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Vanessa and Annie sent us this solution:

Fixing $z_1$ then moving $z_2$ so that $z_3$ is on the x-axis:

We tried this for several different values of $z_1$. Here is what we found:

We also tried keeping $z_3$ on the $y$-axis. We tried this for the same values of $z_1$. Here is what we found:

Sam sent in this algebraic explanation:

Set $z_1=a+bi$, $z_2=c+di$.

Then $z_3=z_1\, z_2=(a+bi)(c+di)=(ac-bd)+(ad+bc)i$.

As $z_3$ lies on the $x$-axis, $Im(z_3)=0$, so let the given point on the $x$-axis be

$m$ (i.e. $Re(z_3)=m$).

Then we have $ac-bd=m$ and $ad+bc=0$.

$\Rightarrow \, c=\frac{-ad}{b}$.

Substitute this into $ac-bd=m$:

$$m=-\frac{a^2 d}{b}-bd = \frac{-(a^2+b^2)d}{b}$$

$$\Rightarrow d=\frac{-mb}{a^2+b^2}.$$

Substitute this into $\frac{-ad}{b}=c$ to get

$$c=\frac{ma}{a^2+b^2}.$$

So for $z_3=z_1\, z_2=m$ and $z_1=a+bi$,

$$z_2=\frac{ma}{a^2+b^2}-\frac{mb}{a^2+b^2}i = \frac{m}{a^2+b^2}(a-bi).$$

So $z_2$ will always be a scalar multiplied by $a-bi$. This means it will have

the same trajectory as $z_1$ reflected about the $x$-axis.

Similarly, if you want to have $z_3=n$, where $n$ is on the $y$-axis, and $z_1=a+bi$, then you need to have $$z_2=\frac{n}{a^2+b^2}(b+ai).$$

Well done everyone!

Fixing $z_1$ then moving $z_2$ so that $z_3$ is on the x-axis:

We tried this for several different values of $z_1$. Here is what we found:

- For $z_1=2+2i$, we think you need $z_2$ to be on the line $y=-x$. So $z_2=c(1-i)$ for any $c$.
- For $z_1=1+2i$, we think you need $z_2$ to be on the line $y=-2x$. So $z_2=c(1-2i)$.
- For $z_1=2-i$, we think you need $z_2$ to be on the line $y=-\frac{1}{2x}$. So $z_2=c(2+i)$.
- For $z_1=4+3i$, we think you need $z_2$ to be on the line $y=-\frac{3}{4} x$. So $z_2=c(4-3i)$.

We also tried keeping $z_3$ on the $y$-axis. We tried this for the same values of $z_1$. Here is what we found:

- For $z_1=2+2i$, we think you need $z_2$ to be on the line $y=x$. So $z_2=c(1+i)$ for any $c$.
- For $z_1=1+2i$, we think you need $z_2$ to be on the line $y=\frac{1}{2}x$. So $z_2=c(2+i)$.
- For $z_1=2-i$, we think you need $z_2$ to be on the line $y=2x$. So $z_2=c(-1+2i)$.
- For $z_1=4+3i$, we think you need $z_2$ to be on the line $y=\frac{4}{3} x$. So $z_2=c(3+4i)$.

Sam sent in this algebraic explanation:

Set $z_1=a+bi$, $z_2=c+di$.

Then $z_3=z_1\, z_2=(a+bi)(c+di)=(ac-bd)+(ad+bc)i$.

As $z_3$ lies on the $x$-axis, $Im(z_3)=0$, so let the given point on the $x$-axis be

$m$ (i.e. $Re(z_3)=m$).

Then we have $ac-bd=m$ and $ad+bc=0$.

$\Rightarrow \, c=\frac{-ad}{b}$.

Substitute this into $ac-bd=m$:

$$m=-\frac{a^2 d}{b}-bd = \frac{-(a^2+b^2)d}{b}$$

$$\Rightarrow d=\frac{-mb}{a^2+b^2}.$$

Substitute this into $\frac{-ad}{b}=c$ to get

$$c=\frac{ma}{a^2+b^2}.$$

So for $z_3=z_1\, z_2=m$ and $z_1=a+bi$,

$$z_2=\frac{ma}{a^2+b^2}-\frac{mb}{a^2+b^2}i = \frac{m}{a^2+b^2}(a-bi).$$

So $z_2$ will always be a scalar multiplied by $a-bi$. This means it will have

the same trajectory as $z_1$ reflected about the $x$-axis.

Similarly, if you want to have $z_3=n$, where $n$ is on the $y$-axis, and $z_1=a+bi$, then you need to have $$z_2=\frac{n}{a^2+b^2}(b+ai).$$

Well done everyone!