It is to show for an $a\in \mathbb{C}^{\ast}$ that $aB_{1}(1)= B_{|a|}(a)$
where B denotes a disc
Okay, maybe this is correct:
$aB_{1}(1) = a(e^{i\phi}) = ae^{i\phi} = |a|e^{i\phi} = B_{|a|}(a)$
But this seems very wrong!
V
It is to show for an $a\in \mathbb{C}^{\ast}$ that $aB_{1}(1)= B_{|a|}(a)$
where B denotes a disc
Okay, maybe this is correct:
$aB_{1}(1) = a(e^{i\phi}) = ae^{i\phi} = |a|e^{i\phi} = B_{|a|}(a)$
But this seems very wrong!
V
let a be a complex number of the form: $a:= u+vi$ and $z:= x+yi$
$B_{1}(1)$ means that $ |z-1| < 1 $ and so $a|z-1| = (u+vi)|z-1| = (u+vi)(\sqrt{(x-1)^{2}+y^{2}} $ so we can write it as $|z-1|u+|z-1|vi< a|1| = |a| = \sqrt{u^{2}+v^{2}}$
This seems to be the wrong route also. ??
V
Well, your idea is OK. But you should improve some things:
First: If B is a disk and you write $e^{i\phi}$, then it is a parametrization of a circle not the whole disk. But actually a disk is a sum of circles plus the middle point, so the strategy is OK.
Second: As @Adam wrote, you can't write $ae^{i\phi}=|a|e^{i\phi}$, because it is not equal if $a \neq |a|$. However it is true that $ae^{i\phi} \in B_0^{|a|}$. So in that way, you can prove that $aB_0^1 \subseteq B_0^{|a|}$ (1). So what remains is that it is not an inclusion but an equality. Since $a \cdot a^{-1}=1$ it is enough to show that $ a^{-1}B_0^{|a|} \subseteq B_0^1$ and proving that is actually the same as (1).
Third: You have to remember that the middlepoint of your circle is $1$ and when you write $e^{i\phi}$ it generates a circle around $0$, so you have to add $1$ to it.
P.S. There is no $a^{-1}$ for $a=0$, so it is a special case, but a very simple one.
Your second attempt also should lead to success, but you shouldn't multiply $a$ by distance between $z$ and $1$ (notice that it leaded you to write an inequality with complex numbers, which makes no sense!) - you should multiply $a$ by $z$ and look what is the distance between $az$ and $a$ assuming that the distance between $z$ and $1$ is smaller than $1$
First let me answer your specific question.
Let $z \in B_{1}(1)$, that is to say, $|z - 1| \lt 1$. We want to show that $az \in B_{|a|}(a)$, that is, we want to show that $|az-a| \lt |a|$. But $ |az - a| = |a| \cdot \underbrace{|z-1|}_{\lt 1} \lt |a|, $ as we wanted, therefore $aB_1(1) \subset B_{|a|}(a)$.
Conversely, let $w \in B_{|a|}(a)$, that is $|w -a| \lt |a|$. We want to show that $w \in aB_{1}(1)$. Since $a \neq 0$ we can write $ |a| \gt |w-a| = |a| \cdot \left|\frac{w}{a}-1\right|, $ so $\left|\frac{w}{a}-1\right| \lt 1$. But this means that $z = \frac{w}{a} \in B_{1}(1)$, so $w = az \in a B_{1}(1)$, hence $B_{|a|}(a) \subset aB_1(1)$.
Putting 1. and 2. together we have $aB_1(1) = B_{|a|}(a)$, as desired.
To make this a bit more useful, we generalize slightly:
Consider $B_{r}(p)$ with $r \gt 0$ and let $a \in \mathbb{C}^\ast$. Then $aB_{r}(p) = B_{|a|r}(ap)$.
Indeed, if $z \in B_{r}(p)$, so $|z-p| \lt r$, then $ |az-ap| = |a|\cdot|z-p| \lt |a|r, $ so $aB_{r}(p) \subset B_{|a|r}(ap)$.
Conversely, if $w \in B_{|a|r}(ap)$ then $ |a|r \gt |w-ap| = |a|\cdot \left|\frac{w}{a} - p\right|, $ so $\left|\frac{w}{a} - p\right| \lt r$, thus $z = \frac{w}{a} \in B_{r}(a)$ and therefore $w=az \in aB_{r}(p)$. The claimed equality $aB_{r}(p) = B_{|a|r}(ap)$ is proved.
To sum up: multiplying by a complex scalar $a \in \mathbb{C}^\ast$ scales all balls by a factor $|a|$ (i.e., multiplies the radii by $|a|$) and moves the centers from $p$ to $ap$.
Since this looks like homework, I'll just give you a hint. You are on the right track, but it is not true that $a e^{i\phi} = |a| e^{i \phi}$. Indeed, remember that $a = |a| e^{i \theta}$ for some $\theta$ (the modulus of $a$). Geometrically, what is going on is that multiplication by $a = |a| e^{i \theta}$ scales things by $|a|$ and rotates things by $\theta$.
Following the second attempt:
So we look at this system of inequalities:
$aB_{1}(1)$: $|az-a|$ and $|z-1|<1 $
what we want to show is that this equals $|z-a|<|a|$
Then stuck.
Following the first attempt:
for $a \in \mathbb{C}^{*} = \mathbb{C}\backslash \{0\}$
we look at $B_{1}(0) : |z|<1 $ so $|az|< |a|$ which is $B_{|a|}(0)$ this must be finished like this.
But I don't know where I have shown it.