Show that the norm of a complex functional is equal to the norm of its "real part" The statement of the exercise is pretty simple to understand, but I'm having trouble to prove it. The exercise is the following:

(Exercise 3, from Introduction to functional analysis, by A. E. Taylor,  page 190).: If $X$ is a complex normed linear space and $X_r$ is the associated real linear space, we write $x'(x)=x_1'(x)-ix_1'(ix)$, where $x_1'\in (X_r)'$ and $x'\in X'$. Show that $\|x_1'\|=\|x'\|$.

Here, $X'$ denotes the dual space of $X$, i.e. the space of the linear continuous transformations $x:X\to \mathbb{K}$, where $\mathbb{K}$ is the field of the scalars. And $X_r$ denotes the same vector space $X$, but considering the scalars to be only in $\Bbb{R}\subset \Bbb{C}$.
Okay, we have to show that $\|x_1'\|=\|x'\|$, i.e.
$$\sup\{\|x'(v)\|\,:\, \|v\|\leq 1\}=\sup\{\|x_1'(v)\|\,:\, \|v\|\leq 1\}.$$
I've thought to show that $\|x_1'\|\leq \|x'\|$ and $\|x'\|\leq \|x_1'\|$. The first one is easy, because, since $x_1'(v)\in \Bbb{R},\forall v \in X_r\approx X$, we have
$$\|x_1'(v)\|\leq \|x_1'(v)-ix_1'(v)\|=\|x'(v)\|.$$
I could not prove the second one. Then I've tried by contradiction: supposed that $\|x_1'\|<\|x'\|$, made some computations, but could not find a contradiction at all...
 A: For given $x$, there is $\alpha \in \mathbb C$, $|\alpha|=1$ with $x_1'(\alpha \, i \, x) = 0$ (why?).
Then,
$$|x'(x)| = |x'(\alpha \,x)| = \ldots.$$
A: Let $x\in X$ be given. We can suppose that $x\neq0$, otherwise it is easy to work with $x=0$.
Define $A_x=\{\alpha i x\,:\,\alpha \in \Bbb{C} \}$. We can consider both $A_x\subset X$ and $A_x\subset X_r$ and, respectively, $\dim_{\Bbb{C}}A_x=1$ but $\dim_{\Bbb{R}}A_x=2$. Then we look to
$$f:=x_1'|_{A_x}:A_x\to \Bbb{R}$$
with $A_x\subset X_r$: then $f$ is then a real linear transformation. Since we are now in finite dimension, we can use the rank-nulity theorem:
$$2=\dim_\Bbb{R}A_x=\dim_\Bbb{R}\ker(f)+\dim_\Bbb{R}\mathrm{im}(f).$$
Since $0\leq \dim_\Bbb{R}\mathrm{im}(f)\leq 1$, we must have $1\leq \dim_\Bbb{R}\ker(f)\leq 2$. 
Therefore, $\ker(f)$ is not the trivial space $\{0\}$. This means that exists $v\in A_x$, with $v\neq 0$, such that $f(v)=0$. But then $v=\alpha ix$, for some $\alpha\in \Bbb{C}-\{0\}$ and $f(v)=x_1'(\alpha ix)=0$.
Now, to obtain such an $\alpha$ with $|\alpha|=1$, just take $\dfrac{\alpha}{|\alpha|}$.
Now, use @gerw suggestion!
