(\mathbf{v}) -  \textrm{Im}(a)\, \textrm{Im}(\mathbf{v}) \right\|_1    +   \left\| \textrm{Im}(a)\, \textrm{Re}(\mathbf{v}) +  \textrm{Re}(a)\, \textrm{Im}(\mathbf{v}) \right\|_1 
\\
&
   \sum_{i=1}^{n} \left| \textrm{Re}(a)\, \textrm{Re}( {v_i}) -  \textrm{Im}(a)\, \textrm{Im}({v_i}) \right|     +   \sum_{i=1}^{n}\left| \textrm{Im}(a)\, \textrm{Re}({v_i}) +  \textrm{Re}(a)\, \textrm{Im}( {v_i}) \right| 
\\
&
   \sum_{i=1}^{n}\left(  \left| \textrm{Re}(a)\, \textrm{Re}( {v_i}) -  \textrm{Im}(a)\, \textrm{Im}({v_i}) \right|     +   \left| \textrm{Im}(a)\, \textrm{Re}({v_i}) +  \textrm{Re}(a)\, \textrm{Im}( {v_i}) \right| \right)
\end{align}
while
\begin{align}
\left|a\right|\,p_o(\mathbf{v}) =
&
    \left|a\right|\,\left( \left\| \textrm{Re}(\mathbf{v}) \right\|_1    + \left\| \textrm{Im}(\mathbf{v}) \right\|_1\right)
\\
&
    \left|a\right|\,\left(\sum_{i=1}^3\left| \textrm{Re}( {v_i}) \right|    +  \sum_{i=1}^3\left| \textrm{Im}( {v_i}) \right|  \right)
\\
&
    \left|a\right|\,\sum_{i=1}^3\left(\left| \textrm{Re}( {v_i}) \right|    +   \left| \textrm{Im}( {v_i}) \right|  \right)
\\
&
    \sqrt{\left(\textrm{Re}(a)\right)^2 + \left(\textrm{Im}(a)\right)^2} \,\sum_{i=1}^3\left(\left| \textrm{Re}( {v_i}) \right|    +   \left| \textrm{Im}( {v_i}) \right|  \right)
\end{align}
I find that, in general,
$$ p_o(a\,\mathbf{v})\neq a\,p_o(\mathbf{v})  $$
Resulting from the fact that the function $p_o$ is not scalable, $p_o$  is not  a norm at all. Therefore, by definition, the function cannot be a named norm. Specifically, within the context of question Q.2, though it is possible to define a function such that the function of $\textbf{x}$ equals to 33, such a function will not be a norm.
Answer to Q.3:
Yes, the 2-norm of $x=([1+i],[3+2i],[6−20i])$ is as given.
[2] https://en.wikipedia.org/wiki/Norm_(mathematics)#Definition
A: The wording on your question seems odd (not sure what a 'named norm' means) but I feel your real question translates to see if
$$\rho(z_1,\dots,z_n)=\sum_{i=1}^n |\textrm{Re}(z_i)|+|\textrm{Im}(z_i)|$$
is a norm on $\mathbb{C}^n$. To see that this isn't true just consider $v=(1+i,0,\dots,0)$ and $\lambda=1-i$. In order to have $\rho$ be a norm it shoud satisfy $\rho(\lambda v)=|\lambda|\rho(v)$, but we actually have $$\rho(\lambda v)=\rho(2,0,\dots,0)=2$$
and $$|\lambda|\rho(v)=2\sqrt 2 $$
This is basically the argument you used in your answer but with a concrete example to show it doesn't always hold.
